CHAPTER 14 SIMPLE HARMONIC MOTION • Simple harmonic motion • Oscillating mass on a spring • Energy in simple harmonic motion • Physical pendulum • Simple pendulum • Torsional oscillator Simple harmonic motion Simple harmonic motion (SHM) is a very basic kind of oscillatory motion. The simplest example is a mass attached to a spring. If a mass is displaced from the equilibrium position, in the absence of frictional forces it will oscillate with SHM. Let us analyze the motion of the mass on the vertical spring. Equilibrium x m m Equilibrium m y Unstretched (no mass)
Transcript
CHAPTER 14
SIMPLE HARMONIC MOTION
• Simple harmonic motion
• Oscillating mass on a spring
• Energy in simple harmonic motion
• Physical pendulum
• Simple pendulum
• Torsional oscillator
Simple harmonic motion Simple harmonic motion
Simple harmonic motion (SHM) is a very basic kind of oscillatory motion. The simplest example is a mass attached to a spring.
If a mass is displaced from the equilibrium position, in the absence of frictional forces it will oscillate with SHM. Let us analyze the motion of the mass on the vertical spring.
Equilibrium x
m
m
Equilibrium m
y
Unstretched(no mass)
If the mass is displaced by y from its equilibrium position, the restoring force acting on the mass is given by Hookes Law, i.e.,
Fy = −ky,
where k is the spring constant.By Newton’s 2nd Law
may = Fy,
i.e., m
d2ydt2 = −ky.
∴
d2ydt2 = −
km
y.
This expression defines SHM, i.e., if the acceleration of an object is proportional to its displacement and is oppositely directed, the object will move with SHM.
A solution to the differential equation is
y(t) = A cos(ωt + δ),where A, ω and δ are constants.
m
Equilibrium
y
If the mass is displaced by y from its equilibrium position,
the restoring force acting on the mass is given by Hooke’s
Law, i.e.,
y(t) = Acos(ωt + δ)
The maximum
displacement, A, is the
amplitude, is the
angular frequency, is
the phase angle, which is
determined by the
displacement at Take the first and second derivatives
of y with respect to t, then
and
So, is a solution provided
The frequency of oscillations is and the periodic
time (for one complete oscillation) is
ω
δ
t = 0.
dydt
= v(t) = −ωAsin(ωt + δ),
d2y
dt2= a(t) = −ω2Acos(ωt + δ) = −ω2y.
y(t) = Acos(ωt + δ) ω2 = k
m.
f = ω
2π
T = 1
f.
∴T =
12π
km
.
Note that the period of SHM is independent of the amplitude. (In music, it means that the pitch or frequency of a note struck on any stringed instrument, for example, is independent of how loudly the note is played.)
We would obtain the same type of solution for a horizontal spring, viz:
x(t) = A cos(ωt + δ).
Question 8: A 0.40kg mass attached to a spring of force constant 12N/m oscillates with an amplitude of 8cm. Find (a) the maximum speed of the mass, (b) the speed and acceleration of the mass when it is at y = 4cm from the equilibrium position, and (c) the time it takes the mass to travel from y = 0 to y = 4cm.
∴T = 2π
ω= 2π m
k and f = 1
T= 1
2πkm
.
Question 14.1: A 0.40 kg mass, attached to a spring with
force constant 12.0 N/m, oscillates with an amplitude of
8.0 cm. Find
(a) the maximum speed of the mass,
(b) the speed and acceleration of the mass when it is at
from the equilibrium position, and
(c) the time it takes the mass to travels from to y = 4.0 cm
y = 0
y = 4.0 cm.
The solution for the displacement is
The mass starts to oscillate at when i.e., at
maximum displacement (amplitude) so
(a) The speed of the mass is so the
maximum speed is
Maximum speed occurs when i.e., when
(b) When we have
y(t) = Acos(ωt + δ).
t = 0 y = 8.0 cm,
δ = 0.
v = dy
dt= ωAsin(ωt) ,
vmax = ωA = Akm
= (0.08 m)12.0 N/m0.40 kg
= 0.438 m/s.
y = 0.04 m,
0.04 m = (0.08 m)cosωt ⇒ωt = cos−1(0.5) = π3
.
∴v = ωAsinωt = (0.438 m/s)sinπ3
⎛⎝⎜
⎞⎠⎟
= (0.438 m/s)3
2= 0.379 m/s.
ωt = (2n +1) π
2,
y(t) = 0.
(c) When
From (b), when
y = 0,
Acosωt = 0⇒ωt = cos−10 = π
2,i.e., t = π
2ω.
y = 4.0 cm, ω ′t = π
3, i.e., ′t = π
3ω.
