Examples of simple Harmonic Motion• Concept strands (4-6)
Damped Oscillations
Forced Oscillations and Resonance
Wave Motion
General Equation of a One Dimensional Progressive Wave• Concept strands (7-8)
Principle of superposition of Waves• Concept strand (9)
Reflection and Transmission of Waves
Transverse stationary Waves in a stretched string• Concept strand (10)
sound
stationary Waves in Air Columns• Concept strands (11-13)
Beats• Concept strands (14-15)
Doppler Effect• Concept strand (16)
Longitudinal Mechanical Waves (sound) as Pressure Waves• Concept strand (17)
COnCEPT COnnECTORs
• 30 Connectors
TOPIC GRIP
• subjective Questions (10)• straight Objective Type Questions (5)• Assertion–Reason Type Questions (5)• Linked Comprehension Type Questions (6)• Multiple Correct Objective Type Questions (3)• Matrix-Match Type Question (1)
IIT AssIGnMEnT ExERCIsE
• straight Objective Type Questions (80)• Assertion–Reason Type Questions (3)• Linked Comprehension Type Questions (3)• Multiple Correct Objective Type Questions (3)• Matrix-Match Type Question (1)
ADDITIOnAL PRACTICE ExERCIsE
• subjective Questions (10)• straight Objective Type Questions (40)• Assertion–Reason Type Questions (10)• Linked Comprehension Type Questions (9)• Multiple Correct Objective Type Questions (8)• Matrix-Match Type Questions (3)
3.2 Oscillations and Waves
A motion which repeats itself at regular fixed interval of time is called a periodic motion.
Revolution of planets around the Sun and uniform circular motion are examples of periodic motion.
A periodic motion takes place along the same path. The smallest interval of time, in which the motion repeats itself, is called the time period T. Its SI unit is second(s). The number of periodic motions per unit time is called frequency n. The SI unit of frequency is hertz (Hz) which is the same as s–1. 2p times the frequency is called angular frequency w. The SI unit of angular frequency is
radian per second (rad s–1).
pw pn
22
T= =
is the relation between w, T and n for all periodic motions.
Oscillatory motion
To and fro motion of a particle about a mean position is called oscillatory motion. A periodic motion, in which a particle moves on either side of an equilibrium (or mean) position, is an oscillatory motion.
Oscillation of simple pendulum, and loaded springs are examples of oscillatory motion.
Thus all oscillatory motions are periodic but not all periodic motions are oscillatory.
Only at the mean position, the net force on the particle is zero. At all positions, other than the mean position, the particle in oscillation has acceleration (due to an unbalanced force acting on it). This force which tries to bring the particle towards the mean position in oscillatory motion is called restoring force.
note:The motion of a ball, bouncing off vertically from the floor, is not oscillatory as there is no mean position during the motion of the ball.
periOdiC MOtiOn
SiMple harMOniC MOtiOn (ShM)
Simple harmonic motion (SHM) is a special case of oscillatory motion. Hence all simple harmonic motions are oscillatory (and also periodic) but all oscillatory motions are not simple harmonic. An oscillatory motion, in which the acceleration of the particle at any position is directly proportional to the displacement from the mean position and the direction of the acceleration at all positions is towards the mean position, is called a simple harmonic motion.
That is, SHM is an oscillatory motion in which the restoring force (the force which brings the particle back to the mean position) at any position is directly proportional to the displacement from the mean position and is directed towards the mean position.
Figure 3.1 shows the sequential positions of a particle executing oscillatory motion. The maximum displacement of the particle, on either side of the mean position is called the displacement amplitude A. If the displacement of the particle from the mean position is x and the force acting on the particle is F , then the oscillation
is SHM if F ∝ - x or F = -k x , where k is a positive constant.
Fig. 3.1
F = -k x
represents a SHM. There are two types of SHM (i) Linear harmonic and (ii) Angular harmonic motion.
Oscillations and Waves 3.3
Differential equation of a simple harmonic (linear harmonic) motion
Consider a particle of mass m, executing SHM along a path X’OX as shown, with the mean position at O. Let the speed of the particle be v0 when it is at position P (at a distance x0) from O. Time measurements com-mence when particle is at P and moving towards right. i.e., t = 0 at P.
At t = 0
X’ X
XX’ O P t = 0
At t = tO Q
x0v0
v
Fig. 3.2
x
(a)
(b) P
At t = t, the particle is at Q, at a distance x from O and moving towards the right with a velocity v. Restoring force F at Q is given by
F = -k x (where k is a positive constant)
But F = m a , where a = acceleration at Q
i.e., m a = -k x ⇒ a = k
xm −
Let us put km
= w2 (Q km
is always positive) so that
w = km
\ a = - w2kx x
m= − — (i)
But a = 2
2
d xdt
⇒ 2
2
d xdt
= -w2 x or since it is one-dimensional
motion ( )x x→
w
22
2
d xx 0
dt+ =
which is the differential equation for linear SHM along the x-axis.
Solution of the differential equation: expression for the displacement of SHM
The differential equation for the SHM has the solution
x = Asinwt
as can be verified by substitution. This solution assumes that if t = 0, x = 0. i.e., we are starting to measure the displacement when the particle is in its mean position (point O of Fig. 3.2(a))
But if we had started the measurement of displacement when the particle was not at the mean position but at the position P as in Fig. 3.2(b) we can write the initial displacement as
x0 = Asinf
The displacement at any later time t (point Q of Fig. 3.2(b)) will be then
x = Asin(wt + f)
This is a general expression for the displacement of a SHM.
We also find that
x = Acos(wt + f) = Asin t2p
w f + + is an equally
valid solution. But then, our starting point of measurement corresponds to a displacement
x0 = Asin2p
f +
The exact solution of the differential equation
a = dv dv dx dv
. vdt dx dt dx
= =
\ (i) ⇒ vdvdx
= -w2x
\ w0 0
v x2
v x
vdv xdx= −∫ ∫
where the lower limits of the integration are the values of position coordinate x and velocity v of the particle at the point P at t = 0 and the upper limits are the corresponding values at time t when the particle is at Q (see Fig. 3.2)
w
0 0
v x2 22
v x
v x2 2
= −
3.4 Oscillations and Waves
⇒ v2 - v02 = -w2[x2 - x0
2]
⇒ v2 = v02 + w2x0
2 - w2x2
⇒ v = w w2 2 2 2 20 0v x x+ −
⇒ v = ww
22 200
vx x
+ −
— (ii)
Put A2 = w
2200
vx
+
(ii) ⇒ v = w 2 2dxA x
dt= −
\ 2 2
dx
A x− = wdt
⇒ w0
x t
2 2x 0
dxdt
A x=
−∫ ∫
⇒ w0
x1
x
xsin t
A− =
Q 1
2 2
dx xsin
AA x−
= − ∫
⇒ w1 1 0xxsin sin t
A A− − − =
— (iii)
Take 1 0xsin
Af−
=
(iii) ⇒ 1 xsin
A−
= (wt + f)
\ xA
= sin(wt + f)
⇒ x = A sin(wt + f) — (iv)
v = dxdt
= Aw cos(wt + f)
a = dvdt
= -Aw2sin(wt + f) = -w2x
Hence the expressions for displacement, velocity and acceleration in linear SHM are
x = Asin(wt + f)
v = Awcos(wt + f) = w 2 2A x−
and a = -Aw2sin(wt + f) = -w2x
Concept Strand 1
A particle is executing linear S.H.M. What are its velocity and displacement when its acceleration is half the maxi-mum possible?
Solution
a = -Aw2sin(wt + f)
amax = -Aw2
maxa2
= -w2A2
= -Aw2sinp6
\ phase (wt + f) = p6
\ v = Awcosp6
= Aw3
2
x = Asinp6
= A2
It should be noted that in SHM, the angular frequency of displacement, velocity and acceleration are all the same each equal to w.
ConCept Strand
Oscillations and Waves 3.5
displacement amplitude
The maximum displacement xmax = ±A {Q sin (wt + f) = ± 1}.A is the displacement amplitude of SHM.
phase
The term (wt + f) is called the phase of the oscillation. It determines how far the particle has moved away from the mean position and whether it is moving towards or away from the mean position. i.e., the phase determines the state of the particle in SHM.
The SI unit of phase is radian (rad). However, phase can also be measured in units of time (s). As w = positive constant ⇒ wt = angle (radian). Hence w has unit of radian per second (rad s-1). i.e., w is the angular frequency of SHM. The term f is called the phase constant or initial phase or epoch.
Phase at the mean position (or equilibrium position O)
At x = 0,
0 = A sin (wt + f)
⇒ (wt + f) = np where n = 0, 1, 2, 3, …..
When n = 0, wt + f = 0
⇒ v = Aw cos(wt + f) = Aw, which is the maximum magnitude of velocity.
Hence the particle is moving towards the right with maximum velocity at the mean position.
a = -w2 x = 0 ⇒ restoring force on the particle is zero.
When n = 1, wt + f = p
⇒ v = Aw cos(wt + p) = -Aw
Hence the particle is moving towards the left with maximum velocity at that instant. Phase of the particle corresponding to n = 2, 4, 6,…..are the same as that for n = 0 and the phase corresponding to n = 3, 5, 7,…..are the same as that for n = 1.
Phase at the extreme positions
At x = A,
A = A sin(wt + f)
⇒ (wt + f) = ( )pp p 2n 15
, ,...,2 2 2
+, where n = 0, 2, 4,……
When n = 0, wt + f = p2
v = Aw cos(wt + f) = Aw cosp2
= 0
i.e., particle is at instantaneous rest at the extreme position, x = A
a = -w2x = -w2A, maximum at the extreme posi-tion.
At x = -A
-A = Asin (wt + f)
wt + f = ( )pp p 2n 13 7
, ,.....2 2 2
+, where n = 1, 3, 5, …..
when n = 1, wt + f = p32
\ v = Awcos(wt + f) = Awcosp32
= 0
i.e., the particle is at instantaneous rest at the other extreme position x = -A
We have phase q = wt + f. Hence as time t increases, the phase increases (considering f is positive). An increase of 2p radian in phase brings the particle to the same state of motion. i.e., a phase of (wt + f) and a phase of (wt + f + 2p) are equivalent (same state of motion).
Time period of SHM (T)
We have x = A sin(wt + f) for the displacement x at any time t in SHM.
If T is the minimum time in which the motion repeats itself, then
x = A sin(wt + f) = A sin[w(t + T) + f]
⇒ sin(wt + f) = sin[w(t + T) + f]
We know that sin(wt + f) = sin(wt + f + 2p)
⇒ w(t + T) + f = wt + f + 2p
⇒ wT = 2p
T = pw2
Since w = km
,
3.6 Oscillations and Waves
we get T = 2p mk
T = 2p × Inertia factor
Force constant
Frequency of SHM
Since f = 1T
,
f = p1 k
2 m
i.e., f = p
Force constant12 Inertia factor
×
Angular frequency (w) of SHM
We have already defined the angular frequency w = 2pf
= km
i.e., w = Force constantInertia factor
Concept Strand 2
Two particles begin executing linear SHM of the same amplitude simultaneously along the x-axis. Particle 1 has
frequency w1 and phase constant f1 = p6
. Particle 2 has
frequency w1
2 and phase constant f2 =
p3
. Will they meet and if so, at what time?
Solution
x1 = Asinp
w1t 6 +
x2 = Asinw p1 t2 3
+
If they are at the same location at time t, x1 = x2
⇒ p
w1t 6 +
= w p1 t2 3
+
\ w p1 t2 6
=
⇒ t = pw13
The time periods of the particles are
T1 = pw1
2
and T2 = pw1
4
Since t < 1 2T T,
4 4, the two particles will meet before they
reach the maximum amplitude.
ConCept Strand
Different forms of equation for SHM
In the derivation of the equation x = A sin(wt + f), we had started the time measurement (t = 0) at x = x0 so that
1 0xsin
A−
= f, the initial phase or epoch. If we had started
the time measurement from the mean position (i.e., x = 0 at
t = 0), then we get x = A sinwt (or f = 0). If we had started the time measurement from the extreme position on right
(i.e., x = +A, at t = 0), then f = p2
⇒ x = A sinp
wt2
+ =
A coswt. Thus the sine form and cosine form of SHM are equivalent. However, the phase constant (f) depends on the form chosen.
Oscillations and Waves 3.7
Graphical representation of SHM
2π 0
π 23π
ωt
Displacement – Time graph
3π 3π2π 25π
x
A
2π 0
π 23π
v
2π
Velocity – Time graph
25πAω
−Aω2 2π0
π 23π
ωt
Acceleration – Time graph
25π
a
Fig. 3.3
ωt
2π 3π
Relation between displacement x and acceleration a
We have a = -w2 x . A graph between a and x is a straight line with negative slope as shown in Fig. 3.4
Fig. 3.4
−A
+ω2A
+A
a
x
–ω2A
note:The acceleration a in SHM leads displacement x by p radian.
Relation between displacement x and velocity v in SHM
We have v = w 2 2A x−
⇒ v2 = w2(A2 - x2) ⇒ v2 + x2w2 = w2A2
\ w
2 2
2 2 2
v xA A
+ = 1
This is of the form 22
2 2
yxa b
+ = 1, which is an ellipse, as shown in Fig. 3.5
Fig. 3.5
v
+A x
−A
−ωA
ωA
Velocity-displacement graph in SHM
When x = ± A, v = 0 and when x = 0, v = ± wA
notes:
(i) When w = 1 rad s-1, vmax = ±A, the shape of the graph relating v and x becomes a circle of radius A.
(ii) The velocity v in SHM, leads displacement x by p2
radian.
Relation between velocity v and acceleration a in SHM
We have a = -w2 x ⇒ x = w2
a
We have v2 = w2[A2 - x2]
= w2w
22
4
aA
−
= A2w2 - w
2
2
a
\ v2 + w
2
2
a = A2w2
or w w
2 2
2 2 2 4
v aA A
+ = 1, which is the equation for an ellipse
Fig. 3.6
v+Aω
+Aω2a−Aω2
−Aω
When a = 0, v = ± Aw and when v = 0, a = ± Aw2
3.8 Oscillations and Waves
notes:
(i) If w = 1 rad s-1, the graph between v and a becomes a circle of radius A.
(ii) The acceleration a in SHM leads velocity v by p2
radian.
Simple harmonic motion and uniform circular motion
Simple harmonic motion can be visualized as the projection of two dimensional circular motion along a diameter. Figure 3.7 shows a particle P in uniform circular motion.
Fig. 3.7
A si
n ( ω
t + φ
)
M
Q
N X
Y
Pωtφ
A cos(ωt + φ)
AA
The particle is at the position P at t = 0 and revolves with constant angular velocity w, along a circle. The projection of P on the diameter along X axis is M. At a later time t, the particle is at Q. Its projection now on the diameter is N. As the particle P revolves around in a circle anti-clockwise, its projection M follows it up moving back and forth along the diameter such that the displacement of the point of projection at any time t is the x component of the radius vector A,
x = A cos(wt + f) — (1)
If, on the other hand, we consider the y component of the radius vector, it varies as
y = A sin (wt + f) — (2)
We thus see that uniform circular motion is the combination of two mutually perpendicular linear
harmonic oscillations. (of the same amplitude A and angular frequency w).
Energy of a simple harmonic oscillator
Consider a particle of mass m, executing linear SHM of angular frequency w and amplitude A. The displacement x, velocity v and acceleration a at any time t are given by
x = A sin (wt + f) v = wA cos (wt + f) = w 2 2A x− and
a = -w2x = -w2A sin(wt + f)
The restoring force F acting on the particle is given by F = -kx, where k = mw2
Kinetic energy of particle KE = 21mv
2
= 12
mw2A2 cos2 (wt + f)
= ( )w2 2 21m A x
2−
KE = 12
mw2A2 cos2 (wt + f)
= 12
mw2(A2 - x2)
Since cos2 (wt + f) = ( )w ϕcos2 t 1
2+ +
KE = 14
mw2A2 [cos2(wt + f) + 1)
Thus KE in SHM is periodic with angular frequency of 2w (i.e., twice the angular frequency of SHM).
When the particle has been displaced from x to x + dx, the work done by the restoring force is
dW = Fdx = -kx dx
Total work done by the restoring force in displacing the particle from x = 0 (mean position) to x = x is
W = x 2
0
kxdW kxdx
2= − = −∫ ∫
= -w2 2m x2
(Q k = mw2)
= - w2m2
A2sin2 (wt + f)
Taking potential energy at x = 0 as zero, change in poten-tial energy at position x is the potential energy at position x.
Oscillations and Waves 3.9
\ PE = negative of work done by restoring force
= w w2 2 2 2m x m A2 2
= sin2 (wt + f)
PE = 12
mw2A2sin2(wt + f) = w2 21m x
2
Since sin2 (wt + f) = ( )w ϕcos2 t 1
2+ −
, the potential
energy in SHM is periodic with angular frequency of 2w (i.e., twice the angular frequency of SHM).
Total mechanical energy of the particle
E = KE + PE
\ E = 12
mw2(A2 - x2) + 12
mw2x2
E = 12
mw2A2
Hence the total energy of the particle in SHM is constant and is independent of the instantaneous displacement:
The relation between KE, PE and time in SHM is shown in Fig. 3.8., which assumes that t = 0 when x = ± A
t
EE = Total energy
21
mω2A2
PE
KE
Fig. 3.8
The variation of KE and PE in SHM with displacement x is shown in Fig. 3.9
E = Totalenergy
21= mω2A2
x = +A
KE (parabolic)
PE (parabolic)x = 0x = −A
Fig. 3.9
Concept Strand 3
A particle of mass 0.1 kg executes SHM of amplitude 0.2 m. When the particle passes through the mean posi-tion, its K.E is 0.032 J. The initial phase is 30°. What is the equation of motion?
Solution
w2 2 2max
1 1m v m A 0.032J
2 2= = ⇒ w2 = 2
0.032 20.1 (0.2)
××
⇒ w = 4 rad s-1 ⇒ y = 0.2 sin (4t +
p6
) m
ConCept Strand
Combinations of simple harmonic motions
Let us consider two simple harmonic motions represented by x1 = A1sinwt and x2 = A2 sin (wt + f) superimposed on a particle of mass ‘m’. The latter is shifted in phase f from the former.
Its resultant displacement x = x1 + x2
= A1sinwt + A2sin(wt + f)
= A1sinwt + A2 sinwt cosf + A2 coswt sinf
= [A1 + A2cosf] sinwt + A2sinf coswt
Put A1 + A2cosf = A cosq and A2sinf = A sinq. Then x = A cosq sinwt + A sinq coswt x = A sin(wt + q)
which is also an SHM of amplitude A.
A = ( ) ( )2 21 2 2A A cos A sinϕ f+ +
A = 2 21 2 1 2A A 2A A cosf+ +
tanq = 2
1 2
A sinA A cos
ff+
3.10 Oscillations and Waves
when f = 0, amplitude A = 2 21 2 1 2A A 2A A+ +
= (A1 + A2)
when f = p, amplitude A = (A1 - A2)
when f = p2
, amplitude A = 2 21 2A A+
The velocity of the particle is
v = wAcos(wt + q)
exaMpleS Of SiMple harMOniC MOtiOn
angular simple harmonic motion
A body free to rotate about an axis can make angular oscillations. For example, if a photo frame or a calendar suspended from a nail on the wall, is slightly pushed from its mean position and released, it makes angular oscillations.
For an angular oscillation to be angular simple harmonic motion, a body which has been shifted from its mean position and released must experience a net torque that is restoring in nature (or trying to bring the body back to the mean position). If the angle of oscillation is small, this restoring torque will be directly proportional to the angular displacement.
t ∝ -q (q = angular displacement and t = net torque on the body)
\ t = -kq (k is a positive constant)
But t = Ia (I = moment of inertia of the body and a = angular acceleration)
\ Ia = -kq
i.e., Iq2
2
ddt
= -kq
\ q
I
2
2
d kdt
= − q = -w0
2q wI0k
=
\ q2
2
ddt
+ w02q = 0
This is the differential equation of an angular SHM.The solution for this is
q = q0 sin (w0t + f)
where q0 = Maximum angular displacement on either side of mean position and w0 = angular frequency of SHM = 2pfAngular velocity at time t is given by
w = q0w0cos(w0t + f)
Simple pendulum
A point mass (usually a metallic bob or some object of small size) attached to a light, inextensible string and suspended from a fixed support, is called a simple pendulum. Usually, the vertical line passing through the fixed support is the mean position of the simple pendulum. The vertical distance between the point of suspension and the centre of mass of the suspended body (when it is in the mean position) is called the length of the simple pendulum, denoted by L
Time period of a simple pendulum of length L
A point mass M suspended from the end of a light inexten-sible string whose upper end is fixed to a rigid support, is a simple pendulum (Fig. 3.10). Let the length of the pendulum be L. For small angles of oscillation, (in the figure, q has been drawn large for better visual effect) the force equation is
T - Mg cos q = 2M v
L,
where, T is the tension in the string and v = speed of bob when angular displacement is q. The torque tending to bring back the bob to equilibrium position is
t = Mg sin q × L = Ia
Where, I is the moment of inertia and a is the angular acceleration. For small angles of oscillations, sinq ≈ q.
Fig. 3.10
L Tθ
Mg cosθ
Mg sin θ
MgA
hB
Oscillations and Waves 3.11
\ I a = - Mg L q
\ a q w qI
20
Mg L−= = − (where w0
2 = I
MgL)
q2
2
ddt
= -w02q
is the equation for a simple pendulum(angular simple harmonic motion). w0 is the angular frequency of SHM given by
w0 = I
Mg L
Using I 2ML= , we get
w0
gL
= ⇒
The time period is T = pw0
2
pL
T 2g
=
Energy of a simple pendulum
From Fig. 3.10, the P.E of the bob at point B with respect to point A is
Mgh = Mg(L - Lcosq) = MgL(1 - cosq)
= MgL q21 1 2sin2
− − = MgL.2 q
2
2
[Q cosq = 1 - 2sin2q2 and sin
q2
= q2
for small angles]
= 12
MgLq2
\ P.E = q21MgL
2
K.E of the bob is
12Iw2 = q
221 d
ML2 dt
= 21ML
2[q0
2w02cos2(w0t)]
= 12
ML2w02[q0
2{1 - sin2(w0t)}]
= 12
ML2gL [q0
2 - q2] [Q q0sinw0t = q]
= q q2 20
1 1MgL MgL
2 2−
\ Total energy E = K.E + P.E
= q20
1MgL
2 = constant =
12
MgL(1 - cosq0)
We see that the energy of the pendulum is conserved and is equal in magnitude to the P.E at the maximum amplitude (where, K.E is zero), an obvious result.
Seconds pendulumA pendulum whose time period is 2 second is called a sec-onds pendulum.
For a seconds pendulum, T = 2 s = pL
2g
⇒
p2
gL 0.993 m= =
notes:
(i) If the temperature of a system changes, then the time period of the simple pendulum changes due to change in length of the pendulum.
(ii) The expression T = 2p Lg
holds good only for small
angular displacements from the mean position. (iii) If a simple pendulum is placed in a non-inertial frame
of reference (e.g., an accelerated lift, horizontally accelerated vehicle, vehicle moving along an inclined plane etc.), the mean position of the pendulum may change. In these cases, ‘g’ is replaced by ‘geffective’ for determining the time period ‘T’.
For example, for a simple pendulum inside a lift moving upwards with acceleration ‘a’,
T = 2p ( )peff
L L2
g g a=
+
If the lift is moving down with an acceleration a,
T = 2p ( )L
g a− For a simple pendulum inside a car moving on a
horizontal road with an acceleration ‘a’,
T = ( )
p 12 2 2
L2
g a+
( )Q 2 2
effg g a= +
(iv) Time period of a simple pendulum in free fall or inside an artificial satellite is infinity
(Q geff = zero in such cases)
3.12 Oscillations and Waves
(v) We have acceleration = -w2 × displacement in SHM. Hence
w = accelerationdisplacement
⇒ T = pw2
= 2p × displacementacceleration
(vi) For simple pendulum of length L comparable to radius of Earth(R), it can be shown that time period
T = 2p1
g gL R +
. Where g = acceleration due to
gravity near the surface of Earth. (vii) For infinitely long pendulum (L >> R) near Earth,
T = 2p Rg
≃ 5078 second (84.6 minute)
Spring-Mass systems
(a) Oscillations of loaded spring
Consider a weightless spring of spring constant k hung from a rigid support and loaded with a mass M at the free end. Let the original length of the spring be l0; the length when the mass is hung, l1; and the length when the mass is pulled down, l2. An additional force F is applied to pull the spring down.
�0
MgMg
y
(a) (b) (c)
Fig. 3.11
�1�2
The force equations corresponding to the situations (b) and (c) of Fig. 3.11 are:
\ F = - ky, where y = displacement from mean position (length l1 for mean position)
The force being proportional to the displacement y and directed opposite to it, the motion is simple harmonic.
The time period is p pl l1 0( )M
T 2 2k g
−= = (us-
ing equation (b))
notes:
(i) The time period T of a spring-mass system is independent of acceleration due to gravity ‘g’ because
T = 2pMk
We had earlier written T = 2pl l1 0
g−
, but as ‘g’
changes, (l1 - l0) also changes and hence l l1 0
g−
is a constant.
(ii) The time period of a spring-mass system is the same in all orientations of the spring pendulum. Different orientations only change the equilibrium (mean) positions but time period remains unaltered.
(iii) If the mass of the spring is not negligible, then some weightage will require to be given for the mass of the spring. If ms is the mass of the spring, it can be shown
that the time period T is given by T = 2psm
M3
k
+
Hence 1 rd3
of the mass of the spring has to be
added to the connected mass M to determine the time period.
(iv) If two or more springs are connected to a block in parallel or in series combination, then the equivalent spring constant of the combination has to be used in determining the time period of SHM.
(b) Series combination of light springs
Fig. 3.12
k1 k2 M
smooth floor
k3
Oscillations and Waves 3.13
As shown in Fig. 3.12, three light springs are connected in series to a mass M. Let the total extension be x and the extension in each spring be x1, x2 and x3. The same restoring force F is acting through all the springs when they are connected in series. i.e., all the springs exert the same force in series combination, because there are no masses in between.
We have x = x1 + x2 + x3 — (1)
If ks is the effective spring constant in series combina-
tion, x = s
Fk
,
x1 = 1
Fk
, x2 = 2
Fk
and x3 = 3
Fk
\ (1) ⇒ s 1 2 3
F F F Fk k k k
= + +
s 1 2 3
1 1 1 1k k k k
= + +
Hence the effective spring constant in series combination is smaller than the smallest spring constant of the given springs.
A spring of N turns and spring constant k, can be considered as a series combination of N springs, each of spring constant k1.
i.e., 1 1
1 1 1k k k
= + +…….(N terms)
i.e., 1
1 Nk k
= ⇒ k1 = Nk ⇒
k = 1kN
i.e., the spring constant of a spring is inversely proportional to the number of turns in it.
k ∝ 1N
But the length of the spring (l) is directly proportional to N i.e., l ∝ N
k ∝ 1N
∝ l1
m n
km kn
Fig. 3.13
If a spring of length ‘l’ and spring constant k is cut into two pieces of lengths in the ratio m : n, the spring con-
stants of the two pieces are given by km = ( )k m n
m+
and kn
= ( )k m n
n+
. i.e., k(m + n) = kmm = knn. Hence if a spring of spring constant k is cut into N identical parts, the spring constant of each piece is k1 = Nk
If a mass of m is hung and allowed to oscillate from each cut piece, the time period of oscillation will be
T1 = 2p p1
m m T2
k Nk N= = ,
where, T = 2pmk
, the time period of oscillation of
the mass attached to the original uncut spring.
(c) Parallel combination of light springs
Fig. 3.14
m
k1
k2
k3
smooth floor
m
k1 k3k2
smooth floor
or
In parallel combination of springs, the extension of each spring is the same, say ‘x’.
Hence the spring forces are F1 = k1x, F2 = k2x, F3 = k3x
Total spring force F = F1 + F2 + F3
If kp is the effective spring constant in parallel combination, F = kpx
i.e., kpx = k1x + k2x + k3x
⇒ kp = k1 + k2 + k3
3.14 Oscillations and Waves
Concept Strand 4
A mass M is hung from two springs suspended from a rigid support as shown. Find the time period, if the exten-sions of the springs are equal.
Solution
At any instant, the displacement is y. The re-storing forces are -k1y and -k2y where k1 and k2 are the force constants of the two springs.
\ F = -(k1 + k2)y
Effective force constant is k = (k1 + k2)
The time period is T = p1 2
m2
k k+
Concept Strand 5
M
k1
k2
A body is attached between two springs fixed to rigid supports as shown. Calculate the time period.
Solution
Let y be the displacement at any instant. The restoring force in the two springs will be in the same direction, because when the mass is displaced, one of the springs is elongated and the other compressed, the restoring force being in the same direction in each of the springs.
\ F = -(k1 + k2)y
Effective force constant k = k1 + k2
\ T = p1 2
m2
k k+
Concept Strand 6
A mass is attached to two springs joined in series. Find the time period.
Solution
If y is the displacement at any instant y = y1 + y2. Where y1 and y2 are the displacements in springs 1 and 2.
\ F1 = - k1 y1; F2 = - k2y2
\ F = - ky = - k(y1 + y2)
= 1 2
1 2
F Fk
k k −
− −
\ 1 2
1 2
F FFk k k
= +
But F = F1 = F2
\ 1 2
1 1 1k k k
= +
\ T = p 1 2
1 2
M k k2
k k+
M
k1 k2
T = T’
T
T
1
Mg = T
∴T = T’ = Mg
T
Mg
2
F.B.D
ConCept StrandS
(d) Two body spring systems
−FF
smooth
x1
x2
m1 m2
Fig. 3.15
k
Consider two blocks of masses m1 and m2 connected by a light spring of spring constant k and made to oscillate as
shown in Fig 3.15. It is desired to find the time period of oscillation T
If L is the length of the spring, the extension of spring x is given by x = (x2 - x1) - L or x + L = (x2 - x1)
The spring force F is the same on each block but in opposite directions.
Applying Newton’s laws of motion separately to each block, we have
m1
21
2
d xdt
= +kx — (i)
m2
22
2
d xdt
= -kx — (ii)
Oscillations and Waves 3.15
(ii) × m1 - (i) — m2
⇒ m1m2
2 22 1
2 2
d x d xdt dt
−
= -kx[m1 + m2]
( )( )
2 21 22 1
2 21 2
m md x d xkx
m mdt dt
+− = −
( ) ( )2
1 22 12
1 2
m mdx x kx
m mdt
+− = −
( ) ( )2
1 22 12
1 2
m m dx x
dtm m−
+ = -kx
( )2
1 22
1 2
m m ddtm m+
(x + L) = -kx
m2
2
d xdt
+ kx = 0
m
2
2
d x kx
dt+ = 0
⇒ w = mk
⇒ T = 2p ( )m
p 1 2
1 2
m m2
k m m k=
+
note:
m = 1 2
1 2
m mm m+
is known as the reduced mass of m1 and m2
the physical pendulum
Simple pendulum has a point mass suspended by a light, inextensible string. This is not achievable in reality. So we say that it is an idealized model.
A physical pendulum is a real pendulum, in which a body of finite shape and size oscillates. From the frequency of oscillation we can calculate the moment of inertia of the body about the axis of rotation.
Consider a body of irregular shape and mass m, pivoted from a point O as shown. The weight mg acts down at the centre of gravity G; at equilibrium, the centre of gravity is directly below the point of suspension O. Figure 3.16 shows the body displaced through a small angle q. If the body is released from this position, a torque t is exerted by the weight of the body to restore it to equilibrium. We then have t = -(mg)(d sinq), the negative sign showing that the restoring torque is clockwise while the displacement is counter clockwise and vice versa.
The equation of motion can be written as
t = Ia = Iq2
2
ddt
= -mg d sin q
where I is the moment of inertia of the body about the axis of rotation. Applying the small angle approximation,
qq
I
2
2
mgdddt
= −
This equation represents SHM with angular frequency
w0 = I
mgd
The time period is
T = 2p Imgd
The formula derived here for the time period can be used to determine the moment of inertia of an irregular solid about its axis of rotation.
torsional pendulum
Torsional pendulum consists of a wire of length L and radius r hung from a rigid support and loaded with a heavy circular disc such that the wire passes through the centre of gravity of the body (Fig. 3.17).
When the wire is twisted through an angle q, the sys-tem acquires potential energy and when released, it executes angular oscillations. These are called torsional oscillations.
θ
Fig. 3.17
3.16 Oscillations and Waves
Conservation of energy requires that
Iw q2 21 1c constant
2 2+ =
where the first term is the K.E. and the second term the P.E. I is the moment of inertia and c is the couple per unit twist,
ph 4rc
2L=
Here r = radius of suspension wire, L length of suspension wire and h = rigidity modulus of material of suspension wire.
Differentiating the total energy expression, we get q
qI
22d d c d
( )dt dt dt
− = ⇒
qq
I
2
2
d cdt
= −
The expression shows that the torsional oscillations are angular harmonic with time period
IpT 2
c=
Oscillations of a liquid column
The liquid in a U-tube will perform SHM if the liquid level in one limb is depressed slightly and released. Let the total
length of the liquid in the U-tube be l. The liquid level in one limb is depressed by a height x. The liquid level in the other limb rises through a height x so that the net difference in height is 2x, as shown in Fig. 3.18
Force on liquid level at A’ = 2x.arg. The force act-ing on the mass of liquid of length l causes acceleration
= rrl l
2xa g 2gx
a− −
=
(Q acceleration is downwards at A; while displacement is upwards)
Force is proportional to x ⇒S.H.M ⇒ w = l
2g
Vertical oscillations of a loaded wire of negligible mass
A mass m is suspended from a wire of cross section A, length L and negligible mass. Let Y be the Young’s modulus of the material of the wire. Then due to the attached mass, the wire is set into oscillations. Let l be the extension. Then
Y = l l
Fstress FLAstrain A
L= =
Therefore, the restoring force
F = -YAL
l
(Q Restoring force is opposite in direction to the displacement)\ Force constant k =
YAL
= mw2
\ w = l
YAmFig. 3.18
x x
A
A’
daMped OSCillatiOnS
In previous sections, we discussed the oscillations of typical systems uninfluenced by external forces. They are called free oscillations. On the other hand, when the medium in which the body oscillates provides frictional forces against the motion, the oscillations get damped in time. The damping force is usually proportional to the speed of the particle, given by
F’ = -bv = dy
bdt
−
where, b is a constant. The equation of motion of the particle becomes
2
2
d y dym b ky
dtdt= − −
The solution of this differential equation can be obtained as
y(t) = b t
2mAe cos[ ' t ]w f−
+
Oscillations and Waves 3.17
Fig. 3.19
where the frequency w' is given by
w' =
12 2
2
k bm 4m
−
.
The factor b t
2me−
makes the amplitude of the oscillation decrease continuously (Fig. 3.19).
fOrCed OSCillatiOnS and reSOnanCe
Suppose an oscillatory time-dependent force of the form F(t) = F0 cos wdt is acting on a damped oscillator. The oscil-lator equation can be written as
w2
0 d2
d y dym b ky F cos t
dtdt+ + =
It can be shown that the solution is y(t) = A cos (wdt + f), where the amplitude A is given by
A = w w w
01
2 2 2 2d d
F
m ( ) b − +
The following observations can be made:
(i) The system oscillates at the forcing frequency wd (ii) The amplitude increases as the forcing frequency
wd approaches the natural frequency w of the oscillator.
(iii) The amplitude depends on the magnitude of the damping constant b, (Fig. 3.20).
ωω = ωd
b3 < b2
b2 < b1
b1
Fig. 3.20
Amp
(iv) If there is no damping, amplitude increases to such large values as would destroy the system. Increase in the amplitude of the oscillation when w = wd is known as Resonance. The phenomenon of resonance has many useful applications. In the field of communication, radio signals are picked up by means of electrical resonant circuits.
S u m m a r y
T = pl
2g a+
T = period of a simple pendulum inside a lift moving upwards with an acceleration ‘a’. l = length of pendulumg = acceleration due to gravity
T = pl
2g a−
T = period of a simple pendulum inside a lift moving downwards with an acceleration ‘a’.
T = ( )p
l1/22 2
2g a+
T = period of a simple pendulum inside a car or vehicle, moving with a horizontal acceleration of ‘a’. (which could be centripetal acceleration also)
3.18 Oscillations and Waves
T = ( )
rp
r sl
2g−
; geff = sr
1 −
g T = time period of a simple pendulum, made of bob (density ‘r’) oscillating inside a non viscous liquid (of density ‘s’)
T = pl
2g
T = 2 s and l ≈ 1 m for seconds pendulum
T = time period of a simple pendulum suspended from a body at rest on an inclined plane of inclination ‘q’ with horizontal.
T = pq
l2
g cos
T = time period of a simple pendulum suspended from a body sliding down on an inclined plane of inclination ‘q’ with horizontal.
Period of oscillation, T = pM
2k
PE of oscillation = 21ky
2Where, y = displacement Maximum PE = Maximum KE
= 21kA
2where, A = amplitude
KE of oscillation = ( )2 21k A y
2−
Restoring force F = ky and k = Fy
If a body of mass ‘M’ is connected to a spring of spring constant k and made to execute SHM, then
M
Time period of simple pendulum in free fall or inside an artificial satellite of Earth
Infinity
eq 1 2 n
1 1 1 1.....
k k k k= + + +
keq = equivalent spring constant of a series combination of ‘n’ springs of spring constants k1, k2, …… kn.
keq = k1 + k2 + ……. + kn keq = equivalent spring constant of a parallel combination of ‘n’ springs of spring constants k1, k2,…….. kn
T = pM
2nk
If a spring of spring constant k is cut into ‘n’ identical parts, time period of oscillation of a mass ‘M’ suspended from one part
T = p
mM32
k
+
Time period of oscillation of a mass ‘M’ from a spring of mass ‘m’ and spring constant ‘k’
T = mp2
k, where m = reduced mass of system
= ( )
1 2
1 2
M MM M+
kSpring of negligible mass
M1 M2
T = time period of oscillation of connected system of two masses M1 and M2 with a spring as shown.
Oscillations and Waves 3.19
T = p1 2
M2
k k+
T1 = p1
M2
k, T2 = p
2
M2
k
2 2 21 2
1 1 1T T T
= +
k = k1 + k2
k1 k2
M
T1 = time period with spring k1,T2 = time period with spring k2,T = time period with parallel combination of springs k1 and k2
k = equivlent spring constant in parallel combination.
T = ph
2g
, where
T = period of oscillation of liquid column in a U-tube, h = height of liquid in each limb of U-tube.g = acceleration due to gravity
h
Period of oscillation of a uniform float in a liquid,
T = psm
2A g
Also T = rp
sh
2g
m = mass of floatA = cross-sectional area of floats = density of liquidg = acceleration due to gravityh = height of float (total)r = density of floatm = Ahr
Period of oscillation of a torsional pendulum T = Ip2
C,
where I = moment of inertia of oscillating body about the axis of rotation,
C = Couple per unit twist of suspension wire = p h
l
4r2
,
where h = rigidity modulus of wire, r = radius of suspen-sion wire, l = length of wire
�
Time period of oscillation of a compound pendulum,
T = Ip2
mgH
•
H
CG
m = mass of bodyI = moment of inertia of body about axis of rotationH = depth or distance of centre of gravity of body from axis
of rotation.g = acceleration due to gravity
3.20 Oscillations and Waves
(i) Mechanical waves (also called elastic waves) Waves which require a medium for their propagation
are called mechanical waves or elastic waves. Waves on strings, sound waves and waves on the
surface of water are examples of mechanical waves. A mechanical wave is produced due to a
disturbance at a point in a medium. The disturbed particle interacts with the neighbouring particle and its energy is handed over to the next particle (due to inertia) and to the next and so on. The disturbed particles return to the equilibrium positions (due to elasticity of medium). Thus for a mechanical wave to propagate through a medium, the medium must possess elasticity and inertia.
T = I
p2MB
T = period of oscillation of a bar magnet in a uniform magnetic field
M = magnetic moment of bar magnetB = external uniform magnetic field (magnetic induction)I = moment of inertia of bar magnet about axis of rotation
T1 = 2p1
Mk
T2 = 2p2
Mk
T = 2p ( )1 2
1 2
M k kk k
+
T2 = T12 + T2
2
1 2
1 1 1k k k
= +
k = ( )1 2
1 2
k kk k+
T1 = time period with spring k1 aloneT2 = Time period with spring k2 aloneT = Time period with series combination of springs k1 & k2
k = Equivalent spring constant of series combination
KE = E0cos2(wt + f)PE = E0sin2(wt + f)
( )w ϕ2PEtan t
KE= +
( )2
2 2
yPEKE A y
=−
E0 = Maximum energy of a particle executing SHMw = angular frequency of SHMf = initial phase of SHMKE = kinetic energy of SHM at instant ‘t’ (displacement y)PE = potential energy of SHM at instant t (displacementy)A = Amplitude of SHM
Wave MOtiOn
A wave is formed due to a disturbance created in a region. A part of the energy associated with the creation of the wave is carried by the wave along the direction of propagation. A wave motion transfers energy from one place to another without the transfer of matter. A wave also possesses momentum. Energy and momentum are transported by the wave in the direction of its propagation.
Classification of waves
(a) Classification based on medium of propagation
Based on medium of propagation, waves are classified into: (i) Mechanical Waves and (ii) Non-mechanical waves
Oscillations and Waves 3.21
wave, it is called a transverse wave. Transverse waves can only travel through a medium which has rigidity modulus or shape conservation. Hence transverse mechanical waves can travel through solids and on the free surface of liquids (which has surface tension for shape conservation). Transverse mechanical waves can be produced on stretched strings by vibrating it perpendicular to its length.
The region of maximum upward displacement is called crest. The region of maximum downward displacement is called trough.
TFree end
CCrest
C C
TTrough
Fixed end
Transverse wave on a string
Fig. 3.21
(ii) Longitudinal waves If the particles of the medium vibrate about their
mean position in a direction parallel to the direction of propagation of the wave, it is called a longitudinal wave. Longitudinal waves can travel through solids, liquids and gases as the medium requires only elasticity of volume (or bulk modulus) for its propagation. Longitudinal wave can be produced on a stretched string by making its free end vibrate parallel to its length. The vibrating prongs of a tuning fork brought near the mouth of a pipe produce longitudinal sound waves in the tube. The longitudinal wave travels through a medium in the form of compressions and rarefactions. The region of high pressure is called compression and the region of low pressure is called rarefaction.
The particles of the medium execute periodic motion about a mean position when the wave propagates through the medium.
(ii) Non-mechanical waves Waves which do not require a medium for their
propagation are called non-mechanical waves. i.e., they can propagate through vacuum also. These waves are transverse in nature.
Electromagnetic waves and matter waves (probability waves) are example of non-mechanical waves. Electromagnetic waves consist of a time varying electric field and a time varying magnetic field at right angles to each other and in phase. Both the electric field and the magnetic field are perpendicular to each other and also to the direction of propagation of the wave. Light waves, radio and television signals, X-rays etc., are electromagnetic waves.
(b) Classification of waves based on mode of propagation
One-dimensional waves are waves traveling only along a straight line.Example, Waves on stretched strings
Two-dimensional waves are waves traveling in a plane.Example, Ripples on water surface.
Three-dimensional waves are waves traveling in space.Example, Sound waves, light waves etc.
(c) Classification of mechanical waves based on vibration of particles of the medium
Based on the mode of vibration of the particles of the medi-um, mechanical waves are classified as (i) transverse waves and (ii) longitudinal waves.
(i) Transverse waves If the particles of the medium vibrate in a direction
perpendicular to the direction of propagation of the
Compression
RarefactionSound wave through air
Compression
Free endRarefaction
Longitudinal wave on a stretched spring
Fig. 3.22
3.22 Oscillations and Waves
Period (T) of a wave is the time taken by any particle of the medium to complete one vibration. Its SI unit is second (s). During a period T, the wave advances a distance equal to its wavelength.
Wavelength (l) is equal to the distance between two consecutive particles of the medium which are in the same state of vibration. It is equal to the distance travelled by the wave in the time interval equal to its period (T).
Frequency (f) of a wave is the number of vibrations made per second by any particle of the medium. Its SI unit is hertz (Hz). It is also denoted by n or n.
f = 1T
The frequency of a wave is a characteristic property of the source producing the wave. Hence the frequency of a wave does not change when a wave travels from one medium to another medium.
Wave velocity v is the distance travelled by the wave in one second Thus wave velocity
lv
T= = lf
Its SI unit is metre per second (m s-1).The velocity of the wave is determined by the me-
chanical properties of the medium through which the wave propagates. Since the wave velocity is measured with re-spect to the medium, the wave velocity changes when the medium is in motion (Example, speed of sound through air changes when wind is blowing).
There are two velocities associated with a wave. One is the wave velocity which is explained above. It is a constant in a given isotropic, homogeneous medium. The other ve-locity is the particle velocity, which is the speed with which the particles of the medium vibrate when the wave passes through the medium. The particle velocity is a varying quantity and has maximum value at the mean position and is zero value at the amplitude.
notes:
(i) When a wave travels from one medium to another, its frequency remains same but wave velocity and wavelength change.
(ii) In a transverse wave, wavelength is equal to the distance between two consecutive crests or two consecutive troughs.
(iii) In a longitudinal wave, the wavelength is equal to the distance between the centres of two consecutive compressions or rarefactions.
notes:
(i) Some waves are not purely transverse or longitudinal. The seismic (or Earthquake) waves produced in the interior of Earth, travel both in the form of longitudinal and transverse waves. Such waves have complex nature. Tidal waves in sea are another example of complex waves.
(ii) In a solid medium, the transverse and longitudinal waves travel with different speeds. The speed of a longitudinal wave through a solid is more than the speed of a transverse wave through the same solid.
(d) Classification of mechanical waves based on transfer of energy
On the basis of energy transfer, mechanical waves can be classified as (i) standing waves or stationary waves and (ii) progressive waves.
(i) Standing waves are those waves which remain confined to a region without any transfer of energy and momentum out of that region of the medium. These are also called stationary waves. The amplitude of vibration for different particles are different.
(ii) Progressive waves are those waves which transfer energy and momentum between the particles of the medium. The wave propagates through the medium and all the particles of the medium have same amplitude of vibration. In an isotropic, homogeneous medium, the wave propagates with the same speed in all directions.
periodic waves
If the disturbance in the medium is created only once, it travels through the medium in the form of a wave pulse. If the disturbance is continuous and is periodic in nature, then the wave produced is termed a periodic wave. A repetitive wave train is an example of a periodic wave.
A periodic wave that is varying sinusoidally is called a sinusoidal periodic wave. The particles of the medium execute simple harmonic motion(SHM) when a sinusoidal periodic wave passes through the medium.
Terminology of progressive wave motion
Amplitude (A) of a wave is the maximum displacement of any particle of the medium from its equilibrium (or mean) position. Its SI unit is metre (m).
Oscillations and Waves 3.23
By the time the wave reaches B from A, the phase of vibration of A has changed. The difference between the states of vibration of A and B is called the phase difference Df.
If we choose two consecutive crests C1 and C2, the path difference between them is l, the time difference is T and the phase difference is 2p. (By the time the wave reaches C2 from C1 , C1 has completed one vibration).
A path difference of l corresponds to a phase difference of 2p. Thus a path difference x corresponds to a
phase difference p
l
2 x
\ Phase difference D f = p
l
2 x
Phase difference = pl
2
× path difference
A path difference l corresponds to a time difference T. Therefore, a path difference x corresponds to a time difference
lx
T× .
note:pl
2k = , is called angular wave number or propagation constant of the wave.
It is measured in rad m-1. It gives the phase difference between two points in the medium which are separated by unit path difference. The relations connecting path difference, phase difference and time difference are given in the table below.
Table 3.1Path difference Phase difference Time Difference
x
pl
2 xl
xT
l2Dfp
Df T
2Dfp
l DTT
Dp
T2
T
DT
Phase or phase angle f represents the state of vibration of the particle of a medium with respect to its mean posi-tion. SI unit of phase is radian (rad).
Phase difference Df indicates the different states of vibration of a particle at two different instants or of any pair of particles at the same instant.
Df = f2 - f1, where f1 = phase of particle 1 and
f2 = phase of particle 2 at a particular instant of time/particle 1 at a different instant of time
Path difference x indicates the distance between two points measured along the direction of propagation of the wave through the medium.
Time difference DT indicates the time taken by the wave to travel from one point to another through the medium.
Relation between path difference and phase difference
Consider a progressive wave advancing in the positive direction of x-axis. (Fig. 3.24)
λ C1
x1
x2
C2
x
y
Displacement
x-axisO A B
Fig. 3.24
Let A and B be two points in the medium through which the wave passes. Path difference between A and B, x = x2 – x1.
Transverse wave
λ C C
time
t
y Displacement
λ
O
λ
Longitudinal wave
CC
λ
Fig. 3.23
T T
3.24 Oscillations and Waves
As the wave travels with constant speed v along the x-direction, we can write y = f(vt - x) — (i)
This is the equation of a wave of any shape traveling along the positive direction of x-axis with a constant velocity v. The exact shape of the wave is determined by the function f. If the traveling wave is a sine or cosine function of (vt – x), then it represents a harmonic traveling wave or a plane progressive wave.
Similarly, the equation for a wave traveling along the negative direction of x-direction could be
y = f(vt + x) — (ii)
Hence the function y = f(vt ± x) or f(x ± vt) represents traveling waves along the x-axis.
The general equation of a wave traveling along the x-axis can be written as
y = f1(vt + x) + f2(vt - x) — (iii)
On differentiating equation (iii) twice partially with respect to t, we get
2
2
yt
∂∂
= v2f1”(vt + x) + v2f2”(vt - x) — (iv)
where f1” and f2” are differentials of the second order of f1 and f2 respectively.
On differentiating equation (iii) twice, partially with respect to x, we get
2
2
yx∂∂
= f1”(vt + x) + f2”(vt - x) — (v)
Comparing equation (iv) and (v), we get
2 22
2 2
y yv
t x∂ ∂
=∂ ∂
This is a general differential equation of a one dimen-sional wave along the x-direction.
Hence any function y(x, t) which satisfies the differ-
ential equation 2 2
22 2
y yv
t x∂ ∂
=∂ ∂
represents a one dimensional
progressive wave along the x-direction, where v = constant speed of wave along the x-direction. In addition to the above condition, the wave function must have finite values at all instants.
Progressive waves can be either longitudinal or transverse but they have the same mathematical representation. In order to represent a wave mathematically, a physical quantity is needed whose value oscillates in space and time about its mean value (or equilibrium value) as the wave propagates through the medium. This oscillating physical quality is called the wave function, which is usually denoted as y = f(x, y, z, t). For a one-dimensional wave along the x-direction, the wave function is of the form y = f(x, t).
For example, for transverse waves along stretched springs, y is the displacement of the particle from equilibrium position. For sound waves, y represents the density or pressure fluctuation from mean value(equilibrium value). For electromagnetic waves (non-mechanical), y represents either the electric field or magnetic field.
Y
X
y
yP
Fig. 3.25
O F
Consider a stretched string along the OX direction. The origin O acts as a source of disturbance, which travels as a transverse wave in the positive direction of the X-axis, with a velocity v. As the wave proceeds, every successive point in the string is set in motion. Time measurement (t = 0) starts at the instant when the particle at the origin O is passing through its mean position. The displacement y of this particle at the origin at any instant t is represented by y = f(t), where f(t) is a function of time. P is a point at a
distance x from O. The wave reaches point P in xv
second
i.e., the particle P will start oscillating xv
second later than
the particle at the origin O. This means that the displacement of the particle at P at any instant t will be the same as the displacement of the particle at the origin
xv
second earlier
(i.e., at t - xv
). Hence the displacement of the particle P is
given by y = x
f tv
−
General equatiOn Of a One diMenSiOnal prOGreSSive Wave
Oscillations and Waves 3.25
(iii) y = A sin pl
2(vt – x)
(iv) y = A sin (wt - kx)
where, f = p
w pl
1 2, 2 f , k ,
T= =
w pl
p l2 f
f vk 2 /
= = =
The equations of a progressive wave traveling in the negative direction of x axis can be written as,
y(x,t) = A sin p2 x
tT v +
(i) y = A sin 2pf (t + xv
)
(ii) y = A sin w (t + xv
)
(iii) y = A sin pl
2(vt + x)
(iv) y = A sin (wt + kx),
note:
(i) Equations y = Asin(wt - kx) and y = Asin(kx - wt) represent two waves propagating along the positive x-direction. These equation are not the same as there is a phase difference of p radian between the two waves.
If y = Asin(wt - kx) represents the displacement of a particle at position x at any instant t,
The particle velocity vp = yt
∂∂
= Awcos(wt - kx)
= A(kv)cos(wt - kx)
[Q w = kv]
= -v[-kAcos(wt - kx)]
i.e., vP = -vyx∂∂
[Q -kAcos(wt - kx)
= yx∂∂
]
particle velocity = -(wave speed) × (slope of wave)
Equation of a sinusoidal progressive wave
Consider a progressive wave traveling in the positive direction of x-axis with a wave velocity v. The particles of the medium vibrate simple harmonically with an amplitude A, period T and frequency f. Let w be its angular frequency,
w = p
p2
2 fT
=
x
x axis
t = t
y
Dis
plac
emen
t
P t = 0 O
y
Fig. 3.26
The displacement y of the particle at O, at time t is given by,
y(x,t) = y(0,t) = A sin wt = A sin p2
tT
.
Consider the particle P at a distance x from O. The
wave from O will reach P in a time. t = xv
. For a wave
travelling in the +ve direction of x-axis, the phase of the
particle P lags behind the phase of O by Df.
Df= p2
tT
=p2 x
T v\ The displacement y of the particle P at time t is given
by,
y(x,t) = A sinp p2 t 2 xT T v
−
y(x,t) = p2 x
A sin tT v −
The equations of the progressive wave travelling in the positive direction of x axis can also be written as
(i) y = A sin 2pf ( t – xv
)
(ii) y = A sin w ( t – xv
)
3.26 Oscillations and Waves
Energy density of a wave (U)
The total energy contained in the wave per unit volume of the medium is called the energy density of the wave.
If w is the angular velocity and A is the amplitude of the wave, maximum particle velocity vmax = wA.
Maximum kinetic energy of a particle
= 12
mvmax2 = w2 21
m A2
Total energy per unit volume = Maximum kinetic en-ergy of all the particles per unit volume.
\ Energy density of a wave = U = 12rw2A2 = 2p2f2A2r,
where r is the density of the mediumSI unit of U is J m-3. J m-3 is equivalent to N m-2 which
is the unit of pressure.Hence the pressure exerted by the wave on a surface is
numerically equal to the energy density of the wave.
Power of a wave (P)
The energy contained in a volume element of the medium in unit time is called the power of the wave. Energy con-tained in an element of length ‘Dx’ and area of cross section ‘a’ in a time ‘Dt’ will be E = U aDx
Concept Strand 7
One end of a rope is tied to a peg on the wall. The other free end is held taut by the hand and periodi-cally shaken. A wave travels down the rope sinusoidally with frequency 2.0 Hz and amplitude 7.5 cm. The wave speed is 12.0 m s-1. Find the angular frequency, pe-riod, wavelength and wave number of the wave. Write the equation for the displacement as a function of time.
Solution
Angular frequency w = 2pf = 4p rad s-1 = 12.6 rad s-1
Time period T = 1f
= 0.5 s
wavelength l = vf
= 12.0 × 0.5 = 6.0 m
wave number k = p pl
2 26
= = 1.05 rad m-1
y = Asin(kx - wt) = (0.075 m)sin[(1.05 rad m-1x - (12.6 rad s-1)t]
ConCept Strand
\ Power of the wave P = DD DE ua xt t
=
P = Uav = 2p2f2 A2r av
(Qwave velocity, v = DD
xt
)
Intensity of a wave (I)
Intensity of a wave is defined as the energy transferred in unit time or power transmitted across unit area held per-pendicular to the direction of propagation of the wave. SI unit of intensity is W m-2.
I = Pa
= Uv
= 2p2f2A2rv. Therefore,
Intensity of a wave = Energy density of the wave × speed of the wave
Hence, intensity of a wave is directly proportional to
(i) the square of its amplitude, (ii) the square of its frequency, (iii) the velocity of propagation and (iv) the density of the medium through which it travels.
Concept Strand 8
In Concept Strand 7, calculate the intensity of the wave, if the linear density of the rope is 250 g m-1.
Solution
I = 12w2A2rv = 1
2(12.6 rad s-1)2(0.075 m)2(0.25 kg m-1)(12 m s-1)
= 1.34 W
ConCept Strand
Oscillations and Waves 3.27
locations. This phenomenon of redistribution of intensity in the region of superposition is called interference.
Sources which produce interference phenomena (i.e., sources which produce waves of same frequency, same wavelength and of constant phase difference) are called coherent sources.
Let two coherent waves of intensity I1 and I2 be repre-sented by y1 = A1sin(wt - kx) and y2 = A2sin(wt - kx + d)
Both these waves are travelling in the +x direction, have same frequency, same speed and same wavelength. Their constant phase difference is d. Hence the waves are coherent.
The resultant wave, as per principle of superposition, is y = y1 + y2
Hence the resultant wave has amplitude A, same angular frequency, wavelength and speed as the interfering waves.
A2cos2f = (A1 + A2cosd)2 = A12 + A2
2cos2d + 2A1A2cosd
A2sin2f = A22sin2d
⇒ A2[cos2f + sin2f] = A12 + A2
2[cos2d + sin2d] + 2A1A2cosd
A = d2 21 2 1 2A A 2A A cos+ +
is the amplitude of resultant wave and
2
1 2
A sinAsinAcos A A cos
dff d
=+
⇒ tanf = d
d2
1 2
A sinA A cos+
gives the phase difference of the resultant wave with respect to the wave y1 = A1sin(wt - kx).
When two or more waves of the same kind travel in a medium simultaneously, then the resultant displacement at any point in the medium due to the waves at any instant is equal to the sum of the displacements independently caused by the individual waves.
Consider a number of waves travelling simultaneously through a medium. By superposition principle, at any in-stant t,
1 2 3y y y y ...........→ → → →
= + + +
y→
= resultant displacement due to all waves passing simultaneously,
1y→
= displacement due to first wave alone,
2y→
= displacement due to second wave alone, and
3y→
= displacement due to third wave alone, etc.
Waves which obey the superposition principle are called linear waves. These have usually small amplitudes.
The phenomenon of superposition is applicable to both mechanical and electromagnetic waves.
Waves which do not obey the superposition principle are called non-linear waves. They have usually large ampli-tudes.
The consequences of superposition principle of waves are (i) Interference (ii) Beats and (iii) Stationary waves (or Standing waves).
interference
When two waves of the same frequency, same wavelength, having constant phase difference and nearly equal amplitudes traveling in the same direction superimpose, the resultant intensity at any point in the medium is not a mere algebraic sum of the individual intensities of the individual waves. It is more than the sum of the intensities of individual waves at some locations and less than the sum of the intensities of the individual waves at some other
prinCiple Of SuperpOSitiOn Of WaveS
3.28 Oscillations and Waves
Concept Strand 9
Two waves traveling in the positive x-direction having same amplitude, same frequency, same speed and phase difference ofp/2 between them superimpose. Find the expression for the resultant wave and sketch the profile of the resultant wave at any instant.
Solution
By formula, the amplitude of the resultant wave
= p2 2 2A A 2A cos 2A2
+ + =
Phase of the resultant wave = tan-1
p
p
Asin 2A Acos 2
+
= p4
radian
ConCept Strand
0
1
R
2
x
8λ
4λ
83λ
2λ
85λ
43λ
87λ λ
1 and 2 are the component waves and R is the resultant.We know that the intensity of a wave I is proportional
to square of amplitude (A) and square of frequency (f).
i.e., I ∝ A2f2
Hence, for waves of the same frequency, I ∝ A2 ⇒ I = pA2, I1 = pA1
2 and I2 = pA22
where p = constant of proportionality pA2 = pA12 + pA2
2 + 2pA1A2cosd (by multiplying the amplitude of resultant wave with p)
i.e., I = I1 + I2 + 2 I I1 2 cosd
is the resultant intensity of the superimposed waves.The term I I1 22 cosd in the resultant intensity
expression is called the Interference factor. The magnitude and sign (positive or negative) of the interference fac-tor depends upon d (the phase difference with which the waves superimpose at a point in the medium). Maximum intensity (Imax) occurs when cosd = 1. Therefore,
Imax = I1 + I2 + 2 I I1 2
or Imax = I I
2
1 2 +
This is called constructive interferenceMinimum intensity (Imin) occurs, when cosd = -1.
Therefore,
Imin = I1 + I2 - 2 I I1 2
or
Imin = ( )I I2
1 2−
This is called destructive interference. If cosd = 0, then I = I1 + I2.
The phenomenon of interference is based on the law of conservation of energy. There is no loss of energy in interference. Energy which disappears in regions of minimum intensity appears in regions of maximum in-tensity.
When waves produced by non-coherent sources su-perimpose, the resultant intensity is the sum of the intensi-ties of the individual waves. i.e., I = I1 + I2 Interference factor 2 I I1 2 cosf = 0 for non-coherent waves, because f can have all values and average value of cosf is zero.
Interference phenomenon can occur in transverse waves as well as longitudinal waves. It can also occur in mechanical waves and electromagnetic waves (Example, light). Two separate lamps of same frequency can never produce interference, as they are not coherent sources. Production of light is an atomic phenomenon (micro-scopic property of matter) and there is no constant phase relationship between the light produced by different sources.
However, two separate sources of sound of same fre-quency may produce interference phenomenon because propagation of sound through a medium is a bulk prop-erty of the medium (macroscopic property of matter) and such sources of sound could be coherent sources.
Oscillations and Waves 3.29
When a progressive wave travelling in one medium (say Medium I) reaches the boundary with another medium (say Medium II), part of the wave travels back in the
The condition for interference minima is
Phase difference d = (2n + 1)p, where n = 0, 1, 2, …..
Or path difference = ( )l2n 12
+ , where n = 0, 1, 2,…..
The condition for interference maxima is
Phase difference d = 2np, where n = 0, 1, 2, 3…..or path difference = nl, where n = 0, 1, 2, 3, …..
refleCtiOn and tranSMiSSiOn Of WaveS
incident Medium I. It is called the reflected wave. Another part of the wave which travels into Medium II is called the transmitted wave (Fig. 3.27). A part of the energy of the wave may be absorbed at the boundary separating the two media. The frequencies of the incident wave, reflected wave and transmitted wave are the same. The wavelengths of the incident wave and reflected wave are same while the wavelength of the transmitted wave is different from that of the incident wave. If the speed of the wave in the transmitted medium is less than that in the incident medium (i.e., transmitted medium is denser and incident medium is rarer), the wavelength of the transmitted wave will be smaller than the wavelength of the incident wave. Further, the reflected wave is inverted i.e., a phase difference of p radian occurs between the incident wave and reflected wave (Fig. 3.28)
If the speed of the wave in the transmitted medium is more than the speed of the wave in the incident medium (i.e., transmitted medium is rarer than the incident medium), the wavelength in the transmitted medium will be more than the wavelength in the incident medium but there will be no inversion of the reflected wave or the transmitted wave. This is shown in Fig. 3.29.
The above rule can also be applied for reflection of transverse wave along stretched strings. The fixed ends (rigid ends) are to be treated as the boundary with denser media and the free ends are to be treated as boundary with rarer media.
Boundary
Transmitted(yt)
Fig. 3.28
Denser Medium Speed of wave v2(<v1)
Rarer MediumSpeed of wave v1
Incident(yi)
Reflected(yr)
Rarer Medium speed ofwave v2(>v1)
Boundary
Denser Mediumspeed of wave(v1)
Incident(yi)
Reflected(yr)
Fig. 3.29
Fig. 3.27
Part absorption ofenergy at the boundary
Medium Ι Medium ΙΙBoundary
Reflected
Incident
Transmitted
3.30 Oscillations and Waves
Bounded medium
A medium which is separated from other media by defi-nite boundaries or distinct surfaces is called a ‘bound-ed medium’. The boundaries of a medium may be rigid or free.
Example, (i) The strings of a guitar or veena are fixed at both ends and have rigid boundaries at the ends.
(ii) The ends of a flute open at both ends have free boundaries.
(iii) A pipe closed at one end and open at the other end has a rigid boundary at the closed end and a free boundary at the open end.
A bounded medium can vibrate only at certain definite frequencies which are characteristic frequencies of that medium. The minimum frequency with which the medium can vibrate is called the fundamental frequency or first harmonic. The higher frequencies with which the medium can vibrate are multiples of this fundamental frequency and are called overtones or harmonics. Some bounded media can have all odd and even harmonics (i.e., odd and even multiples of the fundamental frequency) while other bounded media can have only odd harmonics (i.e., odd multiples of fundamental frequency).
Stationary waves
When two identical progressive waves in a bounded medium travelling in opposite directions with the same speed along the same line superimpose, the resultant wave pattern formed appears to be stationary and is known as stationary wave or standing wave. The resultant wave pattern is not advancing and hence there is no transfer of energy along the length of the medium. Stationary waves produced in the vibrating strings of a sonometer, guitar, veena etc. are transverse stationary waves. Stationary waves formed in open and closed organ pipes, in air columns of bugle, whistle etc., are longitudinal stationary waves.
A bounded medium is an essential requirement for the formation of a stationary wave. P is a progressive wave travelling in a bounded medium and Q is its re-flected wave from a rigid boundary (Fig. 3.30). These two progressive waves travelling in opposite directions superimpose to produce a standing wave R (stationary wave) in the medium. The resultant wave at a time t = 0 is shown by the thick line.
Incident wave (P)
Reflected wave (Q)
Fig. 3.30
R
(a) In Fig. 3.31, points marked 1, 2, 3, 4, 5, 6, 7, 8, and 9 are the particles of the medium in a straight line which are all equally separated. The distance between particle 1 and 5 is equal to the wave length (l) so that the time taken for the wave to travel from 1 to 5 is the time period T.
At time t = 0, let the two waves of equal amplitudes superimpose so that they are out of phase. The resultant displacement of all the particles of the medium is equal to zero and is represented by the thick line.
1 2 3 4 5 6 7 8
9
Resultant wave at t = 0
Fig. 3.31
(b) At time t = T4
, the forward wave P has travelled a
distance l4
to the right, the backward wave Q has
travelled a distance l4
to the left. The resultant wave pattern is shown by the thick line in Fig. (3.32). Particles 1, 3, 5, 7, 9 are displaced to its maximum value and particles 2, 4, 6, 8 have zero displacement.
12
34
56
78 9
Resultant wave at t =4T
Fig. 3.32
Oscillations and Waves 3.31
(c) At time, t = T2
(Fig. 3.33), the waves become again out
of phase and the resultant displacement of particles of the medium become zero.
2 3 4 5 6 7 8 9
Resultant wave at t =2T
Fig. 3.33
1
(d) At time t = 3T4
, the waves become in phase and the particles 1, 3, 5, 7, 9 have maximum displacement and the particle 2, 4, 6, 8 have zero displacement. The resultant wave pattern is shown by the thick line in Fig. 3.34.
1 2 3 4 5 6 8 9
Resultant wave at t =4T3
7
Fig. 3.34
(e) At time t = T, the waves become out of phase and the displacement of the particles become zero. The resultant wave pattern becomes same as in Fig. 3.31
It can be seen that particles like 2, 4, 6 and 8 in the medium do not undergo any displacement from their initial position at any time. Hence these particles are al-ways at rest and the position of such particles in the me-dium are called nodes. Particles like 1, 3, 5, 7 and 9 vibrate about their mean positions with maximum amplitude. The positions of these particles are known as antinodes. The distance between two consecutive nodes or consecutive
antinodes is l2
. The distance between a node and the
nearest antinode is l4
.
theory of stationary waves
Let a progressive wave P passing through a medium be represented by,
y1 = A sin (wt- kx) …….(1)
If this wave gets reflected at the free boundary of the medium with the same amplitude, the reflected wave ‘Q’ is represented by,
y2 = A sin (wt +kx) ……..(2)
When these two waves superimpose on each other in-side this medium, as per superposition principle the resul-tant wave is given by,
y = y1+ y2 ……….(3)
y = A sin (wt -kx) + A sin (wt + kx)
= A [sin (wt - kx) + sin ( wt + kx)]
=A ( )w w w wt kx t kx t kx t kx2sin cos
2 2 + + − + − +
= 2A sin w t cos kx
i.e., y = (2A cos kx) sin wt ……(4)
Equation (4) represents a stationary wave. Note that in this expression the factors kx and wt are not in the form kx-wt, but are separated. It is a simple harmonic wave of amplitude (2A cos kx) and frequency same as that of the component waves. The values of x which give maximum amplitude represent the antinodes and values of x which give zero amplitude represent the nodes.