∴Δt = 1ω
π2− π
3⎛⎝⎜
⎞⎠⎟= π
6ω= π
60.40 kg
12.0 N/m= 0.096 s ( = 96 ms).
a = d2y
dt2= ω2Acosωt = vmaxωcos
π3
⎛⎝⎜
⎞⎠⎟
= (0.438 m/s)12 N/m0.40 kg
× 0.5 = 1.20 m/s2.
The acceleration is
Question 14.2: Two identical springs each have a spring
constant A mass, is connected to
the springs in two different configurations, as shown
above. What is the periodic time for each configuration?
k = 20 N/m.
M
M
(a)
(b)
M = 0.3 kg,
(b) Give the mass a
displacement x. One
spring is stretched a distance x, the other is compressed a
distance x and so they each exert a force of magnitude
on the mass in a direction opposite to the
displacement. Hence, the total restoring force is
Again, the effective spring constant which is
the same as in (a). So, the periodic times are the same.
(a) Consider what happens when the mass
is given a displacement y (>0). Since
each spring is stretched a distance y, the
net restoring force (by Hooke’s Law) is
So, the effective spring constant is Hence, F = −(20 N/m)y − (20 N/m)y
= −(40 N/m)y. ′k = 40 N/m.
T = 2π M
k= 2π 0.3 kg
40 N/m= 0.54 s.
M
(a)
M
(b)
(20 N/m)x
F = −(20 N/m)x − (20 N/m)x = −(40 N/m)x.
′k = 40 N/m,
A
A
UE = 1
2kA2 : K = 0
UE = 0 : K = 1
2mv2
UE = 1
2kA2 : K = 0
UE = 0 : K = 1
2mv2
UE = 1
2kA2 : K = 0
x
Mechanical
Energy E = UE + K = 1
2kA2
The total energy is the initial energy, i.e., E = 1
2kA2.
Energy in simple harmonic motion
Question 14.3: A block of mass M is connected to a
horizontal spring on a frictionless table. The spring is
compressed a distance A and released so the mass executes
simple harmonic motion.
(a) Deduce an expression for the velocity of the mass at
any displacement x.
(b) When the displacement is , what fraction of the
mechanical energy is kinetic energy?
(c) At what displacement, as a fraction of A, is the
mechanical energy half kinetic and half elastic potential
energy?
A
2
(b) Since the total energy is when the
displacement is , the elastic potential energy is
(c)
E = 1
2kA2,
UE = 12
k A2( )2 = 1
8kA2 = 1
4E.
∴K = E − UE = 34
E.
UE = 1
2kx2 = 1
2E = 1
4kA2.
∴x = A
2.
(a) The total energy of the system is
But the total energy is the initial elastic potential energy, E = 1
2kx2 + 1
2Mv2.
i.e., E = 12
kA2.
∴ 12
Mv2 = 12
kA2 − 12
kx2,
i.e., v = kM
A2 − x2( ).
A
2
Question 14.4: A 3.0 kg mass on a frictionless horizontal
surface attached to a spring, oscillates with an amplitude
of 8.0 cm. If its maximum acceleration is
(a) what is the total mechanical energy?
(b) What is the maximum kinetic energy of the mass?
3.5 m/s2,
(a) The total energy is We need to
determine the spring constant k. The force exerted by the
spring is
so the maximum acceleration occurs when
but when
(b) Maximum kinetic energy occurs when
E = 1
2mv2 + 1
2kx2.
F = ma = −kx, i.e., k = − ma
x,
x = −A.
∴E = 1
2mv2 + 1
2mamax
A
⎛
⎝⎜⎞
⎠⎟A2,
v = 0 x = ±A.
∴E = 12
mamaxA
= 12
(3.0 kg)(3.5 m/s2)(0.080 m)
= 0.42 J.
x = 0.
∴Kmax = 12
mvmax2 = 0.42 J
i.e., vmax = 2(0.42 J)3.0 kg
= 0.53 m/s.
Physical pendulum
Any rigid object that is free to rotate will oscillate about its equilibrium position when displaced and released. We show now that when such an object is released, its
motion will be SHM. Using Newton’s 2nd Law for rotation
I! α =! τ ,
we have
Id2θdt2 = −(Dsin θ)Mg .
If θ is small, sin θ→ θ, then Id2θdt2 = −MgDθ,
i.e., d2θdt2 = −
MgDI
θ.
Note that this equation satisfies our definition of SHM. A suitable general solution is
θ = Acos(ωt + δ).
cmθ
M! g
D
θ(t) = θmax cos(ωt + δ).