Antinodes
At the antinodes, the amplitude is maximum.
2A cos kx = ± 2A, cos kx = ± 1
kx = n p, where wave number, k = pl
2
pp
l2
x n=
x = ln
2, where n = 0, 1, 2, 3. …..
\ Amplitude is maximum at the antinodes where
x = 0, l ll
3, ,
2 2, …….,
ln2
Distance between two consecutive antinodes is l2
.
3.32 Oscillations and Waves
Nodes
At the nodes the amplitude is zero.2A cos kx = 0, cos kx = 0
kx = (2n + 1) p2
, where k = pl
2; ( )p pl
2x 2n 1
2= +
x = ( )l2n 14
+ , where n = 0, 1, 2, 3, ………
\ Amplitude is minimum at the nodes where
x = l l l3 5
, ,4 4 4
, ……. ( )l2n 14
+
Distance between two consecutive nodes is l2
.
note:The stationary waves can be expressed in various math-ematical forms. They are
(i) y = (2A cos kx) sin wt, produced by super position of waves y1= A sin (wt - kx) and y2 = A sin (wt + kx)
(ii) y = -2A (sin kx) cos wt produced by super position of waves y1 = A sin (wt - kx) and y2 = - A sin (wt + kx)
(iii) y = (2A cos kx) cos wt, produced by super position of waves y1 = A cos (wt - kx) and y2= A cos (wt + kx)
(iv) y = (2A sin kx) sin wt, produced by superposition of y1 = A cos (wt - kx) and y2 = -A cos (wt + kx)
Characteristics of stationary waves
1. A stationary wave produces a disturbance which is localized in a medium and does not travel in any direction inside the medium. Since stationary waves do not advance through the medium, there is no transfer of energy from one particle to another
2. In a stationary wave, some particles of the medium are always at rest. The positions of such particles are called nodes. The distance between any two consecutive nodes is
l2
, where l = wave length of each of the superimposing progressive waves.
3. All particles at different points of the medium in be-tween the nodes are vibrating with same period but different amplitudes, period of vibration is equal to the period of the component waves.
4. Certain particles of the medium vibrate with maxi-mum amplitudes, the position of such particles are called antinodes. The distance between consecutive antinodes is
l2
.
5. The nodes divide the medium into various segments. All particles of a medium in a segment between two nodes vibrate in phase.
6. Between two nodes exists an antinode or between two antinodes exists a node. The distance between a node and the neighbouring antinode is
l4
.
7. All particles of a segment and all particles of the neighbouring segments (on either side) are 1800 out of phase.
Differences between progressive waves and stationary waves
Table 3.2Progressive Wave Stationary wave
1. A progressive wave transports energy through a medium, in the direction of propaga-tion of the wave
1. There is no transport of energy across the medium
2. The wave travels through the medium with a velocity called wave velocity
2. The wave remains station-ary between two boundar-ies in the medium
3. The progressive wave is produced due to a dis-turbance which transfers energy from one particle to another.
3. Stationary wave is pro-duced due to super posi-tion of two identical waves travelling in opposite directions along the same line.
4. The typical equation for a harmonic progressive wave is y = Asin (wt -kx) i.e., y = f(x, t)
4. The typical equation for a stationary harmonic wave is y = A coskx sinwt i.e.,y = f(x) g(t)
5. All the particles of the medium vibrate.
5. Particles at the nodes are always at rest
6. The amplitude of vibra-tion of all the particles is the same
6. The amplitude of vibration of particles at the nodes is zero. Also the amplitude of particles between nodes is different.
7. At any instant, the phase of vibration is different for different particles of the medium.
7. At any instant, the phase of vibration of particles in one segment is the same. But particles in adjacent segment is out of phase
8. The maximum particle velocity, while crossing the mean position, is same for all particles
8. The maximum particle velocity while crossing the mean position, is differ-ent for different particles, increasing from a node to an anti-node and then de-creasing till the next node.
Oscillations and Waves 3.33
Consider a uniform string of length L stretched horizon-tally between two fixed supports. m is the mass per unit length of this string and T is the tension in the string. If v is the speed of a transverse wave along the stretched string, for small amplitudes of the wave, it can be shown that
v = mT
The velocity of mechanical waves through a medium is a bulk property of the medium. The velocity of a me-chanical wave (either transverse or longitudinal) is deter-mined by the inertial and elastic properties of the medium, the tension in the string acting to restore the string to the equilibrium configuration (elastic property) while the mass provides the inertia resisting the return to equilibrium.
If the string is plucked at the centre and released, a transverse wave is produced. It will travel with a wave ve-locity v along the string, gets reflected at the fixed ends and form a stationary transverse wave. Nodes are formed at the fixed ends. In the fundamental mode of vibration of the string, there will be one antinode in between two nodes. This is shown in Fig. 3.35.
2L λ1=
A
N N
Fundamental frequency (first harmonic)
Fig. 3.35
Hence in the fundamental mode, the string vibrates in one segment and the length of string,
L = l1
2, l1 = 2L
The fundamental frequency of vibration,
f1 = l1
v
\ f1 =
m1 T
2L
mm
QT
v , where massper unit length
= =
In the second harmonic mode of vibration, the string vibrates in 2 segments (i.e., with 2 antinodes and 3 nodes).
L = λ2
N
N
N
A A
First overtone (second harmonic)
Fig. 3.36
If l2 is the wavelength of vibration in the second mode (first overtone), then the length of the string, L = l2, l2 = L
The frequency of vibration in the second mode f2 = l2
v
m21 T
fL
=
m21 T
f 22L
= × = 2f1
When the string vibrates in the third harmonic mode, there are three identical segments of the string, with 3 anti-nodes and 4 nodes.
Fig. 3.37
L =2
3λ3
N
AN N
A
N
A
Second overtone (third harmonic)
The length of the string, L = l332
or l3 = 2L3
. The
frequency of vibration in the third mode,
f3 = l3
v
f3 = m
3 T2L
= 3f1 mQ
Tv
=
Thus, when a string fixed at both ends vibrates, all odd and even harmonics are present.
(i) If a string of length L, fixed at both ends, vibrates in the nth mode, there will be n identical segments in the string and n antinodes.
(ii) The nth mode of vibration of the string, is the (n–1)th overtone.
(iii) In the nth mode of vibration, the wavelength of the
wave, ln =2Ln
, where L = length of string between
fixed ends. The frequency of vibration
fn =
1nv
nf2L
= =m
n T2L
,
where, n = 1,2,3, … and T is the tension in the string and m is the mass per unit length of the string.
(iv) If d is the diameter of the wire and r its density, the
mass of wire = p
r2d L4
and mass per unit length
m = p r2d
4.
Concept Strand 10
A string fixed at its ends is 70 cm long. The speed of trans-verse wave through it is 49 m s-1. If the string is vibrating in seven identical segments
(i) How many nodes and antinodes are present in its vibration?
(ii) What is the wavelength of the wave? (iii) What is the frequency of vibration? (iv) What is the fundamental frequency and wavelength
of fundamental vibration of the string? (v) What is the wavelength and frequency in the second
overtone? (vi) What is the wavelength and frequency in the fourth
mode of vibration?
Solution
Length of string, L = 70 cm = 0.7 m Speed of transverse wave, v = 49 m s1
Number of segments = mode of vibration of the string n = 7
(i) Number of antinodes = mode of vibration n = 7 Number of nodes = n +1 = 7 +1 = 8
(ii) The wavelength of the wave, ln = 2Ln
\ Wavelength ln = 2 0.7
0.2 m7×
= (iii) The frequency of vibration,
fn = nv2L
\ fn = 7 49
245 Hz2 0.7×
=×
(iv) fn = n f1, Hence the fundamental frequency of string, f1 =
nf 24535 Hz
n 7= =
wavelength of the fundamental
mode, l1 = 1
v 491.4 m
f 35= =
(Alternatively
f1 = nv 1 49
35 Hz2L 2 0.7
×= =
×
l1 = 2L 2 0.7
1.4mn 1
×= = )
(v) In the second overtone, the mode of vibration is n = 3
ln = 2Ln
and
fn = nf1
l3 = 2 0.7
0.467 m3×
=
The frequency of the second overtone,
f3 = 3 f1 = 3 ×35 = 105 Hz
(vi) In the fourth mode of vibration, n = 4. Frequency in the fourth mode of vibration,
f4 = 4f1 = 4 ×35 = 140 Hz
The wavelength of wave in the fourth mode,
l4 = 2L 2 0.7
0.35 mn 4
×= =
ConCept Strand
Oscillations and Waves 3.35
laws of transverse vibrations in stretched strings
1. Law of length
The fundamental frequency of transverse vibration of
a stretched string is given by f = m
1 T2L
where T is the
tension in the string and m is the mass per unit length of the string. Therefore,
The fundamental frequency of transverse vibration of a stretched string f is inversely proportional to the length of the string L, when the mass per unit length of the string m and tension in the string T remain constant.
f ∝ 1L
2. Law of tension
The fundamental frequency of transverse vibration of a stretched string f is directly proportional to the square root of the tension in the string T, provided the length of the string L and mass per unit length of the string m remain constant.
i.e., f ∝ T
3. Law of mass
The fundamental frequency of transverse vibration of a stretched string f is inversely proportional to the square root of its mass per unit length m, provided the tension in the string T and its length L remain constant.
i.e., f ∝ m
1
SOund
Sound is a form of energy arising due to mechanical vibrations. Hence sound requires a medium for its propagation. Sound cannot travel in vacuum. Sound is propagated as longitudinal mechanical waves through solids, liquids and gases. When sound propagates through a fluid, it produces pressure fluctuations and density fluctuations in the fluid.
Classification of sound
Audible frequencies are sound frequencies that can be heard by human beings. It ranges from 20 Hz to 20,000 Hz.
Most audible frequencies to human ear is from 2000 Hz to 2500 Hz.
Sound waves of frequencies more than 20,000 Hz are called ultrasonic waves. Bats have the ability to produce and detect these waves.
Sound waves of frequency less than 20 Hz are called infrasonic waves. Large vibrating bodies produce infrasonic waves. Infrasonic waves produced by earthquakes are detected by animals. The disturbance produced by a simple pendulum is not heard by human beings as its frequency is infrasonic.
Sound waves of frequencies above 500 × 106 Hz are called hypersonic waves.
newton’s formula for speed of sound in solids and liquids
Newton showed that the speed of sound in a medium
v = rE ,
where, E is the appropriate modulus of elasticity of the me-dium and r is the density of the medium.
Therefore, the speed of sound in solids, v = rY ,
where Y is the Young’s modulus of the solid and r is the density of the solid.
Speed of sound in steel is about 5921 m s-1
Speed of sound in a liquid, v = rB , where B is the
bulk modulus of the liquid and r is the density of the liquid.Speed of sound in water at 20°C is 1482 m s-1
Speed of sound in gases
Newton considered the propagation of sound through gases as an isothermal process (pV = constant) as the medium is not getting heated up when sound is passing through it. So, Newton used the pressure of gas p in determining the speed
3.36 Oscillations and Waves
of sound, using the formula v = rp . (It should be noted
that isothermal bulk modulus of a gas is the pressure p). However, there was a large discrepancy in the speed of sound determined by this formula with the experimentally determined values. Hence a correction to this formula was applied by Laplace.
According to Laplace, the propagation of sound in gas takes place adiabatically, so that instead of pressure, the adiabatic bulk modulus of gas gp has to be used, where
g = P
V
CC
and Cp and Cv are respectively the specific heats
at constant pressure and constant volume of the gas. Hence
speed of sound in gas, v = grp
Heat is liberated in regions of compression and ab-sorbed in regions of rarefaction. Each region is subjected to alternate compression and rarefaction so quickly that no heat exchange with the surroundings takes place. i.e., the process of sound propagation through gas is adiabatic. The value obtained by Newton Laplace formula is in excellent agreement with the experimental value.
factors affecting the speed of sound in gases
(a) Effect of pressure
If pressure is increased at constant temperature, by Boyle’s law, pV = a constant for a fixed mass of gas. i.e.,
rp
is a
constant for a fixed mass of gas, r is the density of the gas
\ v = grp = a constant
Hence, change in pressure does not affect the speed of sound through a gas.
(b) Effect of temperature
The velocity of sound in a gas, v = grp . But density
r = MV
, M is the mass of the gas occupying the volume V,
\ v = gpVM
,
pV = RT, if M is the mass of 1 mole of gas.
Velocity of sound in a gas,
v = gRTM
v T∝ , where T = absolute temperature of gas.
The speed of sound in a gas is directly proportional to the square root of the absolute temperature of the gas. So, as the temperature increases, velocity of sound in gas in-creases.
If v1 is the velocity of sound at T1K and v2 that at T2 K,
1 1
2 2
v Tv T
=
If v0 is the velocity of sound at 0°C and vt at t°C
t
0
v 273 tv 273
+=
\ vt = (v0 + 0.61 t) m s-1
Speed of sound through air at 0°C = 330 m s-1. The speed of sound increases by 0.61 m s-1 for every degree Cel-sius rise of temperature.
The increase in speed of sound per degree celsius (or per Kelvin) rise of temperature is known as temperature co-efficient of speed of sound. Its value in air is 0.61 m s-1. Speed of sound in a gas depends on the atomicity of the gas molecules. g = 1.67, for monoatomic gas and 1.4 for di-atomic gas.
(c) Effect of density of gas
Velocity of sound v = grp . Consider gases of H2 and O2
at the same pressure, volume and temperature v ∝ r
1.
Sound travels faster in H2 as its density is less.
(d) Effect of humidity
Under the same conditions of temperature and pressure, the density of water vapour is less than that of dry air. The presence of moisture decreases the effective density of air. Hence, sound travels faster in moist air than in dry air.
note:Moist hydrogen gas is heavier than dry hydrogen gas. The presence of moisture (or water vapour) increases the effec-tive density of hydrogen gas.
Oscillations and Waves 3.37
unable to vibrate, hence a node is formed at the closed end. The particles of the medium at the open end can have maximum displacement; hence an antinode is produced at the open end of the pipe.
The air column inside the tube can vibrate at different frequencies. The lowest possible frequency of vibration is called the fundamental frequency or the fundamental note or first harmonic. The frequencies which are integral multiples of the fundamental frequency are called harmonics. If the fundamental frequency (or first harmonic) is f0, the second harmonic is 2 f0, the third harmonic is 3 f0 and so on.
The higher frequencies at which an air column can vibrate are also called the overtones. The frequency of vibration just above the fundamental frequency (or first harmonic) is termed as first overtone, the next higher frequency as second overtone, the next one, the third overtone and so on. A closed air column of length L vibrating in different modes is shown in Figs. 3.38-40.
So, sound travels faster in dry hydrogen than in moist hydrogen.
(e) Effect of wind
Wind simply adds its velocity vectorially to that of the sound waves in a medium.
If the component vW of wind speed is in the direction of sound wave, resultant speed of sound = v + vW.
If the component vW of the wind speed is in the opposite direction of sound wave resultant speed of sound = v - vW.
If the wind is blowing perpendicular to the direction of sound wave, it will not affect the speed of sound.
(f) Effect of change in frequency or wavelength of sound wave
Change of frequency or wavelength do not affect the speed of sound in a medium. In a homogenous, isotropic medium, sound travels with the same speed in all directions.
i.e., v =lf = a constant.
When the sound passes from one medium to another the frequency remains constant, the wavelength and velocity change.
(g) Effect of amplitude
Velocity of sound v = grp for small amplitudes.
Amplitude generally does not affect speed of sound in gas. However, a very large amplitude may affect the speed of sound.
Relation between speed of sound in gas and the rms speed of gas molecules
We have speed of sound v = grp =
gpVM
rQMV
=
i.e., v = gRTM
(Q pV = RT for 1 mole of ideal gas)
The rms speed of gas molecules (which will be explained in a later chapter) is
vrms = g
3RT 3M
= . gg
RT 3M
= v. Therefore,
vrms = g3 × speed of sound through the gas
StatiOnary WaveS in air COluMnS
Stationary waves can be set up in the air column inside cylindrical hollow tubes or pipes. There are two types of pipes. (i) open pipe which has both ends open and (ii) closed pipe which has one end closed and other end open. If a longitudinal wave is set up at the open end of a pipe, it travels through the medium in the pipe and gets reflected at the other end of the pipe. The incident wave and the reflected wave are identical but travel in opposite directions. When these waves superimpose, stationary waves are produced inside the pipe. The closed end acts as a rigid end and the open end acts as a free end for reflection of longitudinal waves.
Modes of vibration of air in a closed pipe
The air column in a pipe can be set into vibration by keeping an excited tuning fork of required frequency at the open end. At the closed end the particles of the medium are
3.38 Oscillations and Waves
Fig. 3.38
N
L =4
λ1
Fundamental frequency (first harmonic)
A
In the fundamental mode of vibration, there is a node at the closed end and an antinode at the open end.
From Fig. (3.38) length of the pipe,
L = l
l11, or 4L
4=
The fundamental frequency of vibration/first har- monic,
f1 = l1
speed vwavelength
= = v
4L
In the next mode of vibration of the closed pipe, there are two nodes and two antinodes.
L =4
3λ2
N
A
N
A
First overtone (third harmonic)
Fig. 3.39
From Fig. (3.39)
L = 2 2 232 4 4l l l
+ = , or l2 = 4L3
The frequency of vibration,
f2 = l2
v =
3v4L
\ f2 = 3v4L
, is the first over-tone or third harmonic.
In the next mode of vibration, there are three nodes and three antinodes. From Fig. (3.40),
L =4
5λ3
N
A
A
A
N
N
Second overtone (fifth harmonic)
Fig. 3.40
L = l3 + l l3 354 4
= ,
or l3 = 4L5
The frequency of vibration,
f3 = l3
v=
5v4L
f3 = 5v4L
, is the second overtone or fifth harmonic.
Thus for closed pipe,
f1 : f2 : f3 = 1: 3: 5: ….:(2n – 1)
\ fn = (2n – 1)f1 = ( ) v2n 1
4L−
and wavelength of superimposing waves, ( )ln4L
2n 1=
−In a closed pipe only odd harmonics are present.
note:If the vibrations in a closed air pipe is having n nodes, it is vibrating in the nth mode. There will be n antinodes also. Frequency of vibration, fn = (2n -1) f1 where fundamental frequency f1 =
v4L
.
Oscillations and Waves 3.39
Concept Strand 11
In a closed pipe of length 1.7 m, vibrations are set up with four nodes. If the speed of the wave through the air me-dium is 340 m s-1 calculate
(i) the mode of vibration of the closed pipe (ii) its fundamental wavelength of vibration (iii) its fundamental frequency of vibration (iv) its wavelength when vibrating with four nodes. (v) its frequency of vibration with four nodes. (vi) its first overtone frequency.
Solution
Length of pipe L = 1.7 mSpeed of wave, v = 340 m s-1
Number of nodes, n = 4
(i) Number of nodes, n = 4 Hence the pipe is vibrating in the fourth mode (i.e., In
the third overtone or seventh harmonic) (ii) Wave length ln = ( )
4L2n 1−
For fundamental mode n = 1
\ l1 = ( )4 1.7 4 1.7
2 1 1 1× ×
=× −
= 6.8 m
(iii) Fundamental frequency of vibration
f1 = l1
v =
3406.8
= 50 Hz
[Alternatively, fn = ( )2n 1 v
4L−
]
(iv) When vibrating with four nodes, n = 4 Wavelength of vibration,
When a longitudinal wave enters an open pipe (open at both ends), it propagates through the medium inside the pipe and gets reflected from the other end. The incident wave and the reflected wave superimpose to produce a sta-tionary wave inside the open pipe. Antinodes are formed at the open ends as the particles in the open end can vibrate with maximum amplitude.
The air column inside the tube can vibrate at differ-ent frequencies. The lowest possible frequency of vibration is called the fundamental frequency or the fundamental note or first harmonic. The higher frequencies are called overtones.
In the fundamental mode of vibration in an open pipe, there will be one node and two antinodes. This corresponds to the minimum frequency of vibration in an open pipe (Fig. 3.41)
Fig. 3.41
� = 21λ
A
A
N
Fundamental frequency(First harmonic)
� = L
Length of the pipe,
L = l1
2, or l1 = 2L
3.40 Oscillations and Waves
The frequency of vibration,
f1 = l1
v =
v2L
\ f1 = v
2L, is the fundamental frequency of vibration or
first harmonic in the open pipe.
In the next higher mode, there are two nodes and three antinodes (Fig. 3.42).
A
N
N
L = λ2
First overtone (second harmonic)
A
A
Fig. 3.42
L = 2l2
2, or l2 = L
The frequency of vibration,
f2 = l 1
2
v v2f
L= = \ f2 = 12f is the first overtone
In the third mode of vibration, there are three nodes and four antinodes.
From Fig 3.43
L = l332
, or l3 = 2L3
The frequency of vibration,
f3 = l3
v 3v2L
=
Second overtone (third harmonic)
A
L = 2
3 3λ
N
N
N
A
A
A
Fig. 3.43
f3 = 3v2L
, is the second overtone or third harmonic.
\f1 : f2: f3 ……: fn = 1:2:3: ……:n
\ fn = nf1 = nv2L
, where n = 1, 2, 3. Wavelength of
superimposing waves, ln = 2Ln
In open pipes, both odd and even harmonics of vibration are present. Hence the quality of sound produced in an open organ pipe is better than that produced by a closed organ pipe.
note:
(i) If an open organ pipe contains vibrations with n nodes, it is in the nth mode. There will be (n+1) antinodes. Frequency of vibration, fn = nf1 where fundamental frequency f1 =
v2L
(ii) For an open pipe and a closed pipe of equal lengths, the fundamental frequency of vibration in open pipe is double the fundamental frequency in a closed pipe.
(f1) open = v
2L = 2.
v4L
= 2(f1)closed if length L is the same for both
Concept Strand 12
An open pipe of length 34 cm vibrates with 5 antinodes. If the speed of sound through the air medium is 340 m s-1, calculate
(i) the number of nodes in the wave (ii) the mode of vibration in the pipe (iii) the fundamental wavelength l1 (iv) the fundamental frequency f1
ConCept Strand
Oscillations and Waves 3.41
Fundamental frequency of closed pipe,
( ) ( )1 closed
vf
4 L e=
+
End correction
We had supposed that the antinode is formed exactly at the open end. In reality, the air molecules just outside the pipe have maximum freedom of vibration and antinodes are formed at that point.
The distance of the actual antinode from the open end of the pipe is called end correction ‘e’. According to Rayleigh, end correction,
e = 0.3d,
where d is the diameter of the pipe.Fig. 3.44 shows the positions where the actual antinode
is formed in open and closed pipes.
Accordingly, the fundamental frequency of open pipe,
( ) ( )1 open
vf
2 L 2e=
+
(v) the wavelength when vibrating with 4 nodes. (vi) the frequency when vibrating with 5 antinodes. (vii) the frequency of vibration in the second overtone of
vibration.
Solution
Length of pipe L = 34 cm = 0.34 mSpeed of sound v = 340 m s-1
Number of antinodes = 5
(i) Number of nodes = number of antinodes - 1 Hence, there will be 4 nodes in the vibrating pipe. (ii) Mode of vibration = number of nodes = 4 Hence the vibration in the pipe is in the 4th mode or
third overtone.
(iii) ln = 2Ln
, for the nth mode
Wavelength in the fundamental mode (n = 1),
\ l1 = 2L
2 0.34 0.68 m1
= × =
(iv) Fundamental frequency
f1 = l1
v 3400.68
= = 500 Hz
(Alternatively, fn = nv2L
\ f1 = 1 340
500Hz2 0.34×
=×
)
(v) Since the vibration is with 4 nodes, mode of vibration is n = 4
ln = 2Ln
The wavelength when vibrating with 4 nodes,
\ l4 = 2 0.34
0.17 m4
×=
(vi) Since the vibration is with 5 antinodes, number of nodes is 4.
Hence the mode of vibrations n = 4
fn = n f1. f4 = 4 × 500 = 2000 Hz
[Alternatively, fn = nv2L
,
\ f4 = 4 340
2000 Hz2 0.34×
=×
]
(vii) Since the vibration is in the second overtone, mode of vibration n = 3
fn = n f1
The frequency of vibration in the second overtone (third harmonic),
f3 = 3 f1 = 3 × 500 = 1500 Hz
A
N
L
e A
A
N
e
L
Fig. 3.44
e
3.42 Oscillations and Waves
The periodic waxing and waning of sound, when two sound waves of nearly equal frequencies traveling in the same direction superimpose on each other, is called beats. One cycle of maximum and minimum intensity is counted as one beat.
Time interval between two maxima is called the beat period. It is also equal to the time interval between two minima.
The number of beats per second is called beat frequen-cy. Therefore, beat frequency, fb = f1 – f2, where f1 and f2 are the frequencies of the waves producing beats. This formula is correct only when two frequencies only are involved to pro-duce beats.
The persistence of hearing for normal ear is 0.1 s. Beats can be heard effectively only if the beat period tb is greater than 0.1 s and beat frequency fb is less than 10 Hz. Thus for the human ear to hear the beats, the two sound waves pro-ducing the beats should be such that, f1 – f2 < 10 Hz.
Graphical explanation of formation of beats
If A1 and A2 are the amplitudes of two waves producing the beats,
II
2max 1 2
_ 2min 1 2
(A A )(A A )
+= .
If the amplitudes are equal, A1 = A2 = A
Imax = 4A2, and Imin = 0
Hence to hear beats clearly, the amplitudes of the two waves must be almost the same.
t2t1 t3 Time t
Fig. 3.45
t4 t5
Consider two sound waves of equal amplitudes A, and slightly different frequencies f1 and f2 travelling along the same direction in a medium (Fig. 3.45). At certain points of the medium the two waves are periodically in phase and out of phase. At times t1, t3, t5 …, the two waves are in phase and the resultant wave has amplitude 2A at such points. At times t2, t4, t6 …, the two waves are out of phase and the resultant wave has zero amplitude at such points.
As time passes, the resultant intensity rises and falls periodically and the time elapsed between two maxi-mum intensities = time elapsed between two minimum intensities.
t3 – t1 = t5 – t3 =.. … = t4 – t2 = t6 – t4 = ….= tb,
Concept Strand 13
The air column in a closed pipe of length 40 cm and di-ameter 2.5 cm is set into vibration. What is the frequency of vibration if it is vibrating in its first mode? Velocity of sound in air v = 340 m s-1
Solution
Diameter of pipe = 2.5 cm = 0.025 mEnd correction e = 0.3 d = 0.3 × 0.025 = 0 0075 m
Length of the pipe L = 40 cm = 0.40 m.Frequency of vibration in the first mode
= fundamental frequency, f1 = ( )v
4 L e+
= ( )340
4 0.40 0.0075+
= 340
4 0.4075×
= 208.59 Hz
ConCept Strand
BeatS
Oscillations and Waves 3.43
Beat period = tb
Beat frequency fb = b
1t
Theory of beats
Consider two sound waves of equal amplitudes A, and slightly different frequencies f1 and f2 travelling along the same direction in a medium. The particles of the medium vibrate simple harmonically. If y1 and y2 are the displace-ments of a single particle at time ‘t’ produced by the two waves separately,
y1 = A sin 2pf1t — (1)
y2 = A sin 2pf2t — (2)
As per principle of superposition, resultant displacement,
y = y1 + y2 — (3)
y = A sin 2p f1t + A sin 2pf2t
= A (sin 2pf1t + sin 2pf2t)
= [2A cos 2p ( )1 2f f
t2−
sin 2p ( )1 2f f
t2+
]
We write
y = R sin 2pft — (4)
where the frequency of the resultant wave,
f = ( 1 2f f2+
) and the resultant amplitude,
R = 2Acos 2p( )1 2f f
2−
t — (5)
Intensity will be maximum if R is maximum.
\ cos 2p ( )1 2f f t
2−
= ±1,
p(f1 – f2) t = np,
t = 1 2
nf f−
, where n = 0, 1, 2, 3 ……
i.e., t = 0, 1 2
1f f−
, 1 2
2f f−
, 1 2
3f f−
…..
\ The time period of maxima, i.e., difference between the times for successive maxima
tb = 1 2
1f f−
— (6)
Intensity will be minimum if R is minimum.
\ cos 2p ( )1 2f f t2−
= 0, p(f1 – f2) t = ( )p2n 12+
,
t = ( )( )1 2
2n 12 f f
+−
, where n = 0, 1, 2, 3 ….
i.e., t = ( ) ( ) ( )1 2 1 2 1 2
1 3 5, ,
2 f f 2 f f 2 f f− − −, …..
\ The time period of minima, i.e., the difference between the times for successive minima
tb = 1 2
1f f−
\ Beat frequency fb = b
1t
= f1 - f2
application of beats
(i) Determination of frequency
A tuning fork has two prongs (known as tines) in the shape of letter U having a stem. The tines when struck with a hammer vibrate like clamped bars. The tuning fork when struck, initially will have fundamental and several harmonics. After a few seconds, however, only the fundamental will persist. The two tines vibrate 180° out of phase in the fundamental mode.
The phenomenon of beats can be used to determine the frequency of a tuning fork. For this we use a tuning fork of known frequency f1. Let fx be the frequency to be determined. The tuning forks are sounded together and x be the number of beats heard per second. Beat frequency
fb = x.f1 ~ fx = x — (1)
\ fx = f1 + x, or fx = f1 – x
To check this, one of the prongs of the tuning fork with known frequency is loaded with a piece of wax. The two tuning forks are sounded together again and beat frequency is determined. On loading the tuning fork, its frequency f1 decreases.
If the beat frequency increases after loading,
fx > f1 and fx – f1 = x fx = f1 + x.
If the beat frequency decreases after loading,
fx < f1 and f1 – fx = x
fx = f1 – x.
note:If a tuning fork is filed, its frequency increases.
3.44 Oscillations and Waves
(ii) Tuning of musical instruments to a particular frequency
Two musical instruments can be tuned to the same frequency. For this, the two instruments are sounded together so that beats are heard. Now the frequencies of the two notes heard are nearly equal. The tension in the string of musical instrument whose frequency to be altered is adjusted so that the number of beats is reduced to zero. Now the instrument is tuned to the frequency of the other.
note:In sonometer experiments, if the tuning fork and the segment of wire are set into vibration together, beats can be heard, when their frequencies are nearly equal. The resonance length of the wire can be determined by reducing the number of beats heard to zero. Now frequency of vibrating segment is equal to the frequency of the tuning fork.
Concept Strand 14
A tuning fork when sounded together with another tuning fork of frequency 256 Hz produces 3 beats per second. If a prong of the fork of unknown frequency is filed, the beats are increased to 5 beats per second. Find the frequency of the fork before filing.
Solution
f ~ fx = x (beat frequency)
f ~ fx = 3 …….. (1)
Due to filing frequency fx increases. Since beat frequency also is increased, fx > f.
\ fx – f = 3, before filing. fx = 3 + f = 3 + 256 = 259 Hz
ConCept Strand
Determination of velocity of sound
A simple method to measure the velocity of sound is by using tuning fork and a long cylindrical tube containing water upto a certain level. When the tuning fork vibrates in its fundamental mode, it resonates with the air column in the tube.
The vibrating tuning fork is held at the mouth of the tube such that the plane of the tines are vertical (Fig. 3.46). The water surface acts as a node and the open end at a distance 0.3d above the mouth of the tube as an antinode for the standing waves that are set up in resonance with
the tuning fork. Resonance occurs when L = 3 5
, ,4 4 4l l l
,
……… The frequency of the tuning fork being determined beforehand, velocity of sound wave can be calculated using the relation
Fig. 3.46
v = nl
Concept Strand 15
An organ pipe is 55 cm long. What change in length is required in order to maintain the frequency same if the temperature falls from 90°C to 27°C?
Solution
Whether closed or open, l ∝ n ∝ T
⇒ l 300 1 1
0.55 363 1.21 1.1= = = ⇒ l =
0.550.50 m
1.1=
ConCept Strand
Oscillations and Waves 3.45
When a fast moving train is approaching a listener on the platform, the pitch of the whistle appears to rise and then fall as the train passes him. If the source producing the sound and the listener are at rest or moving in the same di-rection with the same velocity, the frequency of the sound heard by the listener is the same as the frequency of vibra-tion of the source. But there is an apparent change in the frequency of the sound heard by the listener when there is a relative motion between the source, the listener and the medium.
The phenomenon of apparent change in the fre-quency of sound produced by a source due to the relative motion between the source and the listener is called Doppler effect.
(source)vssound
(wind)
θv
vw
(listener) vL
Fig. 3.47
If relative motion causes increase in the distance between the listener (L) and the source (S), apparent frequency will be less than the actual frequency.
If relative motion causes decrease in the distance between the listener (L) and the source (S), apparent frequency will be more than the actual frequency.
note:Doppler effect is a wave phenomenon applicable to longitudinal as well as transverse waves.
General expression for apparent frequency
Consider a source S producing sound waves of frequency f, i.e., Number of waves emitted by the source in one second = f. If the source, the medium and the listener L are at rest, the frequency of sound as heard by the listener is also f. The velocity of the wave = v. Let the wave travel in the positive direction of x-axis. In a period T seconds the wave travels a distance SL = l, the wavelength of the wave (Fig. 3.48).
i.e., True wavelength l = vf
LS
λ
vL= 0
sound velocity v
vs = 0
Fig. 3.48
Let there be a wind of velocity vw in the direction of the wave, and the source also moves in the same direction with a velocity vs. The apparent velocity of sound with respect to the source = v + vw – vs.
v + vW
L S
λ’
vL= 0
sound velocity
vs
S’
Fig. 3.49
The apparent wavelength of the wave
l' = w sv v vf
+ −
Now let us consider the velocity of the listener vL in the direction of the wave. The motion of the listener will not affect the apparent wavelength.
v + vW
L
λ’
vL
sound velocityvs
S’ L’S
Fig. 3.50
dOppler effeCt
3.46 Oscillations and Waves
The apparent velocity of the sound with respect to the listener,
v' = v + vw - vL
\ The apparent frequency of sound as heard by the lis-
tener f ' = lv '
'
f ' = ( )w L
w s
f v v v(v v v )
+ −+ −
The general expression for apparent frequency when all the velocities under consideration are in the same direc-tion is given by,
f ’ = ( )w L
w s
f v v v(v v v )
+ −+ −
If the wind velocity vw = 0,
Apparent frequency f ’ = ( )( )
L
s
v vf
v v−−
f = true frequency of sound, v = velocity of sound.
Special cases
(1) Source of sound moving towards a stationary listener.
Here vL = 0, vS = +vS
f ' = f ( )s
vv v−
, f ' > f
(2) Source of sound moving away form a stationary listener
Here vL = 0, vS = -vS
f ' = ( )s
vf
v v+ , f ' < f
(3) Listener moving towards a stationary source
Here vL = -vL, vS = 0,
f ' = ( )Lv v
fv
+, f ' > f
(4) Listener moving away from a stationary source
Here vs = 0, vL = +vL,
f ' = f LV VV−
, f ' < f
(5) Source and listener moving towards each other
Here vS = +vS, vL = -vL,
f ’ = f( )( )
L
s
v vv v
+−
, f ' > f
(6) Source and listener moving away from each other
Here vS = -vS, vL = +vL,
f ' = f( )( )
L
s
v vv v−+
, f ’ < f
(7) Source approaching the listener and the listener is moving away from source
Here vS = +vS, vL = + vL,
f ' = f L
s
V VV V
− −
The apparent frequency depends on the values of vL and vS.
It can be shown that the apparent frequencies of sound heard are different in the cases of the source moving towards the listener or the listener moving towards the source. i.e., Doppler effect in sound is asymmetric
The change in frequency (f ~ f ’) due to Doppler effect in sound is well noticeable.
(8) Apparent frequency with medium moving
If the medium (usually air) itself is moving with respect to the frame of reference(usually Earth), the apparent fre-quency is given by
f ' = f M L
M S
v v vv v v
± ± ± ±
Remember that the speed of a wave v in a medium is with respect to the medium (i.e., stationary medium). If the medium is moving with a velocity vM, then the relative velocity of the wave will became (v + vM), if medium moves in the direction of the sound or (v - vM), if medium moves in the opposite direction of sound.
The effect of motion of source is to change the wavelength of sound as perceived by the listener and the effect of motion of listener is to change the relative speed with which the wave intercepts the listener.
Oscillations and Waves 3.47
positive x direction and the displacement of a particle at the undisturbed position x, at an instant t is equal to y, then we have
y = y0sinwx
tv
− — (i), where y0 = displacement am-
plitude
Partially differentiating (i) with respect to x, we get
ww0
dy xy cos t
dx v v− = −
A longitudinal wave in a fluid can be described in terms of the displacement of the particles of the medium. It can also be expressed in terms of the excess pressure generated in the medium due to compression. As far as the sound (lon-gitudinal mechanical wave) is concerned, description of sound in terms of pressure wave is more appropriate.
Consider a duct of uniform cross-sectional area A containing a fluid of bulk modulus B. Let this duct be placed in the positive x direction and let the speed of sound (longitudinal wave) through the fluid be v. If a sound wave of angular frequency w passes through this duct in the
notes:
(i) If either the source or the listener is moving with a speed greater than the speed of sound through the medium, no Doppler effect will occur.
(ii) If the velocity of source vs, or velocity of listener vL or velocity of wind vW is not along the line joining the source and listener, then the components of these velocities along the line joining the source and listener are to be used in the expression for apparent frequency.
Applications of Doppler effectDoppler effect is used in(a) RADAR for determining the speeds of aeroplanes and
automobiles.(b) SONAR for determining the speeds of submarines.(c) tracking satellites by the Earth stations.(d) determining the speed of rotation of Sun about its axis
by observing the Doppler effect of light coming from its diametrically opposite ends.
(e) medical diagnostics
Concept Strand 16
A train moving at a speed of 72 kmph sounds a whistle of frequency 500 Hz. Calculate the apparent frequency of the whistle as heard by a man on the platform when the train (i) approaches him and (ii) recedes from him. Given speed of sound = 340 m s-1.
Solution
Velocity of sound v = 340 m s-1
Velocity of source vs = 72 kmph
= 72 5
18×
= 20 m s-1
(i) Apparent frequency of sound as heard by the listener.
f ' = ( )( )
L
s
v vf
v v−−
The train is approaching a stationary listener, vL = 0, vs = + 20 m s-1
v = 340 m s-1, f = 500 Hz
f ' = ( )( )
L
s
v vf
v v−−
f ' = ( )
( )500 340 0
340 20−
− =
500 340320×
= 531.25 Hz
(ii) The train is receding from a stationary listener. vL = 0.vs = 20 m s-1
v = 340 m s-1, f = 500 Hz
f ' = ( )( )
L
s
v vf
v v−−
f ' = ( )
( )500 340 0
340 20−
+ =
500 340360×
= 472.2 Hz
ConCept Strand
lOnGitudinal MeChaniCal WaveS (SOund) aS preSSure WaveS
3.48 Oscillations and Waves
⇒ dy = y0w
w dx
cos t xv v− −
Increase in volume of that element at time t is DV = Ady
= Ay0w
w dx
cos t xv v− −
Initial volume V = Adx
\ Volume strain d dDd d
A y yVV A x x
= =
i.e., dd
yx
= volume strain = -w
w0y xcos t
v v −
⇒ corresponding stress (i.e., excess pressure developed in the element at x at time t),
p = BDVV
−
D
Qp
BV
V
− =
⇒
p = w
w0By xcos t
v v −
— (ii)
i.e., p = p0coswx
tv
− ,
wherep0 =
w0Byv
is the pressure amplitude
since wv
= k (k = pl
2 = angular wave number) we can
write p0 = By0k
Hence displacement amplitude y0 = lp
0 0p pBk 2 B
=
From equations (i) and (ii), it can be seen that the pressure wave has a phase difference of
p2
radian with the displacement wave.
Hence the pressure maxima occur when the displacement is zero and the displacement maxima occur when the pressure is at its normal level.
Intensity of sound
The intensity of sound at a given point in a medium is defined as the energy transferred per unit time across unit area held normal to the direction of propagation of the sound at that point. We know that energy density of a wave in a medium, U = rw2 21
A2
, where A = Displacement amplitude.
Intensity I = Energy density × speed of wave through medium
= Uv = 12rw2A2v
In the case of sound, U = 12rw2y0
2 (Q y0 = displacement
amplitude)
= rw2
2 0p12 Bk
= l
rwp
22 0p1
2 2 B
. Therefore
U = r l2 2 2
02
f p12 B
or U = r 2 2
02
p v12 B
= r
r
20
2
p1 B.
2 B
r rQ 2B B
v for longitudinal wave v
= ⇒ =
U = 20p1
2 B
Intensity, I = Uv = 20p v1
2 B\ Intensity of sound,
I = 20p v
2BThe SI unit of intensity of sound is W m-2.
Intensity level
The human ear can detect a wide range of intensities of sound from 10-12 W m-2 to 1 W m-2.
Threshold of hearing is the faintest sound that a healthy human ear can detect. It has an intensity of 10-12 W m-2 at a frequency of 1000 Hz.
Threshold of pain is the loudest sound that a healthy human ear can tolerate. It has an intensity of 1 W m-2 at a frequency of 1000 Hz.
Intensity level is a term used for comparing sound intensities. Since human ear can detect a wide range of frequencies, it is convenient to use a logarithmic scale for comparing the sound intensities.
Intensity level of a sound is defined as the logarithm of the ratio of the intensity (I) of the sound to an arbitrarily chosen intensity (Io). The logarithm is taken to the base 10 (common log).
Oscillations and Waves 3.49
Intensity level is measured in bel (B) or decibel (dB).
Intensity level, IL = log (IIo
) , bel (B)
IL = 10 log (IIo
), decibel (dB)
where Io = 10-12 W m-2 (threshold of hearing)
If I = 10 I0, IL = 1 bel.
\ bel is defined as the intensity level of a sound whose intensity is ten times the threshold of hearing.
Practical unit of intensity level is decibel (dB) which is 10 times the intensity in bel.
IL = 0 dB, for I = 10-12 W m-2 corresponds to the inten-sity level of the least audible sound.
IL= 120 dB for I = 1 W m-2 corresponds to the intensity level of threshold of pain.
The range of intensity level of audible sound at 1000 Hz frequency is from 0 dB to 120 dB. However, this range is different for different frequencies.
Concept Strand 17
Find (i) the intensity level corresponding to sound inten-sity of 10-8 W m-2 and (ii) the intensity of sound of inten-sity level 50 dB. Given threshold of hearing = 10-12 W m-2.
Solution
(i) Sound intensity I = 10-8 W m-2
Threshold of hearing I0 = 10-12 W m-2
Intensity level IL = 10 logII0
,
dB = 10 log8
12
1010
−
−
= 10 log 104 = 40 dB
(ii) Intensity level IL = 50 dB Intensity of sound = I I0 = 10-12 W m-2
IL = 10 logII0
dB
50 = 10 logII0
logII0
= 5, II
5
0
10=
I = 105I0, I = 105 × 10-12
i.e., I = 10-7 W m-2
ConCept Strand
Loudness of sound
Loudness is a psychological sensation produced by sound in the ear of a person and it depends on the intensity of the sound and the sensitivity of the ear. It is a subjective quality, varying from person to person.
Intensity and loudness are not the same. Intensity depends on frequency and amplitude, but loudness depends mostly on amplitude. Hence loudness depends on intensity.
According to Weber-Fechner relation, loudness L of a sound and its intensity I are related as,
L = k log I,
where, k is a constant which depends on the sensitivity of the ear.
The unit of loudness is sone. One sone is the loudness experienced by a listener with normal hearing when a
sound of intensity level 40 dB at 1000 Hz is incident on his ears.
The unit of loudness level is phon. The loudness level of sound from a given source is said to be one phon if the inten-sity level of an equally loud pure tone of frequency 1000 Hz is 1 dB.
Noise and musical sound
Audible sounds can be classified as noise and music. Non-periodic vibrations of generally low frequency, having rapid variations in the amplitude and hence rapid variation of intensity, producing an unpleasant effect on the listener is called a noise. There is no regularity or rhythm in noise.
Example, Sound produced when a metal plate is pulled on another.
Periodic vibrations of generally high frequency, hav-ing smooth variations in the amplitude and hence produc-
3.50 Oscillations and Waves
ing a pleasing effect on the listener, is called a music. It has regularity and rhythm.Example, Sound produced by a musical instrument.
Characteristics of musical sounds
The characteristics of musical sound are pitch, loudness and quality (or timbre).
(a) Pitch
Pitch is the characteristic of musical sound, which distinguishes between a sharp sound and a dull sound. A sharper sound (penetrating sound) like the drilling of a hole on a metal piece is said to have higher pitch. Pitch of a musical sound is mostly dependent on its frequency and partly on its intensity. The sound produced by a mosquito is of high pitch. It is of high frequency but of low intensity (amplitude of vibration small) and is audible if the mosquito is flying close to the ear. Higher the frequency, higher will be the pitch. Pitch and frequency are not the same, pitch is a psychological sensation whereas frequency is a physical quality.
(b) Loudness
Loudness is a psychological sensation produced by sound in the ear of a person and it depends on the intensity of the sound and the sensitivity of the ear. It is a subjective quality, varying from person to person. The roaring of a lion is louder than the bleating of a goat.
(c) Quality or Timbre
Quality is that characteristic of a musical sound by virtue of which, it can be distinguished from another musical sound of same pitch and intensity. It depends on the number of overtones present, their order and relative intensities.
The sound produced by an instrument contains the fundamental frequency and its over tones. The overtones of one instrument differ from another instrument either in intensity or in the number of overtones or both. We are able to distinguish the voices of different persons by the characteristic quality or timber.
S u m m a r y
( )r r w22
max1 1
E v A2 2
= =
= ( )r pn p rn2 2 2 21
2 A 2 A2
=
E = Energy density (or energy per unit volume of medium) of a wave
A = amplitude of oscillation of particles of mediumw = angular frequency = 2pnn = Frequencyr = density of mediumvmax = maximum particle speed
I = p rn2 2 22 A v I = Intensity of wave (energy per unit area normal to the direction of propagation per unit time) v = velocity of propagation of wave
mT
v = = p r2
Tr
, v = Speed of transverse wave is a stringT = tension in the string andm = linear density of string (mass per unit length)r = density of wire andr = radius of wire
rY
v =v = speed of sound (longitudinal wave) in solidsY = Young’s modulus of the solidr = density of solid
Oscillations and Waves 3.51
rB
v = v = speed of sound in liquidB = bulk modulus of liquidr = density of liquid
rB
v = = grP = g g
rRT RTV M
=v = Speed of sound in gasesB = adiabatic bulk modulus = gP
g p
v
CC
= of gas, P = gas pressure
T = absolute temperature of gas, V = volume of gasR = universal gas constantM = molecular weight of gas
Dx = t2 Tl l
Df Dp× = ×
Df = p p
D Dl
2 2x t
T× = ×
Dt = T Tx
2Df D
p l× = ×
Relations between path difference (Dx), phase difference (Df) and time difference (Dt) in terms of wavelength ‘l’ and period ‘T’.
n1 0
s
v w vv w v
± ±= ± ±
n
note:If q is the angle between v & w, use ‘wcosq’ in-stead of ‘w’ in the above expression
n1 apparent frequency due to Doppler effectn = actual frequency of soundw = wind speed (+w, when wind travels in same direction of sound= −w, when wind travels in opposite direction of sound)v = Speed of sound in the mediumvO = Speed of listener (=+vO, if listener is moving towards sound−vO, if listener is moving away from soundIf listener is not moving, vO = 0vS = speed of source of sound(=+vS, if source is moving away from listener−vS, if source is moving towards listener).vs = 0, if source is not moving.
Two coherent waves of amplitudes a1 and a2 and intensities I1 and I2 interfere.AR = resultant amplitude at phase difference of ‘d’IR = resultant intensity at phase difference ‘d’
harmonicn1 : n2 : n3 : ……. = 1 : 2 : 3 : ……In this case, all odd and even harmonics are possible.Open ends correspond to antinode.Frequency of nth harmonic = Frequency of (n - 1)th overtone. When open pipe vibrates in nth mode (i.e.,Pth harmonic), number of nodes = nnumber of antinodes = (n + 1)number of loops = p
note:A loop is formed between two adjoining nodes.lP = wavelength in the nth modenP = frequency in the nth mode v = speed of sound through gas volumeL0 = length of open pipe
e = end correctiond = diameter of piper = radius of pipel1 = length of pipe in 1st mode of vibrationl2 = length of pipe in 2nd mode of vibration (1st overtone)l3 = length of pipe in 3rd mode of vibration (2nd overtone).n = mode of vibrationsn = 1, fundamental mode or first harmonicn = 2, second mode or second harmonic = 1st overtonen = 3, third mode or third harmonic = 2nd overtonen = nth mode or nth harmonic = (n - 1)th overtone
Oscillations and Waves 3.53
(B) Closed Pipe (open at only one end)
Frequency nP = ( )c
v2n 1
4L− or
lP = ( )c4L
2P 1−n = 1, 2, 3, ……..
n1 = c
v4L
n2 = c
3v4L
= 3n1
n3 = c
5v4L
= 5n1
n1 : n2 : n3 : ……. = 1 : 3 : 5 : ….In this case, only odd harmonics are pos-sible.Closed end corresponds to node and open end corresponds to antinode.Frequency of nth overtone = frequency of (2n + 1)th harmonicIn the nth mode (i.e., (n - 1)th overtone or (2n - 1)th harmonic), there are n nodes and n antinodes.
Lc = Length of closed pipen = mode of vibration = (n - 1)th overtone = (2n - 1)th harmoniclp = wavelength in pth modenp = frequency in pth mode
(C) End correction in open pipe and closed pipee = 0.3d = 0.6 rFor closed pipe,
e = ( ) ( )l ll l 3 12 1 53
2 4−−
=
For open pipe,
e = ( )l l2 12
2−
Resonance tube in an example of closed pipeTransverse vibrations of a string fixed at both ends
Frequency of vibration of the string in nth mode,
nP = m
n T2L
v = mT
n = mode of vibrationn = 1, 2, 3, …..L = length of stringT = tension in the stringm = linear density of string
= p r
p rl
l
22mass r
rlength
= =
r = radius of stringr = density of string
3.54 Oscillations and Waves
v = nP lP ; nP = lP
v
lP = 2LP
;
n1 : n2 : n3 : ……. = 1 : 2 : 3 : …..All odd and even harmonics are present
v = velocity of transverse wave in stretched stringlP = wavelength in nth nodeNo. of loops = nNo. of nodes = (n + 1), fixed ends are nodesNo. of antinodes = n
Melde’s experimentIn transverse mode, vibrations of tuning fork are at 90° to the length of the string. Frequency of vibration of string = frequency of tuning fork
nT = m
n T2L
In longitudinal mode, vibration of the prongs of the tuning fork are along the length of the string.
Frequency of vibration of string,
= Frequency of tuning fork
2
nL = m
n TL
, nT = nL
2
nT = frequency of transverse vibration of stringn = mode of vibration of stringL = length of stringT = tension in stringm = linear density of string (mass/length)nL = frequency of longitudinal vibration of string
Melde’s law
n T = constant or 2
1 22
12
n TTn
= , L = ( )1 2
1 2
2L LL L+
Frquency of tuning fork is given by
n = ( )( )
1 2
1 2
L L xL L
+−
n = mode of vibrationT1, T2 are the tension in the string in mode 1 and mode 2 (n = 1 and n = 2) If a tuning fork of frequency ‘n’ produces ‘x’ beats with length ‘L1’ and with another length ‘L2’ of the same string, length of the wire in unison with tuning fork ‘L’ is given by
Intensity of sound I = 20p1
2 Bv
p0 = pressure amplitudeB = bulk modulus of mediumV = speed of sound through medium
Energy density of sound
U = 20p1
2 B
p0 = pressure amplitudeB = bulk modulus of medium
Y = Asin(wt - kx) is the equation of a progressive wave along +X direction
A = amplitudeY = wave function (displacement for transverse wave)w = angular frequencyk = propagation constant =
pl
2
wk
= v, speed of wave is medium
Y = Asin(wt + kx) represents a progressive wave along -X direction
Oscillations and Waves 3.55
ConCept ConneCtorS
Connector 1: The displacement y in a simple harmonic motion is given by y(t) = 5sin (20t + 0.6) metre. What is the amplitude, frequency, time period and initial phase?
Solution: Comparing with y(t) = A sin (wt + f) Amplitude = 5 m, wt = 20 t \ w = 20 rad s-1
Frequency = wp p
202 2
= = 3.18 Hz.
T = n1
= 0.314 s
Initial phase f = 0.6 rad
Connector 2: The time period of a body executing SHM is 4 s. Find the time taken by the body to travel from mean position to the position of half the amplitude.
Solution: y = A sinwt = A sinp2
tT
when y = A
,2
pA 2
Asin t2 4
= ⇒ sinp 1
t2 2
= or p ot 302
= = p
rad6
t = 13
s
Connector 3: A particle executes SHM along a straight line of length 6 cm. Its velocity when it passes through the centre of line is 24 cm s-1. Find the period of oscillation.
Solution: We have y = A sin (wt + f)
Instantaneous velocity v = dydt
= A w cos (wt + f) ⇒ vmax = Aw
Given vmax = 24 cm s-1
A = 62
= 3 cm; \w = 243
= 8 rad s-1
\ Period of oscillation = pw2
= 2 3.14
8×
= 0.78 s
Connector 4: A particle executing linear SHM has speeds v1 and v2 at distances y1 and y2 from the equilibrium position. What is the frequency of oscillation of the particle?
Solution: v = w 2 2A y− ⇒ v2 = w2(A2 – y2)
(A2 – y2)= w
2
2
v ⇒ \A2 =
w
2
2
v+ y2 — (1)
A2 = w w
2 22 21 21 22 2
v vy y+ = + ⇒
w
2 22 21 22 12
v vy y
−= − ⇒ w2 =
2 21 22 22 1
v vy y
−−
Frequency n = p1
2
12 2 21 22 22 1
v vy y − −
3.56 Oscillations and Waves
Connector 5: A particle of mass 0.1 kg is executing SHM with amplitude 0.1 m and has initial phase of p6
rad. When
the displacement is one half of the amplitude, its kinetic energy is 6 × 10–3 joule. Obtain the equation of motion.
Solution: y = A sin(wt + f)
m = 0.1 kg, A = 0.1 m, f = 30o = p6
rad
K.E. = 12
mw2(A2–y2); At y = A2
, K.E. = w2
2 21 Am A
2 4
− = 6 × 10–3 J
w2 = 16, w = 4 rad s-1
The equation of SHM is given by y = 0.1 sinp
4t6
+ m
Connector 6: A particle is executing SHM of amplitude A. (a) What fraction of the total energy is kinetic when the displacement is half the amplitude? (b) At what displacement is the energy is half kinetic and half potential?
Solution: K.E. = 12
mw2(A2–y2); P.E. = w2 21m y
2
Total energy E = 12
mw2A2
(a) At y = A/2, K.E. = 12
mw2
22 A
A4
−
= 38
mw2A2 = w2 23 1m A
4 2×
⇒ 75% of total energy is kinetic
(b) K.E. = P.E. ⇒ 12
mw2(A2-y2) = 12
mw2y2
⇒ y = A2
Connector 7: Two S.H.M’s are represented by y1 = A sin p
wt4
+ and y2 = A (sin wt + 2 cos wt). Find the ratio of the
amplitudes of the two S.H.M’s.
Solution: y1 = A sinp
wt4
+ — (1)
Amplitude is A y2 = A (sin wt + 2 cos wt) — (2) The second equation contains two S.H.M’s A sin wt and A 2 cos wt
Their amplitude = ( )22 2A A 2 3A 3A+ = =
\ Ratio of amplitudes A 13A 3
=
Oscillations and Waves 3.57
Connector 8: Two identical springs of spring constant k are connected in (i) parallel and then (ii) in series, with the same rigid support. Calculate the time period when a body of mass m is connected to each of the combinations in turn.
Solution: (i) Parallel: ke = k1 + k2 Here k1 = k2 = k; \ ke = 2k
period T = 2pm2k
(ii) Series: ke = 1 2
1 2
k kk k+
; Hence k1 = k2 = k
\ ke = 2k
2k =
k2
⇒ Period T = 2p2mk
Connector 9: A massless spring of spring constant 100 N m-1 is cut into two identical pieces. (a) Calculate the spring constant of each piece. (b) The two pieces are suspended in parallel to support a mass m. If the system vibrates with a period of 0.314 s, find the value of m.
Solution: (a) When a spring is cut into n equal pieces, each piece has spring constant = nk Here n = 2. \The spring constant of each piece = 2 × 100 = 200 N m-1
(b) Effective spring constant when the springs are connected in parallel, ke = k1 + k2 = 200 + 200 = 400 N m-1
⇒ Period T = 2pe
mk
;
\ m = p
2e
2
T k4
= ( )
( )
2
2
0.314 400
4 3.14
×
× = 1 kg
Connector 10: The scale of a spring balance which can read from 0 to 142.4 N is 4 cm long. A package suspended from the balance is found to oscillate vertically with a frequency of 2 oscillations per second. How much does the package weigh?
Solution: Given n = 2 Hz ⇒ T = 0.5 s; The spring constant k = 2
mg 142.4x 4 10−=
× N m-1
Period T = 2pmk
, where m is the mass of package;
\ m = p
2
2
T k22.5 kg
4=
⇒ Weight of the package = mg = 221 N
m1
k
m2
Connector 11: Two masses m1 = 1 kg and m2 = 0.5 kg are suspended together by a massless spring of spring constant k as shown in figure. When the masses are in equilibrium m1 is removed without dis-turbing the system. Calculate the amplitude of oscillations.
(g = 10 m s-2, k = 12.5 N m–1)
Solution: The extension y produced in the spring due to m1 and m2 is given by
ky = (m1 + m2)g or y = ( )1 2m m
gk+
3.58 Oscillations and Waves
Let y' be the extension in length due to m2 alone ky' = m2g ⇒ y' = 2m gk
y – y' = ( )1 2 2 1m m m g m g
gk k k+
− =
This is the amplitude of oscillation when m1 is removed.
A = 1m g 1 100.8 m
k 12.5×
= =
Connector 12: A massless spring of spring constant 600 N m-1 is used to suspend a body of mass 1 kg. Another body of mass 0.5 kg moving vertically upwards with velocity 3 m s-1 hits the suspended body and gets embedded in it. Find the frequency and amplitude of oscillation.
Solution: n = ( ) ( )p p p1 k 1 600 10
Hz2 M m 2 1 0.5
= =+ +
M
mv
Using the conservation of linear Momentum mv = (m + M)V
V = ( ) ( )mv 0.5 3
m M 1 0.5×
=+ +
= 1 m s-1
Just after collision, the system will have K.E. = ( ) 21m M V
2+ at equilibrium position.
(K.E)max = (P.E.)max ⇒ ( ) 2 21 1m M V kA
2 2+ = \ A = V
m Mk+
=5 × 10–2 m = 5 cm
L h k
m
sConnector 13: A simple pendulum of length L and mass m has a spring of force constant k connected to
it at a distance h below its point of suspension as shown in figure. Find the frequency of vibrations of the system for small amplitudes.
Solution: If the pendulum is given a small angular displacement q, the spring will also stretch. let x = htanq
The restoring torque about S will be due to both force of gravity and elastic force of the spring t = –[mg(L sinq) + k(h tanq)h]
h k
m
S
θ x
When q is small, tan q ≈ sinq ≈ q ⇒ t = –(mgL + kh2)q (i.e.,) t a – q Since the restoring torque is linear, the motion is SHM.
t = Ia = mL2q2
2
ddt
⇒ mL2 q2
2
ddt
= (mgL + kh2)q
⇒ q2
2
ddt
= –w2q where w = 2
2
mgL khmL
+; \f =
wp p
2
2
mgL kh12 2 mL
+=
Connector 14: A vertical U-tube of uniform cross section contains water upto a height of 49 cm. The water in one side is depressed and released. Calculate the frequency of oscillations.
Solution: h = 49 cm = 0.49 m, g = 9.8 m s-2
⇒ T = 2p hg
= 2 × 3.140.499.8
= 1.4 s
Frequency n = 1T
= 1
1.4 = 0.714 Hz
Oscillations and Waves 3.59
Connector 15: A disc of mass M and radius r oscillates about a point on its rim. What is the time period?
Solution: For a physical pendulum, T = 2p IMgd
Here I = 2Mr
2 + Mr2 =
32
Mr2
d = r
\ T = 2p 3r2g
Connector 16: A travelling harmonic wave is represented by the equation y = 0.1 sinp
p20 x
200 t17
− where, y is the
displacement in metre, t in second and x is the distance from a fixed origin in metre. Find (i) the frequency, wave length and speed of wave (ii) Phase difference between a point 0.25 m from origin and another point 1.10 m from the origin.
Solution: (i) y = 0.1 sinp
p20 x
200 t17
− , comparing with y = a sin ( )p
l2
vt x −
200p = p
pnl
2 v2= or n = 100 Hz and
p pl
20 217
= or l1710
= = 1.7 m
Velocity v = nl = 1.7 × 100 = 170 m s-1
(ii) Phase difference = pl
2× path difference
Df = p2
1.7[1.10 – 0.25] = p rad.
Connector 17: A simple harmonic wave is expressed by y = 0.04 sin p
p50 t x6
− metre where t = time in second, x in
metre. Determine the frequency and wavelength of this wave. Another wave has the equation y = 0.01 sin p p
px
50 t6 3
− +
. Determine the phase difference and ratio of intensities of the two waves.
Solution: Compare with y = A sin (wt – kx) w = 50p ⇒ 2pn = 50p ⇒ n = 25 Hz
and k = p p p
l2
6 6⇒ =
⇒ l = 12 m
Phase difference = p3
radian
2
1 12
2 2
I AI A
= (Qw is same and v same)
= ( )( )
2
2
0.04 1610.01
= .
3.60 Oscillations and Waves
Connector 18: A 160 cm long string has two adjacent resonances at frequencies of 85 Hz and 102 Hz respectively. What is the fundamental frequency of the string? What is the length of a segment at 85 Hz resonance? What is the speed of waves in the string?
Solution: L = 160 cm = 1.6 m
The fundamental frequency f1 = v
2L The nth harmonic frequency nf1 = 85 Hz — (1) (n + 1)f1 = 102 Hz — (2) Subtracting (1) from (2) f1 = 17 Hz
For 85 Hz, resonance is at n = 8517
= 5
For n = 5, the length of each segment = 160
5 = 32 cm = 0.32 m
We have f1 = V2L
⇒ V = f1 × 2L = 17 × 2(1.6) = 54.4 m s-1
Connector 19: Transverse waves are generated in two uniform steel wires A and B of diameters 10–3 m and 0.5 × 10–3 m respectively, by attaching their free end to a vibrating source of frequency 500 Hz. Find the ratio of the wavelengths if they are stretched with the same tension.
Solution: The density of a wire of Mass m, length L and diameter d is given by r = m
p p2 2
4m 4d L d
= where m is the linear
density, m = p r2d
4
Now, vA = mA
T vB =
mB
T \
mm
A B B
B A A
v dv d
= =
n = 500 Hz; \vA = nlA and vB = nlB
\ll
3A B
3B A
d 0.5 10d 10
−
−
×= = = 0.5
Connector 20: Calculate the Young’s modulus of brass if the velocity of mechanical waves in it is 3068 m s-1 and its density is 8500 kg m–3. What is the wavelength of the wave in brass if the frequency of the sound is 240 Hz?
Solution: r = 8500 kg m–3, v = 3068 m s-1, n = 240 Hz ⇒ v =rY or Y = v2r
Y = (3068)2 × 8500 = 8 × 1010 N m–2
l = nv 3068
240= =12.8 m
Connector 21: If the frequency of a tuning fork is 400 Hz, find how far the sound travels when the fork makes 30 vibra-tions. The velocity of sound in air is 330 m s-1.
Solution: T = n1 1
400= = 2.5 × 10–3 s
Time for 30 vibrations t = 30 × 2.5 × 10–3 = 0.075 s
Distance travelled by sound = v × t = 330 × 0.075 = 24.75 m
Oscillations and Waves 3.61
Connector 22: The ratio of the densities of oxygen and nitrogen is 16 : 14. At what temperature the speed of sound in oxygen will be equal to its speed in nitrogen at 14oC.
Solution: v = gRTM
Speed of sound in oxygen at toC = speed of sound in nitrogen at 14 oC.
⇒ ( ) ( )g g
0 N
R 273 t R 273 14M M
+ +=
⇒ 0 N
273 t 287M M
+= ⇒ 0
N
M273 t 16287 M 14
+= = \ t = 55 oC
Connector 23: A closed pipe produced a note of 512 Hz at 15 oC. What is the length of the pipe? Speed of sound at 0 oC is 335 m s-1.
Solution: vt = ?, v0 = 335 m s-1, t = 15 oC
vt = v0273 t
273+
12
o 015 C
273 15 288v v 335
273 273+ = =
= 344 m s-1 If ℓ is the length of the closed pipe,
ℓ = l4
, l = 4ℓ
Frequency n = l lv v
4= ; \ℓ =
nv 344
4 4 512=
× = 0.168 m
Connector 24: Find the ratio of the length of a closed pipe to that of open pipe in order that the second overtone of the latter is in unison with fourth overtone of former.
Solution: Frequency of second overtone of open pipe is n0 = l0
3v2
Frequency of fourth overtone of the closed pipe nc = l c
9v4
l
l l lc
0 c 0
3v 9v 32 4 2
= ⇒ =
Connector 25: The fundamental frequency of an open organ pipe is 300 Hz. The frequency of the first overtone of another closed organ pipe is the same as the frequency of the first overtone of open pipe. What are the lengths of the pipes? (speed of sound = 330 m s-1)
Solution: n0 = 0
v2L
= 300 \L0 = 330
0.55 m2 300
=×
Frequency of 1st overtone of closed pipe = frequency of 1st overtone of open pipe.
C 0
3v v4L L
= \LC = 03L 3 0.554 4
×= = 0.412 m
Connector 26: Two open organ pipes of length 50 cm and 50.5 cm produce 7 beats in two second. Calculate the velocity of sound in air.
Solution: ℓ1 = 50 cm, ℓ2 = 50.5 cm Let the velocity of sound be v
n1 = l1
v v2 100
=
n2 = l2
v v2 101
=
3.62 Oscillations and Waves
Beats per second = 3.5 = n1 – n2
⇒ v v
100 101− = 3.5 or v = 353.5 m s-1
Connector 27: Two trains travelling in opposite directions at 100 kmph each, cross each other while one of them is blow-ing a whistle. If the frequency of the note is 800 Hz, find the apparent frequency as heard by an observer in the second train,
(a) before the trains cross each other (b) after the trains have crossed each other. Velocity of sound in air = 340 m s-1
Solution: (a) nL1
S
v v 340 27.78f 800
v v 340 27.78 + + = = × − −
= 942.3 Hz
1L S
100 5v v 27.28 m s
18−×
= = =
(b) nL2
S
v v 340 27.78f 800
v v 340 27.78 − − = = × + +
= 679.2 Hz
Connector 28: Two cars situated 1 km apart sound their horns of frequency 330 Hz. A man is moving away from one car towards the other with a speed of 2 m s-1. What will be the beat frequency heard by the man? (velocity of sound = 330 m s-1)
Solution: f1 = Lv vf
v−
=
330 2330
330−
= 328 Hz
f2 = Lv vf
v+
=
330 2330
330+
= 332 Hz
Beat frequency = (f2 – f1) = 332 – 328 = 4 Hz
Connector 29: A stationary source of sound P is producing sound of frequency 165 Hz. Another source of sound Q is moving towards P on a straight path, with a uniform speed of 22 m s-1, producing a sound of frequency 231 Hz. A listener in a vehicle L, which is in between P and Q, is moving towards P along the straight path QP. What is the speed of vehicle L for which the listener in it hears no beat? Given, speed of sound in air = 330 m s-1.
Solution: Let the speed of vehicle L be vL
Then apparent frequency of sound from P as heard by listener, fP' = fP( )Lv v
v+
= ( )L330 v
330+
× 165 Hz — (i)
Apparent frequency of sound from Q as heard by listener, fQ' = fQ( )L
Hence speed of vehicle L is 66 m s-1 (i.e., 237 kmph)
Connector 30: A whistle emitting a sound of frequency 240 Hz is tied to a string of 2 m length and rotated with an angular velocity of 30 rad s-1 in the horizontal plane. Determine the range of frequencies heard by an observer stationed at a large distance from the whistle. Speed of sound = 320 m s–1.
Solution: v = rw = 60 m s-1
ω = 30vr = 2m
v
v' = 320
240 295 Hz320 60⋅ =
−
v'' = 320
240320 60⋅
+ = 202 Hz
Range = 202 Hz to 295 Hz
3.64 Oscillations and Waves
Subjective Questions
1. Two pulleys with centers fixed at distance l between them are rotating with the same angular velocity as shown. A bar is placed on top of them, slightly off center. The coefficient of friction between bar and pulleys is m. Show that the bar will perform SHM. Find the time period.
2. Determine the natural angular frequency (w) of the system. Neglect the mass of pulleys and friction.
2m
θ
mk
3. A mass m1 = 2 kg is attached to a spring of natural length l0 = 20 cm whose other end is fixed at O. Spring constant k = 200 N m-1. The system rests on horizontal smooth plane.
Now the mass m1 is moved 4 cm towards O and released at time t = 0
(i) Write the equation relating x, position of mass m1 from origin at O vs time, (taking direction towards right as positive) up to time t before collision.
At t = p3s
40, a mass m2 = 1 kg moving with velocity 2 m s-1 towards m1 and collides with it completely inelasti-
cally and subsequently the two bodies move together. (ii) Write the new equation x vs t, after collision.
4. A uniform solid cylinder of mass m and radius r is released at the position shown (q0 is small) and it rolls without slipping on the circular track (Radius R)
(i) Show that it performs SHM (ii) Determine the time period (iii) What is its angular velocity when it passes the bottom most point on the track?
5. Assembly of two light rods (length l) each carrying a point mass m is pivoted as shown, with included angle a
(i) Determine the period of small oscillations of the assembly. (ii) Discuss the cases (a) a = 0, (b) a =180°
6. A plane transverse wave propagating in a solid medium of density 250 kg m-3, produces a maximum shear strain of 2p × 10-4 and an energy density of 0.45p2 J m-3 in the medium. Also particles located along direction of propagation (X axis) at 0.1 m from each other are out of phase by 18°. Treating phase = 0 at x = 0 and t = 0, write the equation of the wave.
�
• •
m1 m2
�0
O
Rθ0
r m
�α
�
mm
TOpIC grIp
Oscillations and Waves 3.65
7. A metal rod 2 m long is clamped as shown at two points each 0.6 m away from its ends. The density and Young’s modulus of the rod are 7000 kg m-3 and 1.75 × 1011 N m-2 respectively.
Calculate the frequency of natural longitudinal oscillations of the rod and the next higher frequency.
8. Two wires of materials A and B are joined end to end and fixed at both ends with tension all along the wires at 1024 N. The lengths, areas of cross section and densities of the wires A and B are 0.6 m, 0.5 m, 10-5 m2, 2 × 10-5 m2, 4000 kg m-3, 8000 kg m-3
respectively. Determine the lowest frequency of a transverse wave which sets up stand-ing waves in the wires with a node at the joint C.
9. A tuning fork of frequency 259 Hz and an open organ pipe of slightly lower fundamental frequency are at 16°C. When sounded together, they produce 4 beats per second. On altering the temperature of the air in the pipe, it is observed that the number of beats per second first becomes zero and then increases again to 4. By how much has the temperature of air in the pipe changed?
10. A source of sound of frequency n is located in between a receiver and a reflector. All three are moving along the same straight line in the same direction with receiver moving towards source. The various speeds are : source; us, receiver: ur, reflector : uf. Speed of sound = v. Determine frequencies and wavelengths of direct and reflected waves received by the receiver. (For direct wave n1 and l1 and for reflected wave n3 and l3 respectively).
Straight Objective Type Questions
Directions: This section contains multiple choice questions. Each question has 4 choices (a), (b), (c) and (d), out of which ONLY ONE is correct.
11. The reading of a spring balance is from 0 to 200 N and is 10 cm long. A body suspended from the spring balance is observed to oscillate vertically at 2 Hz. The mass of the body is, (Take p2 = 10)
(a) 22.5 kg (b) 12.5 kg (c) 37 kg (d) 45 kg
12. A particle of mass 0.1 kg executes SHM of amplitude 0.2 m. When the particle passes through the mean position, its kinetic energy is 0.032 J. The equation of motion of this particle, if the initial phase is 30°, is (y is in metre).
(a) y = 0.2 sin 8pt (b) y = 0.2 cos 4t (c) y = 0.2 sin p
4t6
+ (d) y = 0.2 cos
p6
sin 4t
13. A mass m is hanging from a light spring and is in vertical oscillation. If during the oscillations the minimum length of the spring is its natural length, then the maximum force exerted by the spring on the mass is
(a) mg (b) 2mg (c) mg2
(d) data insufficient
14. An object of specific gravity 1.2 is hung from a thin steel wire. The fundamental frequency for transverse standing waves in the wire is 300 Hz. The object is partially dipped in water. The weight of water displaced is 420 g and the part of the body above water has mass 400 g. The new fundamental frequency in Hz is
(a) 220 Hz (b) 240 Hz (c) 360 Hz (d) 270 Hz
15. In a sinusoidal wave, the time required for a particular point to move from maximum displacement to zero displace-ment is 0.170 s. The frequency of the wave is
(a) 1.47 Hz (b) 0.73 Hz (c) 2.94 Hz (d) 0.36 Hz
2 m0.6 m0.6 m
CA B
3.66 Oscillations and Waves
Assertion–Reason Type Questions
Directions: Each question contains Statement-1 and Statement-2 and has the following choices (a), (b), (c) and (d), out of which ONLY ONE is correct.
(a) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1(b) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1(c) Statement-1 is True, Statement-2 is False(d) Statement-1 is False, Statement-2 is True
16. Statement 1 The time period of vertical SHM of a spring-mass system inside a lift remains same whether the lift is accelerating
uniformly or in uniform motion. and
Statement 2
pm
T 2k
= , spring constant k is same for all frames, inertial or non-inertial.
17. Statement 1 When a wave propagates from one medium to another, its intensity will remain constant. and
Statement 2 Intensity is proportional to square of frequency and frequency is a property of the source and does not depend on
medium of propagation.
18. Statement 1 The two waves Y1 = A sin (wt - kx) and Y2
= A sin (kx - wt) cannot produce a standing wave. and
Statement 2 The phase difference between the two waves is p radian.
19. Statement 1 When a standing wave is produced by two waves represented by Y1 = Asin(wt - kx) and Y2 = Asin(wt + kx) the short-
est distance from origin where a person hears minimum intensity is at x = l/4 [l → wave length].
and
Statement 2 Resultant wave Y’ = Y1 + Y2.
20. Statement 1 If a source performs uniform circular motion, a listener at centre will register actual frequency. and
Statement 2 Apparent frequency will be different from actual frequency only when relative velocity is non-zero.
Oscillations and Waves 3.67
Linked Comprehension Type Questions
Directions: This section contains 2 paragraphs. Based upon the paragraph, 3 multiple choice questions have to be an-swered. Each question has 4 choices (a), (b), (c) and (d), out of which ONLY ONE is correct.
Passage I Period of angular oscillations of bodies with complicated shapes
A rigid body of arbitrary shape suspended from a point O, will execute SHM when displaced by a small angle q and released.
Now consider the assembly of 3 rods each of mass m and length l, joined at one end as shown. (q = 45°)
O
B
�/2
A
NPMθθ
C
x�/2
21. The distance of the centre of mass of the assembly from point of suspension O is
(a) l2
(b) ( )l2 2
2+ (c)
( )( )
l 2 2
2 2
+ (d) ( )l
2 16
+
22. The period of oscillations of the system about O is
(a) ( )pl
2 2 1g
− (b) ( )pl
2 2 2 1g
− (c) ( )l 1g 2 2 1−
(d) l2 2
g
23. If the rod OB is now removed, the time period of oscillation is
(a) pl
2g
(b) pl2
2 4 2 23g
− (c) pl2
23 g
(d) pl
2 2.g
Passage II
The equation for a wave travelling in the positive x-direction can be written as y+ = A sinwx
tv
− , where w is the angular
frequency (2pf) and v is the wave velocity.
24. Consider a wave y = 10 sin(wt + kx) reflected at a rigid obstacle at x = 0 with reflected wave having 81% of the incident intensity. Equation for the reflected wave at the boundary is:
(a) y = 8.1sin(wt – kx) (b) y = 9 sin(wt – kx) (c) y = –9sin(wt – kx) (d) y = –8.1 sin(wt – kx)
25. In Q. No. 24 if the resultant is expressed as a combination of standing wave and a travelling wave in the positive direc-tion, then standing wave is represented by:
26. In the above case, the travelling wave is expressed as: (a) –1sin(wt – kx) (b) sin(wt – kx) (c) sin(wt + kx) (d) –19sin(wt + kx)
C
O
mg
θL
d
3.68 Oscillations and Waves
Multiple Correct Objective Type Questions
Directions: Each question in this section has four suggested answers of which ONE OR MORE answers will be correct.
27. A spring of spring constant k and having mass m has a point mass m attached to its end and its other end is fixed and the system is kept on a smooth table. If T is the period of oscillations with a massless spring with same spring constant k connected to a point mass m, and T’ is the period of oscillation with the given spring, then
(a) T' = 23
T
(b) T' = T2
(c) At any instant KE of spring = 14
of total KE of spring-mass system
(d) At any instant KE of spring = 34
of total KE of spring-mass system
28. Consider a pendulum hanging from the ceiling of a cabin. The cabin can be accelerated in 4 different directions as shown. Let T be the period of the pendulum when the cabin is stationary and Ta’. Tb’, Tc’ and Td’ be the 4 other periods in the four cases respectively
case c4
g
case a
case b
case d
4g
4g
4g
(a) Ta' is minimum (b) Tb' = Td' (c) Tb' ≠ Td' (d) Tc' is maximum
29. A stationary listener and a source moving towards the listener, both possess an audio source of frequency f. The listener hears 8 beats per second when source approaches the listener at speed v' =
v100
(v → velocity of sound)
(a) f = 792 Hz (b) When source moves away with same velocity v’, same number of beats are heard by listener per second (c) When source moves away from listener with speed v’, less beats are heard per second. (d) If source and listener move towards each other with velocity v’ no beats are heard by listener
Matrix-Match Type Question
Directions: Match the elements of Column I to elements of Column II. There can be single or multiple matches.
30. A block of mass M attached to a spring of force constant k is in simple harmonic motion having amplitude A1 and energy E1 M
k
Oscillations and Waves 3.69
(I) At the instant when the block passes through the equilibrium position, a lump of putty with mass m is placed on the block and the putty sticks.
(II) In a second experiment, at the instant when the block is at its maximum displacement, the lump of putty of mass m is placed on the block.
Column I Column II (a) Amplitude in case (I) (p) A1
(b) Time period in case (I) (q) 1
2 ME
k M m +
(c) Amplitude in case (II) (r) 2p M m
k+
(d) Time period in case (II) (s) 1M
AM m+
3.70 Oscillations and Waves
I It aSSIgnment exerCISe
Straight Objective Type Questions
Directions: This section contains multiple choice questions. Each question has 4 choices (a), (b), (c) and (d), out of which ONLY ONE is correct.
31. The mean K.E. and P.E. of a non-dissipative harmonic vibrating system are (considering PE at mean position as zero) (a) Unequal (b) Equal (c) Zero (d) In the ratio 1 : 2
32. Two bodies x and y of equal masses are suspended from two separate massless springs of spring constants k1 and k2 respectively. If the two bodies oscillate vertically such that their maximum velocities are equal, the ratio of the ampli-tude of vibration of x to that of y is
(a) 1
2
kk
(b) 1
2
kk
(c) 2
1
kk
(d) 2
1
kk
33. For a particle executing SHM, the kinetic energy is K0 cos2wt. The maximum value of potential energy is (given po-tential energy at mean position is zero)
(a) K0 (b) Zero (c) 0K2 (d) 2K0
34. The maximum acceleration of a body moving in SHM is a0 and maximum velocity is v0. The amplitude is given by (a) v0
2/a0 (b) a0v0 (c) a02/v0 (d) 1/a0v0
35. The equation of motion of an object executing SHM is x = 2.0 cosp
0.50t4
+ m. Its maximum acceleration is
(a) 0.05 m s–2 (b) 0.50 m s–2 (c) 2.0 m s–2 (d) 2.25 m s–2
36. A mass m is vertically suspended from a spring of negligible mass. The system oscillates with a frequency n. If a mass 4 m is suspended from the same spring, the frequency of the system will be
(a) n/4 (b) n/2 (c) 2n (d) 4n
37. The kinetic energy of a particle executing SHM is 16 J when it is in its mean position. What will be the time period of oscillations (in s) if amplitude is 25 cm and mass of the particle is 5.12 kg?
(a) p5 (b) 2p (c) 5p (d) 20 p3
38. The displacement of a particle executing SHM is given by x = 0.01 sin100p(t + 0.05) m. Its time period is (a) 0.01 s (b) 0.02 s (c) 0.1 s (d) 0.2 s
39. For a particle of mass ‘m’, executing SHM along the X-axis with mean position at the origin and having angular fre-quency ‘w’, the unbalanced force on it, when its displacement is x , is equal to
(a) –mw2 x (b) +mw2 x (c) -mw x (d) +mw x
40. A particle starts SHM from the mean position. Its amplitude is a and total energy is E. At an instant when its kinetic energy is 3E/4, its displacement is
(a) a2
(b) a2
(c) a32
(d) 2a
Oscillations and Waves 3.71
41. For a particle vibrating simple harmonically, if the displacement is 8 cm when the velocity is 6 cm s-1 and the displace-ment is 6 cm when the velocity is 8 cm s-1, then the time period is(in second)
(a) 4p (b) 2p (c) p (d) p/2
42. The displacement of a particle executing SHM is y = a sin(wt – f). The velocity of the particle at t = ϕw is
(a) wcosf (b) aw (b) –awcos2f (d) wcos2f
43. A 2 kg mass executes SHM with a frequency of 5 Hz. The restoring force acting on the body when the displacement of the body is 2 cm is
(a) 3.95 N (b) 39.5 N (c) 51 N (d) 48.5 N
44. A body of mass 5 g is executing SHM about a point O with an amplitude of 10 cm. Its maximum velocity is 100 cm s–1. Its velocity will be 50 cm s-1 at a distance
(a) 5 cm (b) 5 2 cm (c) 5 3 cm (d) 5 5 cm
45. The displacement y (in cm) of a vibrating particle is given in terms of time t in second as y = 3 sin 314t + 4 cos 314t. The amplitude of the SHM is
(a) 3 cm (b) 4 cm (c) 5 cm (d) 7 cm
46. If y, v and a represent displacement, velocity and acceleration vectors at any instant of a particle executing SHM, then, always:
(a) y and a are in opposite direction (b) y and a have same direction
(c) y and v have same direction (d) y and v have different direction
47. If two orthogonal SHMs of same frequency, equal amplitude and having initial phase difference of p2 radian act simultaneously on a particle which is free to move, the particle can move in a
(a) parabola (b) straight line (c) helix (d) circle
48. The displacement y of an oscillating particle varies with time t in s, according to the equation y = p t 1
sin2 2 3 +
where y is in cm. The maximum acceleration of the particle is
(a) 0.62 cm s-2 (b) 1.71 cm s-2 (c) 3.62 cm s-2 (d) 5.2 cm s-2
49. A particle performs harmonic oscillations along a straight line with a period of 6 s and amplitude 4 cm. The mean velocity of the particle averaged over the time interval during which it travels a distance of 2 cm starting from the extreme position is
(a) 1 cm s–1 (b) 2 cm s–1 (c) 4 cm s–1 (d) 8 cm s–1
50. A body executes SHM under the influence of a spring force with a time period of 3 s and the same body executes SHM with a time period of 4 s under the influence of another spring force. When both the springs act in parallel and simultaneously along the same direction the time period of the same body is
(a) 2.4 s (b) 2√3 s (c) 5 s (d) 7 s
51. A simple harmonic motion has an amplitude A and time period T. The time required to travel from x = A to x = A/2 is (if x = 0 is the equilibrium position)
(a) T6 (b) T
4 (c) T3 (d) T
2
52. A seconds pendulum on Earth is taken to outer space, where g = 0. The period of oscillation is (a) 2 s (b) 0 (c) Infinity (d) 1 s
53. For a simple pendulum, graph between its length and time period will be a (a) circle (b) parabola (c) straight line (d) hyperbola
3.72 Oscillations and Waves
54. A simple pendulum is suspended from the roof of a trolley, which moves in a horizontal direction with an acceleration a.
Then the time period T is given by pl
2g '
, where g’ is
(a) 2 2g a− (b) g - a (c) g + a (d) 2 2g a+
55. Two simple pendulums of length 5 cm and 20 cm are given small linear displacements in one direction at the same time. The number of oscillations of the smaller pendulum after which they will be in phase again for the first time is
(a) 1 (b) 2 (c) 3 (d) 5
56. A hollow sphere is filled with water through a small hole in it. It is then hung by a long thread and made to oscillate with the hole at the bottom. As the water slowly drains out of the hole at the bottom, the period of oscillation will
(a) continuously decrease (b) continuously increase (c) first decrease and then increase (d) first increase and then decrease to become the initial value ultimately
57. The equation y = A sin2(kx –wt) represents a wave motion with
(a) Amplitude A and frequency wp2
(b) Amplitude A2
and frequency wp
(c) Amplitude 2A, and frequency wp4
(d) Amplitude A2 and frequencywp4
58. A smoked glass plate is held vertically in front of a vibrating fork, provided with a stylus. The plate is allowed to fall freely under gravity, so that the stylus traces a wavy line upon it. If the number of waves marked in a distance d be n, then the frequency of vibration of the fork is
(a) ndg
(b) n2dg
(c) n 1
dg
+ (d) 2d
g
59. Two sound waves are given by y = A sin(wt – kx) and y = A cos(wt – kx). Then the phase difference between the waves (in radian) is
(a) Zero (b p2
(c) p (d) p34
60. Which among the following represents a traveling harmonic wave? (a) y = A (wt – kx) (b) y = A sin wt (c) y = A coskx (d) y = Asin(t–bx + c)
61. Two harmonic waves travelling in the same medium have frequency ratio 1 : 2 and intensity ratio 1 : 36. Then their amplitude ratio will be
(a) 1 : 3 (b) 1 : 6 (c) 1 : 8 (d) 1 : 32
62. Two waves of same frequency, constant phase difference but with different amplitudes superimpose. Then (a) There will be no interference (b) Intensity of sound at any point will vary with time (c) There will be interference with non zero minima (d) There will be interference with zero minima
63. Two waves of same amplitude and frequency superimpose to produce a resultant disturbance of the same amplitude. The phase difference between the two waves is (in radian)
(a) Zero (b) p3
(c) p2
(d) p23
Oscillations and Waves 3.73
64. The ratio of the amplitudes of two superimposing waves is 3 : 2. The ratio of the maximum and minimum intensities of resultant wave is
(a) 9 : 4 (b) 25 : 1 (c) 13 : 5 (d) 5 : 1
65. Two waves of same amplitude A, and equal frequency travel in the same direction through a medium. The amplitude of the resultant wave is
(a) Always < A (b) A (c) 2A (d) Between 0 and 2A
66. The same musical note played on guitar and veena differ in (a) Quality (b) Pitch (c) Both quality and pitch (d) Neither quality nor pitch
67. A stretched string of length 2 m vibrates in 4 segments. The distance between consecutive nodes is (a) 0.5 m (b) 0.25 m (c) 1 m (d) 0.75 m
68. A vibrating rod fixed between two points carries a transverse wave of wavelength l. The length of the rod for the first harmonic should be
(a) l/2 (b) l (c) l/4 (d) 2l
69. If v is the velocity of wave in a string of length ‘l’ clamped at both ends, the frequency of the fundamental note emitted will be
(a) lv
(b) l
2v (c)
lv
2 (d)
lv4
70. Standing waves are produced in a 10 m long stretched string. If the string vibrates in 5 segments and the wave velocity is 20 m s-1, the frequency of vibration is
(a) 2 Hz (b) 3 Hz (c) 10 Hz (d) 5 Hz
71. Among the following materials, the speed of sound is highest in (a) Steel (b) Air (c) Water (d) Vacuum
72. The velocity of sound is generally greater in solids than in gases because, for a solid (a) Density is high and elasticity is low compared to gases (b) Density is very high compared to gases (c) Density is low and elasticity is high compared to gases (d) The elasticity is very high compared to gases
73. The fundamental frequency of a string stretched between two points is 100 Hz. The frequency of the third overtone will be
(a) 200 Hz (b) 300 Hz (c) 400 Hz (d) 600 Hz
74. With the increase in temperature the frequency of sound from an organ pipe (a) Decreases (b) Increases (c) Remains unchanged (d) Changes irregularly
75. The fundamental frequency of a closed pipe is 50 Hz. The frequency of the second overtone is (a) 100 Hz (b) 150 Hz (c) 200 Hz (d) 250 Hz
76. Air column in a closed pipe is in resonance with a tuning fork of frequency 680 Hz. The length of the air column is (v = 340 m s-1)
(a) 0.25 m (b) 0.125 m (c) 0.375 m (d) Could be (b) or (c)
77. When beats are produced by two waves of nearly the same frequency and amplitude (a) The beat frequency depends on the position where beats are heard (b) Beat frequency decreases as time passes (c) Particles vibrate simple harmonically with a frequency equal to the difference of two frequencies (d) Amplitude of vibration at any point changes simple harmonically with a frequency equal to the difference of the
two frequencies
3.74 Oscillations and Waves
78. Two tuning forks of frequencies 256 Hz and 258 Hz are sounded together. The time interval between two consecutive maxima heard by a listener is
(a) 0.5 s (b) 2 s (c) 50 s (d) 252 s
79. Three waves from different sources but having same intensity but frequencies (n - 1), n and (n + 1) superimpose. The number of beats per second will be
(a) 0 (b) 1 (c) 2 (d) 32
80. A police car, sounding a siren of 1200 Hz, moves towards a high vertical cliff, at a uniform speed of v
20 m s-1. The
frequency of the echo as heard by the driver of the police car is (speed of sound = v m s-1) (a) 1255 Hz (b) 1326 Hz (c) 326 Hz (d) 1417 Hz
81. Two light springs of force constants 3.6 N m-1 and 6.4 N m-1 as well as a block of mass 0.4 kg are in one line AB on a smooth horizontal table so that one end of each spring is fixed on rigid supports and the other end is free. The distance PQ between the free ends of the spring is 1.2 m. If the block moves along AB with a velocity of 0.9 m s-1 in between the springs, the time period of oscillation of the block is
(a) 7.9 s (b) 5.0 s (c) 2.8 s (d) 4.5 s
82. A particle of mass 0.2 kg is subjected to simple harmonic oscillation of amplitude 0.1 m. When the particle goes through the mean position, its kinetic energy is 8.1 × 10-2 J. The equation of motion of this particle when the initial phase of oscillation is 36° is given by x =
(a) ( )p0.1cos 5t 5+ m (b) ( )p0.1sin 5t 5− m (c) ( )p0.1sin 9t 5+ m (d) ( )p30.2sin 3t5 + m
83. The mass M attached to the spring does SHM of period p4
10 10 s. If the 1 kg mass is not to
lift off the floor, the amplitude of SHM should be less than (g = 10 m s-2) (a) 0.04 m (b) 0.06 m (c) 0.1 m (d) 0.14 m
84. A uniform cylinder of weight W (in N) and of radius r is suspended with its length vertical from a fixed point by a light spring such that the cylinder is half submerged in a liquid of density r at equilibrium position. When the cylinder is given a small downward push and released, it starts oscillating, with a small amplitude. If the spring stiffness is k, then the motion is with time period of oscillation given by
(a) pW
T 2k
= (b) pr p 2
WT 2
g r=
(c) p
r 2
WT 2
2 gr= (d) p
p r2
WT 2
g(k r g)=
+
85. Two blocks A and B of masses 0.3 kg and 0.4 kg are stuck to each other one below the other as in the figure by a glue. They hang together at the end of a spring with force constant equal to 200 N m-1 and they are at rest. The block B suddenly detach itself from A and falls down free. Then, the total energy of the subsequent oscillation is (Take g = 10 m s-2 and p2 = 10)
(a) 0.04 J (b) 0.05 J (c) 0.06 J (d) 0.07 J
k1 m
B A P QR S
k2
k =100 N m−1
M
1 kg
AB
Oscillations and Waves 3.75
86. A particle of mass 0.2 kg does SHM of amplitude 0.1 m. Its kinetic energy is same at t, t + 1, t + 2, (second) and is not equal to maximum KE at these instants. Its minimum total mechanical energy (in joule) could be (Take p2 = 10)
(a) 0.001 (b) 0.0025 (c) 0.01 (d) 0.025
87. The maximum restoring force on a SHM oscillator is 16 N. Its oscillation energy is 1.6 J. The force constant k of the oscillator (in N m-1) is
(a) 40 (b) 80 (c) 160 (d) 320
88. A spring mass system has time period T. Now the spring and the mass are cut into two. Now the period of one spring and one mass is T1. The period of the other spring and mass is T2. If the spring and masses are connected as shown the new time period is
(a) 1 2T T T2
+ + (b) (T T1T2)
1/3
(c) 1 2T TT
(d) 1 2T T T2
+
89. A rough horizontal platform moves horizontally in SHM; the time period is T and the maximum speed is u. A mass is on the platform and it is always at rest relative to the platform. The least value of coefficient friction between the mass and the platform is
(a) gTu
(b) pgT
u (c)
p2 ugT
(d) 2pgTu
90. A simple pendulum is made by attaching a bob of mass m suspended from a massless cylindrical wire of natural length L, diameter d and modulus of elasticity Y. The time period of small oscillations of the pendulum is
(a) pp 2
Lmg2 1
d y− (b) p
L2
g (c) p
p 2
L 4mg2 1
g d Y
+
(d) p 2
Lmg2
Yg d
91. The frequency of oscillation of a simple pendulum of length 0.4 m, suspended from the roof of a box sliding down a smooth inclined plane of inclination 37° is (g = 10 m s-2)
92. A boat is travelling in a river with a speed of 10 m s-1 along the stream flowing with a speed of 5 m s-1. From this boat, a sound transmitter is lowered into the river with the aid of a solid support. The wavelength of the sound emitted from the transmitter inside the water is 14 mm. Frequency detected by receiver kept inside the river downstream is (Bulk modulus of water = 1.96 × 109 Pa)
. If the maximum particle velocity is equal to 4 times
the wave velocity then the wavelength is
(a) p 0y
3 (b)
p 0y4
(c) p 0y
2 (d)
p 0y5
94. A sinusoidal wave of period 1ms has phase velocity of 0.7 km/s. The wavelength and the distance between any 2 points
with a phase difference of p4
radian at any particular time are respectively
(a) 0.55 m, 6.9 cm (b) 0.7 m, 8.75 cm (c) 1.4 m, 6 cm (d) 1 m, 12.5 cm
95. A wire of linear density 10 g per metre passes over a frictionless light pulley fixed on the top of a smooth inclined plane, which makes an angle of 37° with the horizontal. Masses M1 and M2 are tied at the two ends of wire such that
3.76 Oscillations and Waves
mass M1 rests on the inclined plane and M2 hangs freely vertically downwards. The entire system is in equilibrium and a transverse wave propagates along the wire with a velocity of 120 m s-1. Then
(a) M1 = 20 kg, M2 = 12 kg (b) 2
1
M2
M=
(c) 2
1
M 1M 2
= (d) 2
1
M0.6
M= and M1 = 24 kg
96. A wave disturbance in a medium is governed by equation y (x, t) = 0.02 ( )pp pcos 50 t cos 10 x
2 +
where x and y are
in metre and t in second. (a) A node occurs at x = 0.15 m, wavelength = 0.1 cm (b) An antinode occurs at x = 0.3 m, and speed = 4 m s-1
(c) Speed of wave 5 m s-1 and wavelength is 0.2 m (d) Wavelength is 0.5 m, and speed of wave is 3 m s-1
97. A transverse wave is travelling in a medium and its equation is given by y = 6 sin (5 t - 0.03 x) where y and x are in centimetre and t in second. If density of the medium is 1.2 g/cc, the amplitude and intensity of wave and other parameters are
(a) amplitude is 6 cm, intensity is 4 kJ m-2 s-1 (b) wave velocity is 1.67 m s-1 frequency is 0.796 Hz (c) amplitude is 6 cm, and intensity of wave is 25 kJ m-2 s-1 (d) w is 5 rad s-1 and intensity of wave is 10 kJ m-2 s-2
98. A wave of amplitude A1 is detected by a sensor. The minimum phase difference between this wave and another wave of amplitude A2 so that when both are present amplitude is less than A1 is
(a) 90° (b) cos-11
2
AA
(c) cos-12
1
A2A
−
(d) tan-11
2
AA
99. In a large room a person receives direct sound waves from a source 100 m away. He also receives sound waves from the same source, which reach him after reflection from the ceiling which is 5 m above him, at a point midway between them. The wavelength for which these sound waves interfere constructively at his location is (Assume reflection causes no change in phase)
(a) 0.6 m (b) 0.45 m (c) 0.4 m (d) 0.5 m
100. In a sonometer experiment, the tension of the sonometer wire is maintained by suspending a 500 kg weight from the free end of the wire. The suspended weight has a volume of 0.18 m3. The fundamental frequency of the wire is 250 Hz. If the suspended weight is completely submerged in water, the fundamental frequency of the sonometer wire will be
(a) 220 Hz (b) 230 Hz (c) 200 Hz (d) 240 Hz
101. Two travelling waves 13
y 4sin (x vt)2 = −
cm and 23
y 4sin (x vt)2 = +
cm are superimposed. The distance between
adjacent pressure antinodes is (in cm)
(a) p3
(b) p6
(c) p8
(d) p23
102. A string of length 0.5 m and mass 20 g, is stretched between two points with tension of 2 N. When a pulse travels along it, the shape of the string is the same at times t and t + Dt. Then the minimum value of Dt is equal to
(a) 0.20 s (b) 0.23 s (c) 0.18 s (d) 0.14 s
103. In an experiment for resonance using sonometer and tuning fork, it was found that 80 cm of sonometer wire vibrating in its fundamental frequency is in resonance with a given tuning fork. If the tension is increased by 10 gram weight, the length has to be increased to 81 cm to get resonance again with the same tuning fork. The tension in the wires in newton is (g = 10 m s-2)
105. A wire of density (linear) 0.1 kg m-1 made of steel, has a tension 48.4 N due to a mass suspended at one of its ends. Resonance is observed with this arrangement at a period of 2.5 ms and it resonates at the next highest frequency of 480 Hz. The length of the wire is
(a) 18 2 cm (b) 13.75 cm (c) 16 cm (d) 20 cm
106. A metal wire is fixed between rigid supports. It is initially of negligible tension. Its Young’s modulus is 2 ×1011 N m-2, density is 9 g cm-3 and coefficient of thermal expansion is 12.5 × 10-6/°C. It is cooled through 1 °C. If the cross sectional area of the wire is 4 mm2, the speed of the transverse waves is
(a) 16.7 m s-1 (b) 20 m s-1 (c) 25 m s-1
(d) 60 m s-1
107. The speed of longitudinal wave is 120 times the speed of transverse wave in a taut brass wire. If the Young’s modulus of the wire is 100 GPa, then the tension (in N) in the wire of 1 mm2 cross section is
(a) 5 (b) 7 (c) 0.7 (d) 0.6
108. An open pipe is suddenly closed at one end, due to which the frequency of the 3rd harmonic of the closed pipe is made higher by 120 Hz, than the fundamental frequency of the open pipe. Then fundamental frequency of the open pipe is
(a) 240 Hz (b) 280 Hz (c) 260 Hz (d) 300 Hz
109. An air column in a pipe, closed at one end, will be in resonance with a vibrating tuning fork of frequency 264 Hz if the length of column is (speed of sound = 330 m s-1)
(a) 34.25 cm (b) 62.5 cm (c) 93.75 cm (d) 125 cm
110. A racing car moving towards a cliff sounds its horn. The driver notices that the sound reflected from the cliff has a pitch one octave higher than the actual horn’s sound (i.e.,frequency is doubled). If velocity of sound in air is 330 m s-1, then the velocity of the car is
(a) 165 2 m s-1 (b) 110 m s-1 (c) 165 m s-1 (d) 200 m s-1
Assertion–Reason Type Questions
Directions: Each question contains Statement-1 and Statement-2 and has the following choices (a), (b), (c) and (d), out of which ONLY ONE is correct.
(a) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1(b) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1(c) Statement-1 is True, Statement-2 is False(d) Statement-1 is False, Statement-2 is True
111. Statement 1 The average energy in half time period in SHM is same irrespective of between which two instants of time the chosen
interval of half time period lies. and
Statement 2 The average value of a sinusoidal function in equal time intervals are equal.
3.78 Oscillations and Waves
112. Statement 1 For a simple pendulum, the acceleration of the bob is zero at the mean position. and
Statement 2 Acceleration in SHM = -w2x, where x is displacement from mean position.
113. Statement 1 If two listeners are located at different points in between two sound sources of slightly differing frequencies on the
straight line joining the sources, then the beat frequency will be same for both, but intensity maxima will be different at different instants of time.
and
Statement 2 Beat frequency is always the difference in frequencies, whereas the time at which intensity maximum occurs is a func-
tion of position x.
Linked Comprehension Type Questions
Directions: This section contains a paragraph. Based upon the paragraph, 3 multiple choice questions have to be an-swered. Each question has 4 choices (a), (b), (c) and (d), out of which ONLY ONE is correct.
Passage IA body executing SHM has both potential and kinetic energies and the sum of these will always be constant in undamped harmonic oscillations.
A 3�
MB
k
k
(a)
3�
�
�
In the arrangement shown in Fig. (a), a light rod AB can rotate about end A and at B a weight M is attached. Two
springs each of spring constant k are attached at distances l3
from the two ends, where l is the total length of the rod. The
other ends of the springs are fixed firmly and the springs are initially in unstretched condition. Assume the whole system is in a horizontal plane, kept on a smooth surface.
114. The period of oscillation of the system is
(a) p9M
45k
(b) p9M
210k
(c) p5M
29k
(d) pM
65k
115. The spring constant such that one single spring connected at the centre produces an equivalent effect on the two springs is (Fig. (b))
(a) 5k9
(b) 10k
7 (c)
19k9
(d) 20k
9
Oscillations and Waves 3.79
116. If the rod has a mass m with the value of spring constant above, the period of oscillations
(a) pM m
65k+
(b) pM 3m
65k+
(c) ( )
p3 m 3M
25k+
(d) p9M m
65k
+
Multiple Correct Objective Type Questions
Directions: Each question in this section has four suggested answers of which ONE OR MORE answers will be correct.
117. A particle is executing SHM about the origin and at t = 0, its displacement is zero and its maximum amplitude is
5 cm. Consider the condition of PE 9KE 16
=
(a) The same condition is repeated 2 times in a cycle (b) The same condition is repeated 4 times in a cycle (c) When this condition happens, the displacement from the neutral position is 3 cm (d) The condition happens for the first time at approximately 0.1 T where T is the period
118. Two pendulums have ratio of their lengths equal to 1.21. The pendulum with smaller length has a period of 1 s. From the neutral point, they are set into small oscillations at the same instant and in the same direction.
(a) The larger length pendulum has a period of 1.21 s (b) They will be at the neutral point moving in the same direction after 10 s (c) They will be at the neutral point moving in the opposite direction after 5.5 s (d) They will be at the neutral point moving in the opposite direction after 16.5 s
119. A metal rod when dropped vertically on a hard floor with its length in the horizontal position, it vibrates with its two ends as vibration antinodes. The length of the rod is 0.6 m and its minimum frequency of 2 kHz.
(a) The velocity of sound in the rod is 2.4 km s-1
(b) When the rod is clamped at centre its fundamental resonance is at 2 kHz (c) When clamped at centre its 2nd overtone is 6 kHz (d) When clamped at one end its fundamental resonant frequency is 1 kHz
Matrix-Match Type Question
Directions: Match the elements of Column I to elements of Column II. There can be single or multiple matches.
120. A body of mass m, undergoing simple harmonic motion, has displacement at any instant t given by y = A sinp
w3
t2
+ ,
time period T, velocity v and acceleration a, kinetic energy Ek and potential energy Ep. Then match the quantities in column I to those in column II. (Consider potential energy at mean position as zero)
Column I Column II
(a) y = -A2
(p) max
| v |v
= 3
2
(b) t = T6
(q) max
aa
= 12
(c) t = T8
(r) k
p
EE
= 1
(d) y = -A2
(s) k
p
EE
= 3
3.80 Oscillations and Waves
addIt Ional praCtICe exerCISe
Subjective Questions
121. (i) A block is on the piston of an engine, which undergoes simple harmonic motion in the vertical plane with a period of 1 s. At what amplitude will the block and the piston separate?
(ii) If the piston has an amplitude of 1
m4.9
what is the maximum frequency for which the block and piston stay in
contact continuously? (Take g = 10 m s-2, p2 = 10)
122. A particle executes simple harmonic motion with frequency of 0.2 Hz about a point O. It passes through a point P with velocity of 0.1 m s-1 along OP (OP = 0.1 m). Calculate (tan-1(0.4p) = 0.9 radian)
(i) the time taken by the particle to cover OP (ii) the time that elapses, when it again crosses P.
123. One end of a spring is fixed to a wall and the other end is attached to a block of mass 1 kg. Stiffness of the spring is 147 N m-1 and block performs simple harmonic motion with amplitude 3 cm. When the block is at the left extreme position, an-other block of mass 2 kg, moving directly towards the 1 kg mass with a velocity of 40 cm s-1, collides and gets attached to the smaller block.
(i) Calculate angular frequency and amplitude of oscillations of the combined body. (ii) Assuming that the instant the collision takes place to be the initial starting time
(t = 0), and right hand direction to be positive x-direction, calculate initial phase of oscillations of the combined body.
Neglect friction.
124. A body of mass m = 100 g falls from a height 50 cm on the pan of mass M = 1 kg of a spring balance having a configuration as shown in figure, and gets stuck to the pan. The force constant of the spring is 110 N m-1. The compound body consisting of m and M start vibrating. Find the amplitude and time period of the oscillations
125. A solid cylinder of mass M = 1 kg is attached to a horizontal massless spring, such that it can roll without slipping along a horizontal surface as in figure. The force constant of the spring is 4 N m-1. If the system is released from rest at a position in which the spring is stretched by 0.30 m,
(i) Calculate the total kinetic energy of the cylinder as it passes through the equilibrium position.
(ii) Give the values of the individual kinetic energies. (iii) Under these conditions, find the type of motion executed by the centre of mass of the cylinder. (iv) What is the period and frequency of this motion?
126. A plane longitudinal wave of angular frequency 500 rad s-1 is travelling along a positive x-direction in a homogeneous medium of density r = 0.5 kg m-3. Intensity of the sound wave is 0.05 nW m-2. If velocity of sound in the medium is 200 m s-1, assuming initial phase of medium particles to be zero, at x = 0, write the equation of the wave.
127. A steel wire of length 1 m, mass 0.1 kg and uniform cross-sectional area is rigidly fixed at both ends. The temperature of the wire is lowered by 25°C. a is coefficient of linear expansion of steel. If transverse waves are set up by plucking the string in the middle, calculate the
(i) frequency of the fundamental mode of vibration. (ii) the frequency of vibration in first overtone.
1 kg 2 kgx
pan
Oscillations and Waves 3.81
(Young’s modulus of steel = 2 × 1011 N m-2
a = 16 × 10-6 K-1
density r = 8 × 103 kg m-3)
128. A stationary wave produced in a certain rod is given by the equation, y = 0.01 sin (12x) sin (200 t) where x, y are in m and t in s. Find, at x = 4.2 cm (Young’s modulus of the rod is 7.5 × 1010 N m-2) (i) the amplitude of oscillation, (ii) maximum particle velocity, and (iii) maximum tensile stress.
129. A metal wire of radius 1
mm2
is held on two knife edges 50 cm apart. The tension in the wire is due to a mass of
10 kg. The wire vibrating in first overtone and a vibrating tuning fork together produces 5 beats/second. The tension in the wire is reduced to 81% of its original value. When the two are now excited, 5 beats are again heard. Calculate
(i) the frequency of the tuning fork. (ii) the density of the material of the wire. (g = 10 m s-2)
130. A sonometer wire carrying a weight of 6.4 kg vibrating in its fundamental mode is in resonance with a vibrating tun-ing fork. The vibrating portion of the sonometer length is 10 cm and its mass is 1 g. The tuning fork is now moved away form the vibrating wire with a constant speed and an observer standing near sonometer wire hears 4 beats per second. Find the speed of the tuning fork if velocity of sound in air is 332 m s-1 at 0°C.
(Temperature is 30°C, g = 10 m s-2 ).
Straight Objective Type Questions
Directions: This section contains multiple choice questions. Each question has 4 choices (a), (b), (c) and (d), out of which ONLY ONE is correct.
131. An SHM is represented by the equation y = 12 sinp p2 t
10 4 +
where, y is in m and t in s. Then the frequency is
(a) 1 Hz (b) 10 Hz (c) 0.1 Hz (d) 0.01 Hz
132. A particle undergoing simultaneously two SHMs of same angular frequency along same line about same equilibrium
position and same amplitude A, but with a phase difference of p2
radian will have a resultant SHM of amplitude
(a) 2A (b) 2A (c) A (d) A2
133. A particle simultaneously undergoing two SHMs of same angular frequency along same line about the same equilib-
rium position and of same amplitude A but with a phase difference of p2
radian will have a resultant SHM of phase difference, with each SHM, of (in radian)
(a) zero (b) p8
(c) p4
(d) p
134. A particle simultaneously executes two SHMs along same line with same frequency but with ratio of amplitudes h and phase difference f. If the resultant amplitude is equal to one of the constituents’ amplitude, then cos f can be
(a) h (b) h1
− (c) h2
(d) h
12
−
3.82 Oscillations and Waves
135. A liquid in a U-tube performs SHM when the liquid level in one limb is depressed slightly and released. The time period, if total length of the liquid in the U-tube is l, is
(a) 2p lg
(b) 2p l2g
(c) 2p l2g
(d) p lg
136. A body oscillates with a time period of 3 s when connected to one spring and 4 s when connected to another. When the springs are connected in series and the body is attached, the period of oscillation is
(a) 2.4 s (b) 1 s (c) 5 s (d) 7 s
137. A block of mass 1 kg is hanging from the ceiling of a lift by a spring of constant 100 N m-1. If the lift suddenly starts moving up with acceleration 2 m s-2, the amplitude of the resultant SHM is
(a) 1 cm (b) 2 cm (c) 3 cm (d) 4 cm
138. A block m of mass 1 kg is connected to the back of a truck by a spring of constant 100 N m-1. It rests on the smooth surface of a truck. If the truck suddenly starts moving with speed 1 m s-1, the maximum acceleration of the block is (m s-2)
(a) 5 (b) 10 (c) 15 (d) 20
139. A body executing SHM along x-axis, with neutral point at origin takes 0.1 s to move from -12 cm to +9 cm. If the maximum amplitude of oscillation is 15 cm, its frequency of oscillation is (in Hz)
(a) 2.5 (b) 3.5 (c) 6 (d) 9.1
140. Two masses of 1 kg each are placed inside a frictionless tube as shown. They are connected by identical springs each of constant 10 N m-1. The time period of small oscillations of the system is
(Take p2 = 10)
k
k
mm
(a) 1 s (b) 2 s (c) 3 s (d) 4 s
141. A disc of mass m and radius R which can rotate about the axis XX’, is restrained by a spring of constant k as shown. The time period of oscillation of the disc is
(a) pm
2k
(b) 2pm2k
(c) pmk
(d) p m
3 k2
1 kg
2 m s–2
k = 100 N m–1
m
v
k
X’
X
m,R
k
Oscillations and Waves 3.83
142. A body of mass m is in the shape of L and is hung as shown. The time period of small oscillations is(approximately)
a
a
(a) pa
2g
(b) 2p2 ag
(c) 2p10 a3 g
(d) 2p 2ag
143. A uniform rod of density r oscillates as a pendulum about one end with time period T. It is taken inside a liquid of density 4r and it oscillates as an inverted pendulum. The new time period is
(a) T4
(b) T2
(c) T3
(d) T2
144. A rope of mass density m is hung as shown. A pulse starts from the point of suspension and travels as shown. If the time taken by the pulse to travel from one end of the rope to the other is T, the time the pulse of the same orientation passes A again is
(a) T (b) 2T (c) 3T (d) 4T
145. A rope of uniform mass density is rotated about one end with a constant angular velocity w. The plot of velocity of transverse waves vs the distance from the axis of rotation is
(a) parabola (b) straight line (c) hyperbola (d) ellipse
146. The equation of a disturbance is given by y = A(x)e-t(6x + 9t) where, A(x) = 2x
0A e− , The speed of the disturbance in (m s-1) is
(a) 1.5 (b) 3 (c) 6 (d) 9
147. A mass of 8 kg is hung by a wire of cross sectional area 1 mm2 inside a lift going up with acceleration 2 m s-2. If the velocity of the wave is 80 m s-1, the relative density of the material of the wire is (g = 10 m s-2)
(a) 5 (b) 10 (c) 15 (d) 20
148. If the ratio of velocities of longitudinal and transverse vibrations of a metal rod is 100, its strain is (a) 10-2 (b) 10-4 (c) 10-6 (d) 10-8
149. Consider the following expressions for displacement: (1) A sin2kx coswt (2) Asin2kx coswt (3) Aln(k2x2 - w2t2) (4) A ( )w2 2 2 2k x te −
The wave motions among them are (a) 1, 2 (b) 1, 3 (c) 3, 4 (d) 1, 2, 3
150. A wave y = ( )2
1
x 3t 3+ +m travels in a stretched string. The velocity of the point at the origin at t = 1 s is
(a) 1 m s-1 (b) 0.5 m s-1
(c) 0.125 m s-1 (d) 0 m s-1
151. The wave shown is on a string along the x axis, and has equation y = 10sin(5x - 3t) m. The points that are moving in the +y direction are
(a) 2, 3 (b) 1, 4 (c) 1, 2 (d) 3, 4
A >
y
x3 4
21
3.84 Oscillations and Waves
152. A sound wave of frequency 550 Hz has two points of displacement magnitude A separated by 20 cm. The maximum amplitude of a point in the wave is (v = 330 m s-1)
(a) A (b) 2A (c) 2A
3 (d) (b) or (c)
153.
One end of a stretched string is attached to a wall while the other is attached to a thinner string of same material. Two pulses are travelling with a speed of 20 cm s-1 towards the right at t = 0. Then
(a) The energy is purely potential at t = 0 (b) Constructive interference occurs when the pulses meet first time (c) Constructive interference occurs when the pulses meet second time (d) (b) and (c)
154. A rod of length l is clamped at one end and at a point 25
l from that end has a fundamental fc. If it is clamped only at
one end its fundamental frequency is f0. The ratio of the frequencies c
0
ff
is
(a) 3 (b) 5 (c) 7 (d) 9
155. A rod of length l is clamped at both ends. Now if it is also clamped at a point l2
5 from one end, the ratio of frequency
of first overtone to fundamental is (a) 7 (b) 5 (c) 3 (d) 2
156. BA
A wire is in the form as shown in figure and is under tension. One part (A) has twice the radius of the other. The ratio of the energies of the incident (from A) and transmitted waves in (B) is
(a) 3 : 2 (b) 2 : 1 (c) 9 : 8 (d) 3 : 2
157. In a mixture of gases, the average number of degrees of freedom per molecule is 6. The root mean square speed of the gas molecules is 2c. Then, the velocity of sound in the gas is
(a) c2
(b) 3c4
(c) 2c3
(d) 4c3
158. The ratio of pressure and displacement amplitude for a particular sound wave is 9 × 105 Pa m-1 while for another it is 16 × 105 Pa m-1. Their frequencies are in the ratio
(a) 9 : 16 (b) 16 : 9 (c) 3 : 4 (d) 4 : 3
159. The energy density of a sound wave of pressure amplitude p in a medium of bulk modulus B is given by
(a) 21 p
2 Bk (b)
p
2
2
1 p2 B
(c) p
2
2
1 p4 B
(d) 21 p
.2 B
160. Consider the following media for wave motion (1) Ideal fluid (2) Viscous fluid (3) Gas (4) Solid
1 m 1 m 2 m
Thinner sectionB A
Oscillations and Waves 3.85
Transverse waves can travel in (a) 2, 4 (b) 1, 2, 4 (c) 1, 2, 3 (d) 4
161. An open pipe has fundamental frequency f1. When it is closed it has fundamental frequency f2. If 2
1
ff
is 9
17, the ratio
of DL
for the pipe is
(a) 5
24 (b)
116
(c) 3
17 (d)
719
162. A tuning fork of frequency 110 Hz creates standing waves in a closed pipe. The minimum distance of a point with
pressure amplitude is 0p2
(where p0 is the maximum pressure amplitude) from the open end is (speed of sound
= 330 m s-1) (a) 0.25 m (b) 0.5 m (c) 1 m (d) 3 m
163. An open tube is in resonance with a tuning fork in the third harmonic. If it initially contains some dust, the number of dust piles formed is
(a) 1 (b) 2 (c) 3 (d) 4
164. A closed tube with a small opening at one end has fundamental frequency f. If it is dipped vertically in water so that half its length is immersed, the new fundamental frequency is
(a) f2
(b) f
(c) 2f (d) 4f
165. A sensor which can measure pressure is used to monitor a machine which produces sound of frequency 160 Hz. The distance of the sensor from the wall for maximum reading of the sensor should be about (speed of sound = 320 m s-1)
(a) 0 m (b) 0.5 m (c) 1 m (d) (a) or (c)
166. A closed pipe of length 87 cm and radius 5 cm is vibrating in the first overtone. If the maximum displacement is 9 mm, the displacement at x = 58 cm from the closed end is
(a) 0 (b) 9 mm (c) 0.3p mm (d) 0.6p mm
167. Two tuning forks A and B produce 4 beats when sounded together. Tuning fork A is now filed so that 6 beats are heard. If the frequency of tuning fork B is 400 Hz, the percentage change in frequency of A is
(a) 0.5% (b) 1% (c) 2.5% (d) (a) or (c)
168. Two sources of sound having the same frequency are moving as shown. The speed of the observer so that no beats are heard is(v = 315 m s-1)
(a) 5 m s-1 (b) 15 m s-1 (c) 10.5 m s-1 (d) 20 m s-1
169. A source and an observer are stationary and a wind blows from source to observer. Then (a) wavelength increases, frequency is unchanged (b) wavelength and frequency both increase (c) wavelength unchanged, frequency increases (d) wavelength and frequency both unchanged
170. A car carrying a siren of frequency 1024 Hz is moving with speed 108 kmph towards a wall moving towards it at 36 kmph. The highest frequency heard by the person shown is (v = 330 m s-1) approximately
(a) 1000 Hz (b) 1100 Hz (c) 1200 Hz (d) 1300 Hz
f0
v1
v2
3.86 Oscillations and Waves
Assertion–Reason Type Questions
Directions: Each question contains Statement-1 and Statement-2 and has the following choices (a), (b), (c) and (d), out of which ONLY ONE is correct.
(a) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1(b) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1(c) Statement-1 is True, Statement-2 is False(d) Statement-1 is False, Statement-2 is True
171. Statement 1 If a particle is simultaneously under two SHMs of equal amplitude, equal frequency, with phase difference
p2
, but in mutually perpendicular directions, its resultant motion is SHM.
and
Statement 2 sin2q + cos2 q = 1
172. Statement 1
� → ∞
Earth
Period of the infinitely long pendulum shown is ∞ and Statement 2
For a simple pendulum, period T = 2p lg
; (l → length of the pendulum)
173. Statement 1 In the vertical oscillations of a vertically hanging spring mass system, if the acceleration of the mass at the maximum
downward position is g m s-2, then the maximum amplitude of oscillations is mgk
[g → acceleration due to gravity] and
Statement 2 The neutral point of oscillations is at
mgk
below the natural length l0 of the spring.
174. Statement 1 If a long straight, smooth tunnel is dug through the earth which opens at both ends, the time required for a body
dropped at one end of the tunnel to reach the other end is independent of the length of the tunnel. and Statement 2
The body executes SHM of period T = 2p eRg
; Re radius of earth
175. Statement 1 When the tension in a vibrating string is increased from T to 4T the wavelength of its first overtone reduces to half
the original value: and
Statement 2 When the tension increases four times the frequency of the first overtone doubles
176. Statement 1 In the system shown in the figure, the wheel W rolling on a rough surface without
slipping and attached to the end of the spring k, is in SHM. At the neutral point of oscillations rotational K.E of the wheel is equal to
13
rd of the maximum spring energy.
KW
Oscillations and Waves 3.87
and
Statement 2 Total energy is constant and rotational KE at any instant is a third of the spring energy at that instant
177. Statement 1 The maximum K.E, a body dropped into any deep tunnel dug into earth, can acquire is
12
m Re g(m → mass of body; Re → radius of earth)
and
Statement 2
The motion of the body dropped into a deep tunnel is part of an SHM of w = e
gR
178. Statement 1 Consider two bodies A and B both starting from initial position P at the end of
a radius of the circle (i.e., OP) aligned along x-axis. A executes SHM given by x = rcoswt. B moves on the periphery of the circle with uniform angular velocity w. Then the relative distance between A and B executes SHM.
and
Statement 2 The projection of a uniform circular motion on any of its diameters is an SHM.
179. Statement 1 When two waves of same frequency generated by two independent sources arrive
at a point with same intensity I the resultant intensity cannot exceed 2I and
Statement 2 The total energy from two sources cannot be boosted by interference.
180. Statement 1 Intensity of sound wave changes when source moves while the listener remains stationary, but does not change if
listener moves and source remains stationary. and
Statement 2 Wavelength changes if source moves but does not change if source is stationary.
Linked Comprehension Type Questions
Directions: This section contains 3 paragraphs. Based upon the paragraph, 3 multiple choice questions have to be an-swered. Each question has 4 choices (a), (b), (c) and (d), out of which ONLY ONE is correct.
Passage IWave motion can be described as the progression of vibration in a medium without the actual transport of the vibrating particles of the medium. The progressive wave along the positive x-direction on a string can be written as: y = a sin w
xt
v −
where v represents the wave velocity and y the displacement of the particle at position x at time t. From mechanical
considerations it can be shown that the wave velocity in a string = mT , where T is the stretching force and m is the mass
per unit length.
y
Or
B
A Px
3.88 Oscillations and Waves
181. A uniform rope of length 12 m and mass 6 kg is hung vertically from a rigid support at end ‘A’ and at the free end ‘B’ a 2 kg mass is attached. What will be the time required for a disturbance on the rope to travel from one end to the other?
(a) 2 s (b) 2.5 s (c) 3 s (d) 4 s
182. In the above case, if the disturbance travelling is a sinusoidal wave, then the ratio of wavelengths at ends A and B, l lA B: will be:
(a) 2 : 1 (b) 2 : 1 (c) 1 : 2 (d) 6 : 1
183. In the same situation as above, if mass of the rope and mass of the weight hanging is m, if l is the length of the rope, then the time of travel for a sinusoidal wave from one end to the other is:
Passage II A source of sound of frequency f0 is moving in a circular path of radius R in the
clockwise sense with constant angular velocity w. Three points A, B, C are in the
plane of the body such that OB = 5
R4
, BC = R, OA = 4R5
. AP is normal to OB and BP is tangent to the circle
184. If f1 and f2 are the minimum frequencies for B and C respectively, (a) f1 < f2 (b) f1 = f2
(c) f1 > f2 (d) depends on R
185. If f1 and f2 are the minimum frequencies for A and B respectively, then (a) f1 < f2 (b) f1 = f2
(c) f1 > f2 (d) depends on R
186. If t1 and t2 are the times between the maximum and minimum frequencies for A and B respectively, then 1
2
tt
is
(a) > 1 (b) < 1 (c) = 1 (d) depends on R
Passage IIIDoppler Effect is not confined to sound. It occurs in electromagnetic waves also, including radar and light waves. It depends
on only the relative velocity between source and observer. The apparent frequency =
22
v1 c fv1 c
−
−. For small
vc
(vehicles), f ’
≈ f
v1c
−
In the case of radar, the approaching plane acts as a mirror reflecting radar radiation and hence mirror image of the source approach the receiver with velocity 2 vp where vp is the velocity of the plane. Hence if fr is the actual frequency of
Radar transmitter, the received frequency will be fr’ = fr 1
pr
p
2vc cf 1
c 2v c c
−
= − − = fr
pv1 2
c
+
Hence D f = pr
2vf
cNow, consider a plane approaching an obstacle, like a hill ahead of it. The plane’s radar is sending radar signals at fre-
quency 1 GHz. The reflected signal from the obstacle is received by the plane’s receiver antenna and amplified and mixed with its radar transmitter signal in equal amplitudes. Then,
O
y
C
BAx
P
Oscillations and Waves 3.89
187. The change in frequency of the received signal (given: the velocity of the plane is 300 m s-1) is (a) 1 kHz (b) 500 Hz (c) 2 kHz (d) 700 Hz
188. When the original radar wave and the reflected wave received by the plane’s antenna are superimposed in equal am-plitudes the resultant will have signal level zero value at intervals
(a) 1 ms (b) 0.5 ms (c) 0.25 ms (d) 2 ms
189. In the above case the signal level of the resultant wave will be maximum at intervals of (a) 1 ms (b) 0.5 ms (c) 0.25 ms (d) 2 ms
Multiple Correct Objective Type Questions
Directions: Each question in this section has four suggested answers of which ONE OR MORE answers will be correct.
190. A body A (m = 1 kg) is attached to the top of a vertically fixed spring of spring constant 1000 N m-1 and then the mass is released at the initial position P with the spring being initially in its natural length l0. When A comes back to original position P another body B of equal mass falls on A with a velocity v = 0.2 60 m s-1 at t = t0 and the collision is perfectly inelastic. Then,
(a) B will move in SHM for a duration of p
15 52
s.
(b) B will detach from A after a duration of p
5 5s after t0
(c) Maximum amplitude of oscillations of the combined mass is 4 × 10-2 m. (d) The neutral point of SHM of the combined mass is 2 cm below P.
191. 1 kgP
Smooth floor
0.5 m x
x2 kg
O
Q
Masses P = 1 kg and Q = 2 kg are attached to a spring of length 0.5 m. P is held aligned at origin as shown and Q is displaced along x-axis by x and both P and Q are released. The floor on which they are kept is smooth. P reaches maximum speed of 2 m s-1 when it is
0.2m
3 from origin during the subsequent SHM of the two bodies. Then,
(a) x = 10 cm (b) vmax of Q = 2 m s-1
(c) k of spring is 600 N m-1
(d) CoM of the oscillating bodies is at 0.4 m from origin
192. Consider the case of the uniform cylindrical rod floating with its length vertical on a liquid. If the rod is slightly depressed it executes SHM. Let d be the depth to which the rod is immersed and l be the length of the rod and m the mass and rs the specific gravity of the material of the rod with respect to the liquid: Consider two cases, case a : rods having same length, case b: rods having same mass (K ↔ constant of proportionality)
(a) T = Krs in case a (b) T = K rs in case a (c) T = K d in case b (d) T = ldK in case b
193. A standing wave on a string of length of 2.1 m has 7 nodes and the tension on the string is 49 N. If mass for unit length of the string is 0.01 kg m-1 then
(a) The frequency of oscillations is 100 Hz. (b) The fundamental frequency of the string is 20 Hz
B
A P
�0
3.90 Oscillations and Waves
(c) If the amplitude of the individual wave is 1 cm the total energy of oscillation is 0.021 p2 J (d) If the same wave is allowed to progress on the string with maximum amplitude of 1 cm the power transmitted
along the string is 1.4 p2 watt
194. Two travelling waves, y = A cos200pp
xt
1400 −
and y = A cos p px
280 t5
− − are progressing together in a
medium. Intensity of each single wave is I. Then at a given point in the medium: (a) In each second, the number of times the intensity is 4I is 40 (b) In each second the oscillating particle crosses the neutral point about 80 times (c) In each second the oscillating particle crosses the neutral point about 280 times
(d) During rd1
3 of each second the intensity is less than I.
195. Helium in an open pipe at T1 = 21°C has fundamental f0 equal to the fundamental of a closed pipe of same length filled with hydrogen at T2°C. Now both pipes are filled with oxygen at T2°C
(a) T2 = 417°C (b) T2 = 427°C (c) First overtone with oxygen in closed pipe = f0
(d) First overtone with oxygen in open pipe = f0
196. Y1
A wave travelling on a thin wire represented by Y1 = 0.01sinx
200t10
− (units in metre and second) gets reflected
at the junction where the thin wire is joined to a thick wire. The intensity of the reflected wave is 81% of the intensity of the incident wave
(a) Velocity of the wave in the thick wire is 73
m s-1
(b) The reflected wave is represented by Y2 = 0.009sinx
200t70
+
(c) The standing wave is represented by Y’ = -0.018sinx
70cos200t
(d) The residual traveling wave is Y3 = 0.001sinx
200t70
−
197. A source of frequency f = 144 Hz is rotating about center O, in circle of radius R =
p2
m at 2 cycles s-1. At a distance OQ = 2R = p m another source of same frequency (144 Hz) is kept and the listener at Q listens to both the sounds (Assume all readings are noted electronically. Velocity of sound = 340 m s-1, and approximate p2 = 10)
(a) Between two zero beats the maximum beats heard can be 8 s-1
(b) Between two zero beats the maximum beats heard can be 9 s-1
(c) The minimum period possible between maximum beats and zero beats is 1
12 s
(d) If the listener is shifted by 1 m to Q’ along the same line OQ the maximum beats heard is 10 s-1
Helium, 21°C
Hydrogen, T2°C
�
+
S
ω
P Q Q’O
2π
Oscillations and Waves 3.91
Matrix-Match Type Questions
Directions: Match the elements of Column I to elements of Column II. There can be single or multiple matches.
198. Four oscillating systems are shown in column I. The time period of natural oscillation on the surface of earth is T0. The systems are (i) accelerated up with uniform acceleration(ii) accelerated down with uniform acceleration (iii) Taken to space. Match the system with all possible time periods in the three cases.
Column I Column II
(a)
�
m
Point mass
(p) T = T0
(b)
Rigid rod
(q) T > T0 finite
(c)
Mass suspended using spring
m
k
(r) T < T0
(d)
Torsional pendulum
T = 2πKΙ (s) T = ∞
199. Match column I and II
Column I Column II (a) Beats (p) slowly varying amplitude (b) Resonanced in a closed (q) nearly equal frequencies involved organ pipe (c) Musical note (r) harmonics involved (d) Echo (s) reflection from a rigid surface
200. A closed pipe of length L and diameter D ( )L10
= has fundamental frequency f0. The pipes in column I have funda-
mental frequency f. Match them to possible values of 0
ff in column II
3.92 Oscillations and Waves
Column I Column II
(a) D
L
(p) 0
ff integer
(b)
D
2L
(q) 0
ff < 1
(c)
2L
2D
(r) 0
ff = 1
(d) D
2L
(s) 0
ff > 1
Oscillations and Waves 3.93
anSwerS KeyS
Topic Grip
1. T = 2pml
2 g
2. k
9m
3. (i) x = 0.2 - 0.04 cos 10t
(ii) x = 0.2 + 0.210
2
sin 10 t3
+
p1 6sin 0.4
4−
− 4. (i) SHM (ii) T = 2p
( )3 R r2g−
(iii) ( )q0
2g R rr 3
−
5. (i) T = 2p ( )al
g cos 2
(ii) (a) : T = 2p lg
; (b) : T = ∞
6. y = 2 × 10-4sin px
300 t300
− m
or y = 2 × 10-4sin(300pt - px) m 7. f0 = 6250 Hz; f1 = 18750 Hz 8. f = 400 Hz 9. 18.4°C increase
At an arbitrary position let center of bar be x2 + l
away
from point of contact A. Since the bar is in rotational equi-librium, taking moments about A,
mg × 2x R .2 + = l
l ⇒ R2 = mg
x2 + l
l
⇒ f2 = mg
x2
m + l
l
Since R1= mg-R2
f1 = mR1 =mg
x2
m − l
l
Let acceleration at that position be a.
Then a = 1 2f fm−
= -2 g
.xml
Hence the motion is SHM, with w = 2 gml
\ T = 22
2 gp
pw m
=l
2. At equilibrium let spring be extended by x0.Force equations are T = 2 mg — (1) 2T + mgsinq = kx0
⇒ 4 mg + mgsinq = kx0- — (2)
At an arbitrary position during motion let mass 2m be displaced by x from equilibrium position and let its ac-celeration be a.
2mg - T’ = 2ma — (3)Then spring’s elongation now is
0x
x2
+ , acceleration of mass m is a2
.
\ Force equation is
2T’ + mgsinq - kx0 - kx a
m2 2
=
— (4)
(3) × 2 + (4)
⇒ 4mg + mgsinq - kx0 - kx ma
92 2
= ⇒ using (2), ma
92
= kx2−
⇒ a = k
.x9m−
(where ,a is acceleration of 2m and x its displacement from equilibrium position )
\ w =k
9m.
3.
(i) w = 1
k 200m 2
= = 10 rad s–1
A = 4 cm = 0.04 m. Taking origin at O’ ; x' = A sin( wt + f ) ⇒ x' = 0.04 sin (10t + f ) At
t = 0, x' = - 0.04 ⇒ f =2p−
radian
⇒ x' = -0.04 cos 10t⇒ (with respect to O) x = 0.20 - 0.04 cos 10t
(x in metre).
HIntS and explanatIonS
3.96 Oscillations and Waves
4. (i)
R r
N
θ
a
mg mg cos θ
mg sin θ f
At arbitrary q, N = mg cosq If a is acceleration of CoM ma = mg sinq - f — (1) For condition of rolling with no slip,
f.r = 2mr
2a and
ar
a =
ma
f2
⇒ = \ (1) ⇒
a = 23
g sin q
for small q, sinq ≈ q and q = x
R r− where x is
displacement of CoM from equilibrium position.
\ a = 2g
.x3(R r)−
.
Hence SHM.
(ii) w = ( )2g
3 R r−
\ Time period = 2p 3(R r)2g−
(iii) Maximum angular velocity
= Maximum speed of CoM
r
( )0 R rA
.r r
qww
−= =
0 2g(R r)r 3q −
= , where q0 = amplitude of angu-
lar displacement
5. Let the body be displaced by an angle q as shown (anticlockwise)
(ii) At t =340p
s, x' = -0.04 cos 30.02 2
4p
= m
v1 = w ( )22A x '−
= 10 ( ) ( )2 20.04 0.02 2− ×
= 0.2 2 m s-1
\ Velocity after inelastic collision
= 2 0.2 2 1 2
3× − ×
= -0.2 2 m s–1
New w’ = 1 2
km m+
= 200
3= 10
23
.rad s-1
Calculation of new Amplitude A’ : Initial energy (just after collision)
= ( ) ( )2 21 13 0.2 2 200. 0.02 2
2 2× × + ×
= (K.E) + (P.E. of spring) = 0.12 + 0.08 = 0.20 J
This must equal 12
(m1 + m2) w2A2
0.2 = 2
21 23 10 A'
2 3
× ×
⇒ A’ =
0.210
m
\ Equation of SHM with respect to O’ as origin is x'
= 0.210
sin 2
10 t '3
ϕ
+
At t =340p
s, x' = 0.02 2 m
\ 0.02 2 = 0.2 2sin 10 t "10 3
φ
+
⇒
f'' = sin–1 60.4
4p−
\ Equation with respect to O as origin is x = 0.2 +
10.2 2 6sin 10 t sin 0.4
10 3 4p−
+ −
Oscillations and Waves 3.97
⇒ from (1) v = 300 m s-1. Phase difference 18° for Dx = 0.1 m
⇒ 18° = 360° × 0.1l
⇒ l = 2 m
\ w = 2pv 300
22
pl= × = 300p rad s-1
\ Putting in wA = 0.06p,
A = 0.06 0.06
300p p
w p= = 2 × 10-4 m.
\ Equation is y = 2 × 10-4 sin x300 t
300p
− m
or, y = 2 × 10-4 sin(300p t – p x) m
7. Velocity of the longitudinal wave = Yr
= 11
3
1.75 107 10
××
= 5 × 103 m s-1
The clamped points will be nodes and the free ends antinodes. Let the number of segments between the clamped region be n2 and between the clamped region and open end be n1 on each side.
⇒ 2.4n2 – 3.2n1 = 1.6 ⇒ 3n2 – 4n1 = 2⇒ n2 > n1, and n2 is even.\ Put n2 = 2 = ⇒ n1 = 1 Using (2), l = 0.8 m
\ Natural frequency is vl
= 50000.8
= 6250 Hz.
(i)
equilibrium position
�
θ
�
α 2
displaced position 2α
Restoring torque (clockwise)
= mglsin2a
q + - mglsin
2a
q −
= mgl. 2cos 2a sin q
= 2 mgl cos 2a . q (for small q)
\ Angular acceleration = 2
2mg cos 22m
a ql
l
(Q a = tI
)
= g
cos .2a
ql
Angular acceleration proportional to q, hence SHM
\ w =g cos 2
a
l
⇒ \T = 2g cos 2
pa
l
(ii) For a = 0; T = 2g
pl (known expression)
For a = 180° T = infinity ⇒ no oscillation (Qneutral equilib-
rium)
6. Let equation be y = A sin xt
vw −
Shear strain = dy A x
cos tdx v v
ww − = −
Maximum shear strain = Avw
= 2p × 10-4 — (1)
Energy density, U = 12r w2A2 = 0.45p2
⇒ w2A2 = 20.9p
r=
20.9250p
⇒ wA = 0.06p
3.98 Oscillations and Waves
Since beats decrease first to zero and then increase to 4, the frequency of the pipe increases. This can hap-pen only if the temperature increases. Since frequency
∝ velocity ∝ ( )absolute temperature K .
Let t be the final temperature in °C
t tC C259 4 259 4 263
2 2− = ⇒ = + =
l l — (2)
Dividing (2) by (1) t
16
C 263C 255
=
or 273 t 273 t 273 t 263
273 16 289 17 255+ + +
= = =+
17 263 263273 t 17.53
255 15×
∴ + = = =
\ 273 + t = 2263
15
= 307.4
\ t = 307.4 - 273 = 34.4°CDt = 34.4 - 16 = 18.4 °C
10.
Direct wave : frequency received:
n1 = r)
s
(v un
(v u )++
Wavelength received: Motion of the receiver affects the velocity of sound (which increases in this case)
l1 = sr
1
v u(v u )n n
++=
Reflected wave:frequency received by reflector = n2 = f
s
n(v u )(v u )
−−
\ frequency received by receiver
= n3 = r2
f
(v u )n
(v u )+
×+
= f r
s f
(v u ) (v u )n
(v u ) (v u )− +
×− +
Wavelength received
l3 = r
3
v un+
= s f
f )
(v u )(v u )n(v u− +
− 11. Since the scale is 10 cm long and the reading of the
spring balance is from 0 to 200 N.
spring constant = 1F 200k 2000 N m
x 0.10D−= = =
Period T = m
2k
p
Frequency = 1 1 1 k
Period T 2 mp= =
For next higher frequency, Try n2 = 4 ⇒ n1 = 2.5, not possible Try n2 = 6 ⇒ n1 = 4
\ l = 1.66
m and n = vl
= 50001.6
× 6
= 18750 Hz
8.A B
C
length l1 = 0.6 m l2 = 0.5 mdensity r1 = 4000 kg m-3 r2 = 8000 kgm-3
area a1 = 10-5 m2 a2 = 2 × 10-5 m2
Tension T = 1024 N (same for both)no. of loops n1 n2
velocity v1 v2
frequency f (same for both)wavelength l1 l2
Then,
v1 = 1 1
Tar
= 3 5
10244 10 10−× ×
= 160 m s-1
v2 = 2 2
Tar
= 3 5
10248 10 2 10−× × ×
= 80 m s-1
l1 = 1
1 1
2 1.2n n
=l
, l2 = 2
2
2nl
= 2
1n
f = 1
1
vl
= 1160n1.2
, f = 2
2
vl
= 80 n2
⇒ 1160n1.2
= 80 n2 ⇒ 1
2
nn
= 0.6 = 6
10 =
35
For minimum f, n1 = 3, n2 = 5
\ f = 1160n1.2
= 160 3
1.2×
= 400 Hz
9. 16Cn
2=
lwhere l = length of the pipe
and C16 = speed of sound at 16°C.
16 16C C259 4 259
2 2− = ⇒ =
l l - 4
= 255 — (1)
Oscillations and Waves 3.99
19. Y’ = Y1 + Y2 = Asin(wt + kx) + Asin(wt - kx)= 2AsinwtcoskxAt origin, x = 0, Y’ is maximum intensityY’ first minimum at coskx = 0 ⇒ kx = 2
p
2.x
2p pl
= ⇒ x = 4l
20. Apparent frequency will be different from actual frequency only when the relative velocity component along the line joining them is non-zero.
21.
A C
B N P
M 45° 45° l/2 O
By symmetry the COM should be along OC. Let it be at P and OP = X. COM of OB and OA will together be, again by symmetrical consideration, at M where
OM = cos452 2 2
=l l
If M is the total mass,
\ MX = 2m m22 2
+l l
[where M = 3m]
\ X = m 1 1 2 23m 2 32 2 2
++ =
l l
= ( )2 16
+l
22. T = 02MgdI
p
I about O = 2
2m3 m
3× =
ll
Substitute ‘d’ from above (i.e., distance between O and COM
T = ( ) ( ) ( )
2m .62 2
3g 2 13m g. 2 16
p p=++
l ll
= ( )2 2 2 1g
p −l
23. d = ( )45cos 22l
= 1 cos45
2 2+ l
= 11
22 2
+l
We have 1 2000
22 mp
=
⇒ 2000
m 12.5 kg16 10
= =×
12. 2 2 2max
1 1mv m A
2 2w= = 0.032
⇒ w2 = ( )2
0.032 20.1 0.2
××
⇒ w = 4 rad s-1 ⇒ y = 0.2 sin 4t6p +
m
13.
�0
a = g
maximum
A neutral
At minimum length l0; it is at maximum displacement from neutral position: Hence maximum acceleration is g. If A is amplitude at neutral position kA = mg.
At maximum extension 2A = 2mg
k spring force =
2Ak = 2mg
14. 1 Tf
2 m=
l;
T = rVg
f ' = 1 T'2 ml
; T’ = rVg - fb
weight of the object in air:rV = 400 + 420 × 1.2 = 904 gIts weight when dipped in water = 904 - 420 = 484 g
f ' 484f 904
=
⇒ f ' = 484
300904
≈ 220 Hz
15. T
0.17s T 0.17 44= → = × = 0.68 s
1 1f 1.47Hz
T 0.68= = =
16. Only difference is mean positions are different
17. Part of the wave will be reflected at the interface
18. Because they travel in same direction
3.100 Oscillations and Waves
\ At any displacement ‘y’ from neutral position
Total energy = E = 2 21 m 1m v ky
2 3 2 + +
= constant at any timeDifferentiate with respect to time:
dy1 m dv 1m 2v. k2y
2 3 dt 2 dt + +
= 0
⇒ dydt
= v, dvdt
= a
\ a = ky
mm3
− +
⇒ SHM ⇒
T’ = 2p4m m 2 2
2 . T3k k 3 3
p= =
(a) is correct
2
2
mvKE spring 16Total KE 44mv
6= = 21 4
TKE mv2 3
= Q
(c) is correct
28.
4g
g Case d
T’ = 2pg 'l Case a ⇒ ga’ = g +
g4
= 5g4
⇒ Ta’ = 2p 2g 5l
Case b and d ⇒ g’b, d = 2
2 gg
16+ =
g17
4
Case c : g’c = g - g4
= 3g4
< g
\ Tc’ > T; Tb’ = Td’ = 2p .417 gl
= 42 .
g 17p
l = ( )1
4
2T
17
\ Tc’ > T > Tb’, Td’ > Ta’
→ (a), (b) and (d) are correct
I = 2m
3l
.2 = 22m
3l
T = 2p2mgdI = 2p
2
12
2 m3
2 12mg
2 2 2
+
l
l
= 2p 2. 4 2 2
3 g−
l
24. Since the intensity is 0.81: Ar2 = 0.81A2
Ar = 0.9A = 0.9 × 10 = 9
Since there should be a phase reversal at reflection to an obstacle (i.e., hard substance, no transmission); Ar = -A\ y = -9sin(wt - kx)
25. {9sin(wt + kx) + 1sin(wt + kx)} -9sin(wt - kx)[i.e., splitting the negative direction wave]We get the standing wave and a traveling wave in the negative x direction which is not possible because of obstacle at x = 0.Hence we can consider as[10sin(wt + kx) - {10sin(wt - kx) -1 sin(wt - kx)}20coswt sinkx + sin(wt-kx)\ standing wave = 20coswt sinkx
26. As given in previous case
27. dx v
x
Consider a small elemental length dx at a distance x from fixed end.
dm = m
dxl
; If mass m is moving with velocity v, the
velocity of this element is v’ = x
.vl
⇒ Hence its kinetic energy
d(KE) = 2 2
22 3
1 m x 1 mvdx. v
2 2=
l l lx2dx
Total KE of spring = 2
30
1 mv2∫
l
lx2dx
= 2 3 2
3
1 mv mv2 3 6
=l
l
Oscillations and Waves 3.101
→ (q) and (s) are correct for (a)
Equation ( )2
2
d xM m kx
dt+ = − ⇒
( )2
22
d x k ka x
M mdt M mw = = − ⇒ = + +
⇒ T = M m
2k
p+
→ (r) correct for (b)(ii) At the maximum displacement,
E2 = E1 = 21 2 1
1kA A A
2⇒ =
→ (p) correct for (c)
But T = M m
2k
p+
→ (r) correct for (d)
IIT Assignment Exercise
31. The KE and PE are sinusoidal with the same amplitude ⇒ equal mean value, provided PE at mean position is zero
32. w1 = 1km
; w2 = 2km
A1 w1 = A2 w2
1 2 2
2 1 1
A kA k
ww
= =
33. PE = K0 sin2 wt ⇒ PEmax = K0
34. a0 = w2 A; v0 = w . A2
0
0
va
= A
35. Maximum acceleration ‘a’ = w2A = (0.50)2 × 2= 0.50 m s-2
36. Frequency of oscillations of loaded spring
n = 1 k
2 mp×
⇒ n ∝ 1m
⇒ ' m
4mnn
=
⇒ n' = 2n
29. When source moves towards listener, the frequency heard by listener is
vf
vv100
− =
100f
99
\ Beats frequency = 100
f f99
− = 8
⇒ f100 99
99−
= 8, f = 792 Hz
→ (a) is correctwhen source moves away; frequency heard is
fv 100
fv 101v
100
=+
⇒ Beats frequency
= f - f100101
= f101 100 792
101 101−
= < 8
→ (c) is correct and (b) is wrong.when both source and listener move towards each other
f ’ = f
vv 101100 fv 99v
100
+=
− ⇒ beats = f
10199
- f
beats = 792 101 99
99−
= 16 → beyond the limit of human audibility(d) is correct
30. (i) At the equilibrium point, energy conservation leads to
2 21 1 1 1 1
1 1 kE Mv kA v A
2 2 M= = ⇒ =
Momentum conservation leads to
( ) ( )1 2 2 1M
Mv 0 M m v v vM m
+ = + ⇒ =+
Just after collision,
( ) ( ) ( )2
2 22 2 1 1
1 1 M ME M m v v E
2 2 M m M m= + = =
+ +
Since E2 = 22
1kA
2
( ) ( )2 2
2 1 11 M M 1
kA E kA2 2M m M m
= =+ +
⇒ 2 12 M M
A Ek M m M m
= = + + A1
3.102 Oscillations and Waves
45. y = ( )2 23 4+
× 2 2 2 2
3 4sin314t cos314t
3 4 3 4
+ + +
= 5[cosfsin314t + sinfcos314t]y = 5. sin (314t + f)
where cos f = 35
(f = 53°)
46. a = - w2 y
47. x = a sin wt
y = b.sin t2p
w + = b cos w t
22
2 2
yx1
a b+ =
If a = b then, circle.
48. y = A sin(wt + q); A = 1 cm
w = 4p
rad s-1
Max. acc = w2A = 2
4p
× 1 = 0.62 cm s-2
49. For SHM starting from extreme position the equation is
x = Acoswt = Acos2Tp
t
2 = 4 cos26p
t
cost
3p
=
24
= 12
= cos3p
\ t = 1 sAverage velocity =
21
= 2 cm s-1
x
4
t 2
time
50. 11
1
k 2 2m T 3
p pw= = =
2
2
k 2 2m T 4
p p= =
37. K.Emax = 12
mw2A2
(wA)2 = 2 16 1005.12 16×
=
w = 2.5 2
10 T s0.25 5
p pw
= ⇒ = =
Equation (M + m) 2
2
d xdt
= -kx
w = 2.5 2
10 T s0.25 5
p pw
= ⇒ = =
38. w = 100p2Tp
= 100p
T = 2
100 = 0.02 s
39. F = m a ; SHM a = -w2 x ⇒ m a = -mw2
x ⇒
40. K.E = 34
total energy
i.e., 12
m w2(a2-y2) = 34
× 12
mw2a2
y2 = 2a
4, y =
a2
41. v = 2 2A yw −
8 = 2 2A 6w −
6 = 2 2A 8w −82 - 62 = w2(82 - 62)
w = 1 ⇒ T = 2pw
= 2p s
42. v = a w . cos(wt - f)
t = ϕw
⇒ v = aw cos0 = a w
43. w = 2 p f= 2p × 5 = 10 p rad s-1
F = m w2y = 2 × (10p)2 × 2 × 10- 2 = 4 p2
= 39.5 N.
44. v = w 2 2A y−
A w = 100 cm s-1 ; A = 10 cm ⇒ w = 10 rad s-1
\ 50 = 10 2100 y−
y = 75 = 5 3 cm
Oscillations and Waves 3.103
by the sphere, the centre of gravity returns to centre when the hollow is totally empty
57. y = A sin2 (kx - wt)
= 1 cos2 (kx t)
A2
w− −
( )A Acos 2kx 2 t
2 2w= − −
Amplitude = A2
Frequency f = ' 2
2 2w w wp p p
= =
58. Time of fall 2d't '
g=
Frequency = No.of waves n
time t= =
n2dg
59. There is a phase difference of 2p
between sine and cosine functions. Hence the phase difference between
the waves is 2p
radian
60. Equation for a travelling harmonic wavey = A sin (wt ± kx + f)
61. I ∝ n2 A22 21
v A2
I r w =
12 2
1 12
222
2
AA
InIn
= ⇒ 1
2
A 1A 3
=
62. Imin ∝ |A1 - A2 |2
A1 ≠ A2 ⇒ Imin > 0
63. A2 = A2 + A2+ 2A2 cos f ⇒ f = 23p
64. ( )( )
( )( )
2 21 2max
2 2min 1 2
A A 3 225
3 2A A
II
+ += = =
−−
65. A’ = 2 22A 2A cosφ+ ⇒ 0 < A’ < 2A
66. Since notes are same, the pitch (frequency) is same. But the accompanying amplitude in their frequencies (quality) depends on the kind of instrument.
67. Distance = 24
= 0.5 m
68. First harmonic 2l
= L
1 22 2
k k2 1 12
T m 3 4p
p+
= = +
2 2 2
1 1 1T 3 4
= +
T 2.4= s
51. x = A sin wt ⇒ A = Asinwt1 ⇒ wt1 = 2p
A2
= Asinwt2 ⇒ wt2 = 6p
phase difference between x = A2
and x = A is 2 6 3p p p − =
⇒ Time = T3 .T
2 6
p
p
=
Aliter:x = Acoswt (t = 0, x = A represents the extreme posi-tion)when x =
A2
, we get A2
= Acoswt1
⇒ coswt1 = 12
⇒ wt1 = 3p
⇒ t1 = T T
3 3 2 6p pw p
= =×
52. T = 2g
pl ⇒ T → ∞
53. T = 2g
pl ⇒ l ∝ T2 ⇒ parabola
54. The bob is under the action of two perpendicular ac-celerations. i.e., horizontal acceleration ‘a’ and vertical downward acceleration ‘g’. The resultant accelerationg’ = 2 2g a+
55. In case of simple pendulum,
T = 52
gp
T’ = 202
gp ⇒ T’ = 2 T
Let the two pendulums be in same phase for first time where shorter one has completed ‘n’ oscillations
nT = m . T’ ⇒ n = 2m or n = 2 (minimum value)
56. As the water slowly leaks out, the centre of gravity of water and sphere falls down from the centre of sphere due to which effective length of the pendulum (l) in-creases. After most of the water leaks out, the length of simple pendulum (l) decreases as it is determined
3.104 Oscillations and Waves
78. nbeat = 258 – 256 = 2 Hz
⇒ T = 12
= 0.5
Time interval = T = 0.5 s
79. y = A sin2p (n – 1)t + A sin 2p nt+ A sin 2p (n + 1) t
= A (1 + 2cospt) sin 2pnt ⇒ beat frequency
= 1 Hz
80. ( )( )
( )( )a
v 0 v v 'f f .
v v ' v 0+ +
=− −
=
v1200 v v20
vv v20
× × + −
= 1200 × v × 21 v 1 2020 v 19v
× ×
= 1326 Hz.
81. Time Period of the 2 springs are 11
mT 2
kp= and
22
mT 2
kp= . The compressed springs will push the
block back so the time taken by block in spaces BQ and AP are
11
1
T m 0.4t s
2 k 3.6 3p
p p= = = =
22
2
T m 0.4t s
2 k 6.4 4p
p p= = = =
During complete cycle between A and B, block moves PQ twice, so that
32 PQ 2 1.2 24 8
t sv 0.9 9 3× ×
= = = =
So total time,T = t1 + t2 + t3 =
8 7 83 4 3 12 3p p p
+ + = +
= 7 22 8 9
4.5 s7 12 3 2×
+ = =×
82. Here y = A sin (wt + f) — (1) and f = 5p
rad
Energy E = 2 21m A
2w
— (2)
\ 2
12
2E 2 8.1 109 rad.s
0.2 0.1 0.1mAw
−−× ×
= = =× ×
So substituting in equation (1), y 0.1sin 9t5p = +
69. v v;
2 2l
nl
= = =ll
70. 10
22 5l
= = 2 2 4ml⇒ = × =
v = lf ⇒ 120 m sv
f4ml
−
= = \f = 5 Hz
71. c ∝ modulus of elasticity
r
In solids c = Yr
(Y → Young’s modulus)
72. v ∝ Er
, even though r is high for solids, E is very
high, so that speed of sound is high.
73. '
L2l
= ⇒ n’ = 100 Hz
4''
L2l
= ⇒ l” = '
4l⇒ n’’ = 400 Hz
Aliter:For a stretched wire harmonic frequencies are
fn = n T2 ml
.
n = 1 is first harmonic (fundamental) frequency 100 Hz. First overtone is second harmonic and third overtone is fourth harmonic i.e., 400 Hz.
74. Wave length remains samev = vlv ∝ speed of sound ∝ Tfrequency increases with temperature
75. The second overtone of the closed pipe is its 5th har-monici.e., n3 = (2k - 1) = 5n1 = 5 × 50 = 250 Hz
76. c = 340 m s-1
L = 3 5
, , ...4 4 4l l l
n = 680 Hz
l = 340
0.5680
= m
L = 0.125 m, 0.375 m
77. y = 1 2 1 2v v v v2Acos2 tsin2 t
2 2p p
− +
Oscillations and Waves 3.105
86. KE vs time plot is as shown.
K.E→
.
time →t t +1 t +2
\ period of KE function = 2 s⇒ Kinetic energy completes one cycle during every
half oscillation\ period of SHM = 4 s \ w =
24p
⇒ oscillation energy = 12
mA2w2
= 21
0.2 0.012 4
p× × ×
= 0.0025 J The other possibility is as shown. Not accept-
able since given that KE is not maximum at these instants.
K.E→
.
time →t t +1
But this gives period of KE = 1 s⇒ period of SHM = 2 s.
\ w = 22p
⇒ oscillation energy
= 0.01J > 0.0025J
87. Maximum restoring force kA = 16 k2A2 = 256
21
kA 1.62
= (Maximum energy)
Dividing the equations we get
2k = 2561.6
k = 80 N m-1
88. T = 2pmk
T1 = 2p 1
1
mk
T2 = 2p 2
2
mk
83. M
T 2k
p= ⇒ 4 M
210010 10
pp= ⇒ M = 0.4 kg
\ Equilibrium compression
= 0Mgx 0.04 mk
= =
The 1kg mass will lift if kx > 1 × 10 ⇒ x > 0.1 m⇒ elongation > 0.1 m\ Amplitude ≤ 0.1 + 0.04 = 0.14 m
84. Let length of cylinder be l
Buoyant force = A g2
r l
where cross sectional area
A = pr2
If x0 is the extension in the spring at equilibrium then we have,
0kx W A g
2r= −
l — (1)
Let the further displacement of cylinder from the mean position, be y, vertically down, then
0W
k(x y) W A y g a2 g
r − + + − + =
l — (2)
Using (1) in (2) k A g
a yW/g
r += −
\ SHM with w2 = k A g
W/gr+
⇒ WT 2
g(k A g)p
r=
+ where A = pr2
⇒ 2
WT 2
g(k r g)p
p r=
+
85. Let x0 = extension of the spring before B falls apart.
\ 00.7 10
x 0.035m 3.5 cm200×
= = =
Let x = extension of the spring at equilibrium with block A
Then 0.3 10
x 0.015m 1.5 cm200×
= = =
\ A = amplitude of oscillation = x0 – x = 3.5 – 1.5= 2 cm
w2 = k 200m 0.3
=
TE = ( )2max
1m v
2= ( )21
m A2
w× ×
= ( )2210.3 2 10
2−× × × ×
2000.3
= 0.04 J
3.106 Oscillations and Waves
Frequency detected by observer
5w rw
w r s
v v 1400 510
v v v 1400 5 10n n
+ + = = + − + −
= 0.1007 MHz
93. Maximum particle velocity = wy0
y = y0 sin t x2
T vTp
−
= y0sin 2 xt
T vp −
\ v = wave of velocity 4v = wy0 ⇒
4fl = 2pfy0
0y
2p
l =
94. v = nl
Frequency 3
1 1T 10
n −= = = 103 Hz
\ 3
v 7000.7 m
10l
n= = =
Phase difference = ( )2 12
x xpl
−
x2 - x1 = Phase difference
2l
p
×
0.70.0875m
4 2p
p= × = = 8.75 cm
95. Velocity of transverse wave = Tv
m=
For equilibrium, M2g = M1g sin 37° = T
Now 3223
M g120 M 10
10 10−= = ××
2M = 14.4 kg
\ 2
1
Msin37 0.6
M= = ⇒ M1 = 24 kg
96. Comparing the given equation, with y (x, t)
= Acos t coskx2p
w +
w = 50 p rad, 2
k 10p
pl
= = rad m-1
⇒ l = 0.2 m, 1v 5 m skw −= =
T’ = 2p ( )1 2
1 2 1 2
m m 1.
m m k k+ +
put m = m1 + m2
Since the original spring is a series combination of the two pieces:
k = 1 2
1 2
k kk k+
⇒T’ = 2p 1 2
1 2
m m k.
m k k
⇒ 1 2T TT' 2. .
2 2 2 Tp
p p p=
⇒ T’ = 1 2T TT
89. The maximum acceleration is
w2 A = w (wA) = ( )2u
Tp
\ 2 u
gTp
m > ⇒ 2 ugTp
m >
min
2 ugTp
m = .
90. The period of this pendulum is given by
L L L LT 2 2 1
g g LD D
p p+ = = +
= 2 2
mg mgL L2 1 2 1
g gr Y d Y4
p pp
p
+ = +
= 2
4mgL2 1
g d Yp
p
+ 91. Now g is the acceleration due to gravity. Let l be the
length of the simple pendulum and a the inclination of the inclined plane. Effective g of the bob isg ' g a g cos a= − =
For the string to be restored to its initial original shape, the pulse travels a distance = 2 × the length of the string, as the pulse gets inverted at each reflection from a fixed end.
\ Required time,
2 0.5t 0.14s
5 2×
= =
103. 1 2 11
1 2 2
T T T 0.11 1 1n
2 m 2 m 2 m+
= = =l l l
1 1
1 2
T T 0.1+∴ =
l l l1 = 0.8 l2 = 0.81
solving T1 = 3.975 N @ 4.0 N
T2 = 4.075 N ~ 4.1 N
104. Every point on the string executes SHM. Consider an element dx of mass dm = mdx
Also, At x = 0.15 m, cos 10 px = 0 At x = 0.3 m, cos 10 px = -1, ⇒ antinode
97. Comparing the given equation with standard equation of wave motion, y = a sin (wt - kx)we get:Amplitude = 6 cm, and w = 2 pn = 5 rad s-1
5n 0.796Hz
2p∴ = =
2k 0.03
p
l= = rad m-1
2209 cm
kp
l∴ = =
velocity of the wave
= 1 15167cms 1.67m s
k 0.03w − −= = =
Density of the medium = 1.2 gm/cc = 1200 kg m-3
\ Intensity = 2 21A v
2rw
( )21
1200 25 0.06 1.672
= × × × ×
2 190J m s− −=
98. A12 + A2
2 + 2A1A2 cos q = A12
cos q = 2
1
A2A−
99.
Source Person
L
Reflected wave 5 m
100 m
Length of the path of reflected wave
= 2 × (502 + 52)1/2 = 1/225
2 50 150
× +
= 21100 1 (0.1) 100.5 m
2 + =
For constructive interference,Path difference = d = nli.e., (100.5 – 100) = nlFor n = 1, l = 0.50m
3.108 Oscillations and Waves
106. Strain = Dll
= at
Stress = TA
= Y. strain
(symbols have usual meaning)T = YA atAlso, mass per unit length = m = Ar
Wave velocity = Tm
= YA t Y t
Aa ar r
=
11 61
3
2 10 12.5 10 116.7 m s
9 10
−−× × × ×
= =×
107. Y T120
r m=
Y = Young’s modulus of wirem = r × A
⇒ 2
YT
(120)mr
= = 6 11
2 2
A . Y 10 10(120) (120)
− ×=
= 9.94 @ 7 N
108. Fundamental frequency of an open pipe
= ov
n2
=l
and the third harmonic
of a closed pipe of the same length = c3v 3
n4L 2
= = n0
Given that 00
n3 n 120
2
= +
\ on120
2= \no = 240 Hz
109. Fundamental frequency is
vn
4=
l, where v = 330 m s-1
330264
4∴ =
l
3300.3125m 31.25 cm
4 264∴ = = =
×l
No such option for first overtone
l = 3v4L
= 3 × 31.25 = 93.75 cm
Option (c)
110. Let u = speed of the carn = frequency of the hornfrequency received, and reflected by the cliff
= 1n
nu1v
=−
(m → mass/unit length). Its velocity is
v = dydt
= Awsinx
cos tLp
w
vmax = Awsinx
Lp
. For SHM, energy of oscillation
= ( )( )2max
1dm v
2
L2 2 2
10
1 xE A sin dx
2 Lp
mw = ∫
L
2 2
0
2 x1 cos1 LA dx
2 2
p
mw
−
= ∫
= 2 2A L2 2
mw
where m is linear density
Here A is amplitude, w is angular velocity.
Similarly : 2E = ( )2 21 L2 A
2 2m w =
2 2
2
2 21
1 L4 AE 2 2 41 LE A2 2
m w
mw
×∴ = =
\ E2 = 4 E1 = 4 × 2 × 10-2 J = 8 × 10-2 J
Aliter:Energy E1 = Energy density × volume
= 12rA2w2 × (AL)
[Q Energy density = 12rA2w2] and volume = AL
Energy E2 = 12rA2(2w)2 × AL
⇒ 2
1
EE
= 4 ⇒ E2 = 4E1 = 4 × 2 × 10-2
= 8 × 10-2 J
105. n T
f2 m
=l
1 31
1 1f 400 Hz
T 2.5 10−= = =×
n T400
2 m∴ =
l
480 n 1or n 5
400 n+
∴ = =
5 48.4400
2 0.1∴ =
l =
55l
55m 13.75 cm
400∴ = =l
Oscillations and Waves 3.109
115. Let the end be displaced by x restoring torque about A in first case
= 2 12
kx kx3 3
+l l
= x 2 2 5k x
k . k x.3 3 3 3 9
+ =l l l
In second case let x' be the displacement and
k’ be the spring constant : x ' x x
x '/2 2= ⇒ =
l l\ Restoring torque about A:
t = k’x'x k 'x
k '2 2 2 4
= =l l l
\x 5k x 20k
k ' k '4 9 9= ⇒ =
l l
116. A B
x v M x'
9k20'k =
When the rod has mass, we have to take into account moment of inertia
22
AmL
ML3
I = + ,
Also x' = x2
and velocity of mass v = wL
Energy equation:
2 2 22
2
1 x 1 mL vk ' ML . E
2 4 2 3 L
+ + =
comparing: 2 21 1kx m'v
2 2+ = E
k ' mk and m' M
4 3 = = +
\ w2 = ( ) ( )
k k ' 3 20k 3m' 4 m 3M 4 9 m 3M
= =+ × +
( )5k
3 m 3M=
+
T = ( )3 m 3M
25k
p+
117.
37 (180 − 37)
(180 + 37)
−37
Frequency observed by the driver
= 2 1
un 1u vn n 1 2n or 2n
uv 1v
+ = + = = −
1v 330u 110 m s
3 3−∴ = = =
111. At any instant energy = 2 21m A constant
2w = . The
average value of sinusoidal function for t = 0 to T
t2
=
is not the same as from t = T 3T
to t4 4
=
112. Centripetal acceleration exists at the mean position. \ Statement 1 is false. Statement 2 is always right.
113. Statement 2 is not correct. Beat frequency fb = f1 - f2, (f1 > f2) only when two frequencies are involved. If there are more than two frequencies (say f1, f2 and f3), then fb is not necessarily (f1 - f2) or (f3 - f2) or (f3 - f1).
114. Let x1 and x2 be the displacements at spring’s ends and x the displacement of the mass.
11
x x 2x x
2 / 3 3= ⇒ =
l l
2
2
x x xx
/ 3 3= ⇒ =
l l⇒
PE = 2 2 2 21 2
1 1 1 4 1kx kx k x x
2 2 2 9 9 + = +
= 25kx
18Considering the energy equation
221 5kx
Mv E2 18
+ = ,
differentiate w.r.t time
1 dv 5kM2v
2 dt 18+
dx2x. 0
dt= , ⇒
dv dxa ;v
dt dt= =
a = 10k 10k
x18M 18M
w− ⇒ = ⇒T = 9M
25k
p
= M
65k
p
note:
If E = k'x2 + 21m'v
2 then w2 =
k 'm'
. Hence from the
energy equation above k’ = 5k9
; m’ = M
⇒ \ w2 = 5k9M
3.110 Oscillations and Waves
119.
C
fundamental
� � C
1st overtone
When dropped on floor, the two ends are antinodes,
l = v v
; f2 2l
l= =
l ⇒ v = 2l.f = 2 × 0.6 × 2 × 103
= 2.4 × 103 m s-1
When clamped at centre:(for fundamental)
l = 2l
; v = f1l ⇒ f1 = vl
f1 = 32.4 10v
2 2 0.6×
=×l
= 2 kHz
1st overtone ⇒ 3
2 4l
=l
⇒ l = 23
l
⇒ f2 = 32.4 10v
2 0.63
l
×=
×
�
C
f2 = 6 × 103 Hzwhen clamped at one end ⇒ fundamental as shown:
l = 4l ⇒ f3 =
32.4 10v4 4 0.6
×=
×l = 1 kHz(a), (b) and (d) are correct
120. y = Asin3
t2p
w + = -Acoswt; |ymax| = A
v = dy 3
A cos tdt 2
pw w = +
= Awsinwt; |vmax|
= Aw
a = 2
2
d ydt
= w2Acoswt = -w2y; |amax| = w2A
Ek = 2 2 2 21 1mv m A sin t;
2 2w w=
Ep2 2 2 2 21 1
m y m A cos t2 2
w w w= =
KE = 12
mA2w2cos2wt; PE = 12
mA2w2sin2wt
PEKE
= tan2wt =9
16 ⇒ tanwt =
34
± ⇒ (wt) = 37°
Considering the equivalent circular motion this can happen in 4 positions of the radius vector as shown in the figure → (b) is correct x = ±A sinwt = ∓ 5 sin 37° = ±3 cm → (c) is correct
The minimum time t = 37
T360
≈ 0.1T
→ (d) is correct
118. l1 → smaller l2 → larger
T ∝ l ⇒ T2 > T1 ⇒ 2 2
1 1
T1.21
T= =
ll
= 1.1 T2 = 1.1T1 ⇒ They will be in the same phase when
w1t = w2t + 2p(Q w2 < w1) = 1 2
2 t 2 tT Tp p−
= 2p radian
1 1
1 1t 1
T 1.1T
− = ⇒ t = T1
1.10.1
= 11T1 = 11 s
They will be in opposite phase when
1 2
2t t
T Tp p2
− = p
⇒ t = 1T1.10.1 2
× = 5.5T1 = 5.5 s
They will repeat this phase at 3p ⇒ t = 16.5 s
Alternate method: If the slower one completes n oscillations and at that instant faster one completes n + 1 oscillations, they are in same phase ⇒ (n + 1)T1 = n 1.1 T1
(n + 1) = 1.1 n ⇒ n = 10 ⇒ t = (n + 1)T1 = 11sFor opposite phase, when the larger one complete
n oscillations the smaller one completes 1
n2
+
oscillations
n + 12
= 1.1n ⇒ n = 5, t = 1
n2
+ T1
= 5.5 s and repeat at 11 + 5.5 = 16.5 s
Oscillations and Waves 3.111
\ f max = maximum frequency
= g1 1 10
12 a 24.9
p p= =
72p
Hz
122. (i) Let a be amplitude of oscillation of particle; w its angular frequency and t, time taken to go first at P
Then v = aw cos wt, and OP = a sin wt,⇒ 0.1 = aw cos wt1 — (1)
⇒ 0.1 = a sin wt1 — (2)
⇒ tan wt1 = w
Put w = 2pf = 0.4 p
tan 0.4pt1 = 0.4p (radian)
⇒ t1 = 0.72 s
(ii) When it again comes at P, velocity is negative
⇒ a w cos w t2 = -v = -0.1, and OP = a sin w t2
= 0.1 ⇒ tanwt2 = -w = -0.4p (radian)
⇒ w t2 = p - 0.9 = 3.14 - 0.9 = 2.24 (radian)
⇒ 22.24 2.24
t 1.8 s1.25w
= = =
123. Since at the time of collision, 1 kg block is at extreme left position and its amplitude of oscillation is 3 cm, so at this instant, the block is at rest and spring is compressed by x0 = 3 cm (i) Let velocity of combined mass (just after col-
lision) be v. Applying law of conservation of momentum,
12 0.4 0.8v m s
1 2 3−×
= =+
leftward.
Due to the velocity of combined body, spring is further compressed till combined body stops. Let the maximum compression of the spring at the instant be a.By law of conservation of energy,
( )2 2 2
01 1 1
ka kx m M v2 2 2
= + +
i.e., 2 2
24
1 1 147 3 1 0.8147a 3
2 2 2 910×
× = × + × ×
147 a2 = 0.1323 + 0.2133
⇒ a = 0.048 m = 4.8 cm
1k 147
49 7rad sm M 1 2
w −= = = =+ +
when y = -A2
, coswt = 12
⇒ wt = 3p
⇒ t = T6
v = 3
A2
w ⇒ max
v 3v 2
=
(p) is true for (a) and (b)
a = 2 A2
w ⇒ max
a 1a 2
=
→ (q) is true for (a) and (b)
( )2k
p
Etan t
Ew= ⇒ 3
→ (s) is true for (a) and (b)\ (a) → p, q, s (b) → p, q, s
t = T8
, sinwt = sin1
4 2p
= ; coswt = 12
\ t = T8
⇒ y = -A2
2k
p
Etan
E 4p
= = 1
\ (c) → r (d) → r
Additional Practice Exercise
121. (i) During downward motion the block and piston will separate when acceleration of piston ≥ accel-eration due to gravity; that is just at the maximum height.
\ w2 a = g Here a is amplitude (since in S.H.M, acceleration = -w2x)
But period, 2
Tp
w=
Where w = angular frequency Taking p2 = 10
2 2
2 2 2 2
2
g Tg g 10 1a
4 4 4T
w p p p×
∴ = = = = = 0.25 m
(ii) The block and the piston will remain in contact continuously if the acceleration due to gravity is greater or equal to the maximum acceleration i.e., g ≥ w2a
For maximum frequency g = w2max a
3.112 Oscillations and Waves
Let x0 be the depression of the spring due to M alone.
Then Mg = k x0 ⇒ x0 = Mgk
; Let x be the maxi-mum compression from the neutral point
⇒ \ amplitude = x Conserving energy between position 1 and 2 of
the combined body (m + M) we have
( ) ( ) ( )22 20 0 0
1 1 1M m v kx k x x M m gx
2 2 2+ + = + − +
= 2 20 0
1 1kx kx kxx
2 2+ + - Mgx - mgx
Substituting for v0 and x0
( )
( )
2
2
m 2gh1M m
2 M m+
+
= 2 Mg1kx kx
2 k+ - Mgx - mgx
⇒ 2m gh
M m+21
kx Mgx Mgx mgx2
= + − −
22 2m gh
kx 2mgx 0M m
∴ − − =+
⇒ x =
2 2 2
2
2mg 4m g 8m ghk k(M m)k
2
± ++
( )2khmg mg
x 1k k M m g
= ± ++
= 0.1 10
110×
+ 0.1 10 2 110 0.5
1110 1.1 10
× × ×+
× ≃0.04 m = 4 cm since negative x value is not admissible.
125. (i) The spring originally at rest is stretched by x = 0.3 m, when potential energy (elastic energy)
= 21kx U
2=
21 1
U kx 4 0.3 0.3J2 2
∴ = = × × ×
= 2 × 0.09 = 0.18 J At the equilibrium, the elastic energy U is con-
verted to translational kinetic energy and rota-tional kinetic energy of the cylinder,
Now Ktrans = 21Mv
2
Krotational = 2 2 2 21 1 1 1MR Mv
2 2 2 4Iw w = =
(ii) Let phase of combined body, just after collision be f. Displacement of combined body from equi-librium position is given by x = a sin (wt + f)
where a = 0.048 m, w = 7 rad s-1
Now at initial starting time, t = 0, x = -0.03 m (as combined body is on left of equilibrium posi-
tion) -03 = 0.048 sin f\ sin f = - 0.03/0.048
= 58
− ⇒ 1 5sin8
φ − − = radian
124.
�0
M
T
1
2 m
x0
M• x
Here h = 0.5 m; k = 110 N/m, M = 1 kg, m = 100 g = 0.1 kgThe period of oscillation of a spring depends on the mass attached and force constant and it is given by
m'T 2
kp=
Here the mass or compounded mass attached = M + m
M m kT 2 or
k M mp w
+∴ = =
+
\ Substituting the numerical quantities given, viz., m = 100 g = 0.1 kg
M = 1 kg k = 110 N/m
M m 1.1T 2 2
k 110p p
+= =
12
100p=
= 2
second10 5p p
=
1110
10 rad s1.1
w −= =
Let v0 be the velocity of the combined body immediately after impact. Then (By conservation of
momentum) (M + m) v0 = m 2gh ⇒ v0 = ( )m 2ghM m+
Oscillations and Waves 3.113
= 8 × 10-16 — (1)1v 200m s−∴ =
⇒ (from equation (1) a = 168 10
200
−×
= 2 × 10-9 mSince initial phase of particles at x = 0 is zero,So equation of the wave is given by (in metre)
xy asin t
vw = −
= 9 x2 10 sin500 t
200− × −
= 2 × 10-9 sin (500 t - 2.5x)
127. (i) Stress = Y × strain = Y × 25 a T (tension) = pr2 × stress = a × 25 a × Y (a = area of the wire)
\ velocity of sound T
cm
=
where m = linear density. If n is fundamental frequency
a Y 251 T 1n
2 m 2 aa
r
× ×= =
l l
= 1 Y
252 1
ar×
= 11 6
3
2 10 16 1052 8 10
−× × ××
= 50 Hz
(ii) The frequency of vibration in first overtone is n’ = 2n = 100 Hz
128. (i) Amplitude = 0.01 sin (12x) At x = 0.042 ⇒ y = 0.01sin0.5 (angle in radian)
≃ 4.8 mm
(ii) Now particle velocity
= ( ) ( )dy0.01 200 sin 12x cos200t
dt= ×
The maximum particle velocity takes place when cos 200 t = 1
Maximum particle velocity = 0.01 (200) sin 12x = 2 sin (12 x) Now x = 0.042 m (given)\ Maximum particle velocity = 2 sin 0.5 = 0.96 m s-1
(iii) Tensile stress = dy
Ydx
(where Y is Young’s modulus)
= 7.5 × 1010 × 0.01 × 12 cos (12x) cos 200 t
\ Total kinetic energy
= 23U Mv 0.18J
4= =
(ii) Now 2
translational
2rotational
1 MvK 221K 1Mv4
= =
K stands for kinetic energy
\ Ktranslational = 2
0.18 0.12J3 × =
Krotational = 1
0.18 0.06J3× =
The two kinetic energies here are translational kinetic energy.
(iii) 34
Mv2 + 12
kx2 = constant. Differentiating
⇒ 3 1
M2v v k2x x 04 2
• •+ =
2 ka x SHM
3 M−
⇒ = ⇒
(iv) 2k 2 4
1.633M 3 1
w×
= = =×
rad s-1
T = 2
3.855pw
= s
f 0.26Hz
2wp
= =
In S.H.M if
E = 2 2 21 1 k 'k 'x m'v ; then
2 2 m'w+ = . Hence en-
ergy equation in (iii): m’
= 32
M; k’ = k
⇒ w2 = k2
3M 126. Let displacement amplitude of medium particles be
a, velocity of propagation vw = 2pn the angular frequency and r the densityof the medium.Then intensity I = 2 p2 n2 a2rv = ( )2 21
2 n a v2
p r
But w = angular frequency = 2 pn
( )9 2
22 2
2 2 5 10 10a v
0.5 500
I
rw
− −× × ×∴ = =
×
= 9 2 310 10 10 2
25
− − −× × ×
3.114 Oscillations and Waves
\ velocity of sound c at 30°C
= 1273 30332 350 m s
273−+
× =
13500.99 u 3.54 m s
350 u−⇒ = ⇒ =
+
131. w = 2 1
f10 2 10p w
p⇒ = = s-1
f = 0.1 Hz.
132. x1 = A sin wt, x2 = A sin (wt + 2p
)
⇒ x1 + x2 = A.2 sin(wt + 4p
) cos 4p
⇒ x1 + x2 = 2 A sin(wt + π4
)
133. x1 = A sin wt, x2 = A sin (wt + 2p
)
It has the same phase difference of π4
with x1
and x2
134. A2 = A12 + A2
2 + 2 A1 A2 cos f⇒ (say A = A1) ⇒ A2 = -2 A1 cos f
⇒ cosφη
= − = −AA
2
121
2 135.
�
x
x A A’
PE(x) = (r.A.x)g
KE = π2
(lAr)v2
KE + PE = constant
⇒ 12
2 12
2ρ ρAg x A v E( ) + ( ) =l
Refer Q 125(iv):
w = 2 2ρ
ρAg
Ag
l l=
⇒ T = 2p l2g
The maximum tensile stress (cos 200 t = 1) = 7.5 × 1010 × 0.01 × 12 cos (12 × 0.042) = 7.9 × 109 N m-2
129. s T 2 T 1 T
n2 m 2 m m
= = =l l l
[Q s = 2, (1st overtone)] T = 10 × 10 = 100 N
1 100n
0.5 m=
81T' 100 81
100= × = ⇒
1 81n'
0.5 m=
n 100 10n n'
n' 81 9= = ⇒ >
since the tuning fork, (frequency = N) beats equally with n and n’, we have n - N = 5 and N - n’ = 5
10 N 59 N 5
+∴ =
− or N = 95 Hz
\n = N + 5 = 95 + 5 = 100
= 1 T
ml = 2 11 100
m 4 10 kg m0.5 m
− −⇒ = ×
⇒ ( )
2
2 23
m 4 10r 0.5 10
rp p
−
−
×= =
× 4 35.1 10 kg m−= ×
130. The frequency f in a sonometer is given by the relationMg1 T 1
f2 m 2 m
= =l l
where l is the length of the
wire,M = mass carried by sonometer wire.m = mass per unit length of the wire.
( ) 2
1 6.4 10f 400 Hz
2 0.1 10−
×= =
As the tuning fork is moved away from the wire, the frequency is decreased, and according to Dop-pler’s formula
cf ' f
u c = +
where c = velocity of sound, and u the
speed of tuning fork
f - f ' = 4 Hz i.e., f ’ = f - 4 = (400 - 4) = 396 Hz
c f ' 396 99u c f 400 100
∴ = = =+
Now c = 332 m s-1
Oscillations and Waves 3.115
140. Let one body move by x. By symmetry, the other body also moves by x conservation of energy gives
2. ( )221 1mv 2. k 2x
2 2+ = constant
⇒ ( ) ( )2 21 1. 2m v 2k.4 x
2 2+ = constant
⇒ w = 8k
2 102m
=
T = 2 2 10
2 10pw
= = 1 s
Aliter:This equivalent to⇒ k’ = 2k
w = k 'm
⇒ m = mm m
m m 2=
+ ⇒ w =
4km
141. 22
21 mR 1.k(R )
2 4 2q q•
+ = constant
⇒ w = 2
2
kR k2
mmR4
=
T = 2 m
kp
pw
=
142. A
a
B C 2
a
22 a +=�2
2a
Mass of each section is m2
I = 22 2
2ma m a m a. a
2 3 2 12 2 2
+ + + ×
= ma21 1 56 24 8 + +
= 25
ma6
T = 2pmgdI
If COM is at P, then x co-ordinate of P is given by:
mx = ( )m m a0
2 2 2+
x = a4
m m k’
P(x, y)
X
Y
B
A
C
136. km
122
3=
π
km
222
4=
π
1 1 1
1 2k k k= +
mk
T mk k
=
= +
2
1 12
1 2π2 2 2T 3 4
2 2 2p p p = +
T = 2 23 4+ = 5 s
137. Initial lift moving
�0 �0
x x + ∆
mg mg ma
D = 1 2m.a
k 100×
= = 2 cm
138. On the truck frame v = 1 m s-1(max)vmax = Awamax = Aw2
⇒ From a reference point of zero displacement the two points can be
20 cm A
A 20
10
30
⇒ A0sin10
260
p
= A
A0 = 2A
3
20 cm 5 cm
302 =λ
OR A0sin
560
2p = A
A0sin6p
= A
A0 = 2A
153. No phase change due to refection at the boundary of the two wires. Then, constructive interference when they meet first time. But refection at the fixed end causes phase change Hence destructive interference when they meet second time
154.
For section 25
l�5
3 �5
2
n 22 5l
= l — (1)
n = 1, 2, 3…..
For section 35l :
1 3m
2 2 5l + =
l — (2)
3.118 Oscillations and Waves
162. l = 330110
= 3 m
The open end has pressure amplitude = 0 (pressure node) ⇒ 0p
2 = p0sinf
⇒ f (phase) = 6p
⇒ 2
x6
p pl
= ⇒ x = 12l
⇒ path x = 3
12 = 0.25 m
163.
Formed at nodes
164.
1 L4l
= L
2 2l
=
l1 = 4L l2 = L
1 2
2 1
f 1f 4
ll
= = ⇒ f2 = 4f
165. l = 320160
= 2 m
Pressure antinode forms at the wall
⇒ Distance = 0 or 2l
= 0 or 1 m
166.
The antinode is offset by 0.3D = 3 cmThe node is offset by 2 cm
34l = L + 3 = 90
l = 120 cm
\ A = A0sin2l .2p
≈ 9mm2
.2120
p
= 0.3p mm
167. Let fB > fA
⇒ fA = 396 when filed, f↑⇒ fA’ = 406
⇒ 10396
× 100 ~ 2.5%
v = T TAm r
= ∝ 1r
2 1r 1 2
i 2 1
1 2
1 1r rA r r 1
1 1A r r 3r r
−− = = = ++
r
i
19
II
=
t
i
89
II
= ≥ i
t
98
II
=
157. Ratio of specific heats, g = 2 2 4
1 1f 6 3
+ = + =
where f = degree of freedom = 6
3p2c
r= and
pv
g
r=
⇒ v 4 22c 3 9 3
g= = =
⇒ v = 43
c
158. p0 = A0.Bk
0
0
p 2B.
Apl
=
⇒ 1
2
ff
= 0 02
1 0 0
p ' A " 9.
A ' p " 16ll
= =
159. I = 1
.2rA2w2.v
⇒ u = 1
.2rA2w2
p = ABk, v2 = Br
⇒ u = 2
22 2
1 p. .
2 B kr w =
21 p2 B
160. Restoring force proportional to displacement required⇒ solid only
161. 1
2l
= L + 2 × 0.3D
l1 = (L + 0.6D) × 2
2
4l
= L + 0.3D
l2 = 4(L + 0.3D)
1 2
2 1
4L 1.2Dff 2L 1.2D
ll
+= =
+ =
179
⇒ D 5L 24
=
Oscillations and Waves 3.119
171. x = A sin wty = A cos wtx2 + y2 = A2; circle, not SHM
172.
� → ∞
A x
B δ
mg θ Re
O
For an infinitely long pendulum, period T = 2p eRg
(Re → radius of earth) (Proof : Assume small oscilla-tion AOB since l → ∞, AOB is almost straight line. Restoring force at B F = mgcosd = mgsinq ≃ mgtanq = mg
e
xR
\ acceleration = e
gx
R
⇒ SHM ⇒ w = e
gR
⇒ T = 2p eRg
173.
(neutral point)
�0
x = k
mg
Statement 2 is always true. Acceleration at maximum downward position = acceleration at maximum up-ward position = maximum acceleration = g(given data). Hence at maximum top position there is no contribution of spring force, i.e., it is at its natural
length l0. Hence x = mgk
is the maximum displace-
ment from neutral point
Let fB < fA
⇒ fA = 404⇒ fA’ = 406
⇒ 2
404 × 100 ~ 0.5%
168. Observer, when stationary will receive the frequency from S2 higher than from S1. Hence for equaliz-ing observer has to move towards S1 (say) with a velocity v0
1 2
0 0
s s
c v c vc v c v+ −
=− −
0 0v v1 1c c10 201 1c c
+ −=
− − ⇒
0
0
v1 cv1 c
+
− =
101c
201c
−
−
⇒ v0 = 10c
2c 30 −
= 10 315
300×
= 10.5 m s-1
169. Let the source emit n waves when the point reaches the observer and let this time be T. The nth wave reaches in time 2T. Thus in time T, n waves are received (T t0 2T) and n waves are emitted (0 t0 T) wavelength will increase Alternatively: When the wind blows from source to observer effective sound velocity is c’ = c + vwind. Received frequency is same, because both source and observer are stationary. But c = fl since c increases, f remains same hence l increases.
170. fwall = 00
s
c v.f
c v + −
= 0
c 10f
c 30 + −
f ’ = s
cc v −
.fwall
= wallc
fc 10 −
= ( )
( )( ) 0
c c 10f
c 10 c 30
+
− −
= 330 340
1024300 320
××
× ≈ 1200 Hz
3.120 Oscillations and Waves
\ Rotational KE = 13
(Total energy)
= 13
(Maximum spring energy)
Statement 2 is obviously wrong.
177. Statement 2 is true, a standard derivation. Based on that, maximum v = Aw; A is maximum when the tunnel is dug up to the centre of earth where
A = Re; v = Re
e
gR
= eR g
\ KE = 12
mReg
178. Position of B is Bˆ ˆR ir cos t jr sin tw w= +
= i (projection of circular motion x-axis) + j (projec-tion of circular motion on y-axis) Projection on x-axis, itselfan SHM, is in syn-chronous with SHM of B, hence no x-displacement between the two. Hence displacement between A and B = j rsinwt → an SHM.
179. Intensity at a particular point can be 4I if the waves arrive in phase and their individual intensity is I. But there will be other places where they meet out of phase and the resultant intensity is zero. The total power is conserved, i.e., it cannot be boosted up in totality, by interference.
180. I changes in both cases, because I depends on apparent frequency which is different from actual frequency in both cases.
181.
12 m
x A
dx
6 kg
2 kg
Consider a small distance dx at a distance x from end A. The value of tension Tx = TA- m.x where, (m) is the mass per unit length
\ Tx = xx
x8T6 x 28 x 8 v112 22
m
−− = − ⇒ = =
= 16 x−
The time to pass through dx; dt = ( )
12
dx
16 x−
174. (a)
O
F
θ
P
x
At a distance x from the centre of the path, F’(downward) = FsinqF depends on the mass m’ of the sphere of radius
r = OP ⇒ m’ = 43pr3r
F =
3
2
4G r m43 G3r
p r
= prrm [m → mass of the
body dropped]
F’ = Fsinq = F.x 4
G m xr 3
pr = ⇒ SHM
w2 = 3
3 2
g4 4 R GMG G
3 3 RR R .Rp r
pr = = = [R → radius of
earth]
T = 2p Rg
⇒ Hence the time required to move one
end to the other end is T2
, i.e., independent of length l of the tunnel.
175.
�
f ’ = 1 T
ml; f ” =
1 4T 2 Tm m
=l l
= 2f ’
⇒ statement 2 is correct. But wavelength of first overtone is always l. Statement 1 is wrong.
176. Total energy = Maximum KE = Maximum spring energyAt neutral point:
Maximum KE = 2 2
22
1 1 mr vmv
2 2 2 r+
= 2
21 1 mvmv
2 2 2+
Oscillations and Waves 3.121
away/towards the body in the direction of sound wave from the body. Hence the source should be at the position where the tangent from the point (B or C) meets the circle and vmax = rwf1 = f2, as relative velocity is v = Rw, away for both.
185. At no position, the velocity vector v can be along the line AP. Hence the maximum receding/approaching velocity towards A is less than v = rw, f2 < f1
+ A
θ v P
v’ = vcosθ
186.
B
P r
A O S
P’
T
v
v
At position T and S, v has no component in the di-rection of sound wave towards A (i.e., TA and SA ), hence no change in frequency. Hence midway angular position of 90°, i.e., at P the approach velocity to A is maximum. So too minimum at P’ Both P and P’ are also the points where v = rw(i.e., maximum component of v) occurs for B. Hence occurrence of maximum and minimum frequency is at same instant for A and B.
187. Let vP be the velocity of the plane. Then the reflected signal as received by the plane’s antenna will have a frequency
f ’ = fc vc v
fc v cc v c
P
P
P
p
+−
=+−
( )( )11
//
1P Pv v
f 1 1c c
− = + −
= P Pv vf 1 1
c c + +
= p2vf 1
C
+
pv
1c
<<
Q
\Df = p 98
2v 2 300.f 10 2000Hz
c 3 10×
= × =×
\ total time t = ( )
12
10 2
dxdt :
16 x=
−∫ ∫
Let X = (16-x) ⇒ dX = -dx, limits: x = 0, x = 16:
X = 12; X = 4
\ t { }4
1 14 2 16
416
16
dX X2 X
1X 12
+ − − = = − +
∫
( )2 16 4= × − = 4 s
Aliter:vx
2 = 16 - xsame as uniformly accelerated motion, so that at x = 0 vx(0) = 4, (at x = 12)vx'(12) = 2
⇒ vav = 4 2
32+
=
\ Time = av
Dis tance 12v 3
= = 4 s
182. vA = AA
vT 8 416 4
0.5 f fl
m= = = ⇒ = =
vB =2
4 20.5
= = ⇒ lB = Bv 2f f
=
\ A
B
4 f. 2
f 2ll
= =
183. Tx = 2m -x m
2 xm m= −ll
(where m mass per unit length)
vx = T 2 x2 x
m mm m
−= = −
ll
\ Time to travel a small distance dx:dt
= dx
2 x−l
\ t = ( )
10 2
dx
2 x−∫l
lIntegration by substitution: let X = 2 l -x; dX = -dxlimits x = 0 → X = 2 l ; x = l → X = l
\ t =
1 12
21
22
2
dX XX 2
1X 12
− + − = − = × − +
∫
l
ll
ll
l
= ( )2 2 2 2 2 − = − l l l
184. For the external point the minimum/maximum frequency is while the source has maximum velocity
3.122 Oscillations and Waves
After collision, the velocity, of combined mass is mv0 = (m + m)v0’
v0’ = 1 0.2 60
0.1 602
×= m s-1
w2 = k 1000
m' 2= = 500
since the combined mass moves in SHM (v0’)
2 = w2(A2 - x2) ⇒ 0.01 × 60 = 500(A2 - x2) — (1) If C is the neutral point, PC = x(say)
Kx = 2m.g ⇒ x = 2 101000×
= 2 × 10-2 m
\ in (1) ⇒ 60 × 10-2 = 500(A2 - 4 × 10-4) A2 = 12 × 10-4 + 4 × 10-4 ⇒ A = 4 × 10-2 = 4 cmConsider the oscillation of the combined body moving up from C to P.At P : x = Asinwt ⇒ 2 × 10-2 = 4 × 10-2 sinwt
sinwt = 12
⇒ wt = 6p
⇒ 2Tp
t = 6p
⇒ t = T12
The bodies will move in SHM starting from P and reach back at P. After that the spring extends beyond l0, retards A and the retardation force cannot be passed
from A to B. B is detached time required t’ = T
2t2
+
= T T2 6
+ = T4 2
T6 3 =
T = 2pmk
= 2p2 2 2
1000 10 10 5 5p p
= =
t' = 2 2
s3 5 5 15 5
p p=
191. Let AP be the maximum amplitude of oscillations of P.
Obviously, AP = 0.23
⇒ mPAP = mQAQ
(AQ → maximum amplitude of Q)
AQ = PP
Q
m 1 0.2A
m 2 3= =
0.13
m
x = AP+ AQ = 0.1 0.23 3
+ = 0.1 m
vmax(P) = APw ⇒ w = 2
30.2
× = 30 rad s-1
w2 = ( )k M m
Mm+
⇒ 900 = k32
⇒ k = 600 N m-1
188. Apply beat frequency principle. Beat frequency = Df
= 2 × 103 ⇒ T = 12
× 10-3 = 0.5 ms
Detailed analysis:
T’
T’
T
Let the original radar wave beA sinwt = Asin2pftLet the reflected wave be represented byA sinw’t +f = A sin [2p(f+Df)t+f]When they are mixed.[A sin 2pt + A sin[2p(f+Df)t + f]
= 2 22 2
22 2
A f f t f tsin cosπφ
πφ
+
+
+
∆ ∆
2 2
2 22
2 2A f t f f tcos sinπ
φπ
φ∆ ∆
+
+
+
Amplitude is varying at a frequency3f 2 10
2 2D ×
= = 103Hz
Interval between zeros T’ = 1
T2
where T is the period of the envelope
= 3
1 1. 0.5ms
2 10=
189. As seen from above T, the interval between peaks is same as interval between zeros.\T’= 0.5 ms.
note
The combined signal will have maximum value when cos x is maximum (i.e +1 or -1). Hence twice in a
cycle of cos(x).
190.
B
A P
A
C
xv0 t
v0
C
Oscillations and Waves 3.123
(b) f0 = 1 T 1 49 1002 m 4.2 0.01 6
= =l
≠ 20 Hz
(c) If 0.01 m is the amplitude of individual wave, maximum amplitude of the standing wave is 2A = 0.02 m
Total energy = 2 2A L
4w m
= ( ) ( )2 22 2 2 24 10 2 10 10 2.1
4
p − −× × × ×
= 0.084p2J
(d) P = 12mw2A2v =
2 21 TA
2w m
m
= 12
× 4p2 × (102)2 × (10-2)2 × 49 0.01×
2p2 × 0.7 = 1.4p2 watt
194.
t t t t
2A
A
The equation can be re-written as
Y1 = Acosx
200 t7
p − and
Y2 = -Acosx
280 t5
p −
Y1 + Y2 = 2Asin
x x200 t 2807 5
2
p p − + −
sin
x x280 t 200 t5 7
2
p p − − −
C D D CcosC cosD 2sin sin
2 2+ −
− =
Y’ = 2Asin6x
240 t35
p − sin
x40 t
35p −
Y’ = 2Asinx 6x
40 t sin 240 t35 35
p p − −
COM remains at the same position where it was origi-nally when Q is released from extended position (i.e., 0.5 + 0.1 from O)
MX = 0 × 1 + 2 × 0.6 ⇒ X = 2 0.6
3×
= 0.4 m
MPvPmax = mQvQmax ⇒ vQmax = 22
× 1 = 1 m s-1
192.
� m
A → Cross-sectional area of the rod
d
rl → density of liquidmg = Adrlg ⇒ m = Adrl
d = m
Arl
⇒ (when the rod is depressed by x, the extra
buoyant force ⇒ F = -Axrlg
⇒ m2
2
d xAx g
dtr= − l
a = -A g
xmrl ⇒ \ SHM
⇒ w = A g
mrl ⇒ T = 2p
mA grl
Case a ⇒ T = 2p2
solid2
rr g
p rp rl
l
= 2pgl . solidr
rl
Since l is constant ⇒ T = k sr
(b) is correct
Case b: ⇒ mass constant T = 2pm
A grl
m = Adrl ⇒ T = 2pAd d
2 K dA g g
rp
r= =l
l
193. (a) 7 nodes → 6 loops
⇒ f = 6 T
2 ml
= 6 49
4.2 0.01
= 6
704.2
× = 100 Hz
3.124 Oscillations and Waves
For Oxygen at T2, v0 = 2RT7 15 32 4
= vHydrogen
Hence its fundamental frequencies in open pipe
= Hydrogenv4 2× l
= Hydrogenv12 4l
= 01
f2
\ Ist overtone in open pipe = f0
First overtone in closed pipe = Hydrogenv 13
4 4l
= 03
f4
196. Incident wave : Y1 = 0.01sinx
200t70
− Amplitude = 0.01 m; velocity = 70 m s-1
Now the reflected wave intensity = 0.81 I (I → intensity of incident wave) Intensity ∝ (Amplitude)2
\ Amplitude of reflected wave = 0.9 × 0.01 = 0.009 m The reflection at the thick boundary will cause
a phase reversal.\ Reflected wave Y2 = -0.009sin
xt
70w +
[Q since wave progresses in the reverse direction] It can be shown that if K1 and K2 are the constants
2K
pl
= for the thin and thick wires then:
Ar = 1 2
1 2
K KK K
−+
Ai ⇒ -0.9Ai = 1 2
2
2 2
Ai2 2
p pl lp pl l
−
+
= Ai1 2
1 2
2 2v v
f f2 2
v vf f
p p
p p
−
+ = 2 1
2 1
v vAi
v v−+
solving v2 = 1v 7019 19
= m s-1 ≠ 73
m s-1
Incident wave can be considered
Y1 = 0.009sinx
200t70
− + 0.001sin
x200t
70 +
Hence the standing wave is formed by :
Y’ = x0.009sin 200t
70 −
-0.009sinx
200t70
+
The maximum intensity is wherever 2Asinx
40 t35
p − becomes maximum, i.e., Imax =
(2A)2 = 4A2 = 4I. The number of times this hap-pens is when sin(2p.20)t = ±1 in second; i.e., 2(frequency) = 40 times ⇒ Alternatively this is same as number of beats = f1 - f2 = 140 - 100 = 40
Sin240pt - 6x35
= sin ( ) 6x2 120 t
35p −
⇒ frequency = 120 Hz Hence the neutral point will be 240 times each
second. Also sin ( ) x2 20 t
35p −
will be zero,
40 times each second. Both oscillations are not synchronized
Hence total number of zero crossing is 240 + 40 = 280. At any point, say x = 0, the wave is represent by Y’ = 2Asin40ptsin240pt ⇒ starting from 0, intensity is less than I till the amplitude
term 2Asinwt = A ⇒ sinwt = 12
⇒ wt = 6p
⇒ t
= T12
. By symmetry in each cycle there are 4 such duration when intensity is less than A.
\ total duration = 4t = 4T T12 3
=
\ Fraction of time during each cycle = 13
, which is same fraction in each second when amplitude is below A (i.e., intensity I’ < I) [Note that there are integral number of cycles per second]
195. Helium in open pipe:
vHelium = ( )1
R 273 21RT 5M 3 4g +
=
f0 = Heliumv2l
=
( )R 294523 44
×
lHydrogen in closed pipe:
VHydrogen = 2RT75 2
⇒ f0 =
27RT10
4l
=
( )R 294523 4
4l(of Helium)
7 2RT10
= 4 × 5 294
R3 4
⇒ T2 = 50
29421
×
= 700 K = 427°C
Oscillations and Waves 3.125
198. (a) and (b)(i) g’ = g + a⇒ T < T0
(ii) g’ = g - a⇒ T > T0
(iii) g’ = 0⇒ T = ∞ (c) and (d) (i), (ii), (iii) T = T0
T = 2pKI
p independent of g
199. (a) Beats are produced when two waves of nearly equal frequencies interfere. The amplitude of beats varies slowly with frequency equal to the difference of the frequencies of the original waves.
→ (p), (q) are correct(b) Resonance in a closed organ pipe involves re-
flection at the closed end. Harmonics also are involved, f0, 3f0, 5f0 etc.
(r), (s)(c) Musical note is the superposition of the fun-
damental and its harmonics, amplitude slowly varies
→ (p)(r) is correct(d) Echo is produced by the reflection of a sound
wave from a rigid surface → (s) is correct
200. (a) For the given closed pipe
0
4l
= L + 0.3 D
l0 = 4L + 1.2 D
2l
= L + 0.3D × 2
l = 2L + 1.2 D
1 < 0ll
< 2 ⇒ 1 < 0
ff
< 2
(b) 4l
= L + 0.3 D
l = 4L + 1.2 D
⇒ 0
ff = 1, integer
Use: sinC - sinD = 2cosC D
2+
sin
C D2−
\ Y’ = (2 × 0.009) xcos200t.sin
70 −
⇒
Y’ = -0.018sinx
70cos200t
Obviously the residual travelling wave is
Y3 = 0.001sinx
200t70
− 197.
From Q draw tangents to the circle meeting it at S and N. At these points the source is receding or ap-proaching the listener with velocity v and that is the maximum approach-recede velocity possible, hence
maximum doppler shift. v = rw = 2p
× 2p x 2 = 2p2
= 20 m s-1[p2 ≃ 10]
At N, source approaching ⇒ f ’ = ( )340
340 20−144
= 153 HzBeats heard = 153 - 144 = 9 s-1
At S, source receding ⇒ f ’’ = 340
144360
× = 136 Hz
Beats heard = 144 - 136 = 8 s-1
At L and T velocity of approach/recede is zero hence beats heard is zero. Hence between two zero beats the maximum beats heard is either 8 s-1 or 9 s-1. Time between maximum and zero beats = time to move from N to Q
⇒ t = 360q
× T ⇒ cosq = ON R 1OQ 2R 2
= = ;
q = 60°
\ t = 60 1 1
s360 2 12
× =
Even at shifted position, the maximum velocity of receding/approach is same v at the tangential position from Q’ to circle, hence maximum beats is only 9.