∴
dθdt
= −ωθmax sinωt
and
d2θdt2 = −ω2θmax cosωt = −ω2θ.
Thus, if θ = θmax cosωt is a solution,
ω2 =
MgDI
, i.e., ω =
MgDI
.
But the swing frequency is f = ω2π and the period for
one complete oscillation is T = 1f .
∴T =
2πω
= 2πI
MgD.
Note that providing θ is small the periodic time is independent of the amplitude.
Note that providing is small, the periodic time is
independent of the amplitude. Note also, that I is the
moment of inertia about the pivot point.
θ
Question 14.5: A uniform right-angle square is suspended
by a thin nail so that it pivots freely, as shown above. Each
arm has length and mass If it is
given a small angular displacement, in the plane of the
square, and released, what is the periodic time of the
oscillations?
ℓ = 25 cm m = 0.25 kg.
ℓ ℓ
The periodic time of oscillations of a physical pendulum is
where I is the moment of inertia about the rotation axis,
is the total mass and D is the distance from the
rotation axis to the center of mass. The moment of inertia
about the nail is
and the distance D is
T = 2π I
MgD,
ℓ ℓ
cm D
I = mℓ2
3+ mℓ2
3= 2mℓ2
3,
M (= 2m)
ℓ2
cos45" = ℓ
2 2.
∴T = 2π 2mℓ22 23(2m)gℓ
= 2π 2ℓ 23g
= 2π 2(0.25 m) 2
3(9.81 m/s2)= 0.97 s.
Question 14.6: A uniform disk with a radius and
a mass of 6.00 kg, has a small hole a distance D from the
center of the disk. If the hole is a pivot point such that the
disk can swing freely, what should be the distance D so that
the periodic time of this physical pendulum in 2.50 s?
D
0.80 m
R = 0.80 m
We know Using the parallel axis theorem,
the moment of inertia of the disk about the pivot point is
Squaring both sides we get
The two roots of the equation are
Clearly, the latter is the appropriate solution.
T = 2π I
MgD.
I = Icm + mD2 = 12
mR2 + mD2.
∴T = 2π
12
mR2 + mD2
mgD= 2π
12
R2 + D2
gD.
D2 − gT2
4π2D + R2
2= 0
= D2 − (9.81 m/s2)(2.50 s)2
4π2D + (0.80 m)2
2
= D2 − (1.553 m)D + (0.320 m).
D = 1.31 m and D = 0.24 m.
Simple pendulum
The simple pendulum is a special case of the physical (or compound) pendulum. In this case, if we ignore the mass of the string, the moment of inertia of the bob of
mass m about the pivot point is
I = mℓ2,
and the distance of the center of mass of the bob fom the pivot point is
D = ℓ. The solution is again
θ = θmax cosωt ,
but with ω =
mgDI
=mgℓmℓ2
=gℓ
. The period for one
entire oscillation is T = 1f = 2π
ω, i.e.,
T = 2π
ℓg
.
Note that we can write the arc length as s = ℓθ, so that
" T
θ
θ
" F r
" F t
m" g
s
ℓ
m
s = ℓθmax cosωt .
• The tangential speed of the bob is
vt =
dsdt
= −ωℓθmax sinωt ,
which is zero when ωt = nπ, i.e., when s = smax = ℓθmax .
The tangential speed is a maximum when sinωt = 1, i.e.,
when ωt =
2n +12
π . That occurs when s = 0.
• The tangential accceleration of the bob is
a t =
d2sdt2 = −ω2ℓθmax cosωt ,
which is zero when ωt =
2n +12
π , i.e., when s = 0. The
tangential acceleration is a maximum when sinωt = 1, i.e., when ωt = nπ. That occurs when s = smax = ℓθmax .
Question 14.7: The pendulum shown above is an
“interrupted” pendulum. There is a peg P below the pivot
point O, so when the pendulum swings to the left, the
string makes contact with the peg, which causes the bob to
swing on a different arc. If the length of the string is
and the distance OP is one-half the length of the
string, what is the periodic time of the pendulum?
78.0 cm
O
P
We can consider the pendulum in two parts … when the
bob swings from the right, the one-quarter period is
and when it swings past the
vertical, the one-quarter period is
Hence, the total period is
The behavior of the interrupted pendulum was discussed
by Galileo … can you think why?
ℓ
ℓ
2
T1 =
14
⎛⎝⎜
⎞⎠⎟
2π ℓg= π
2ℓg
,
T2 = 1
4⎛⎝⎜
⎞⎠⎟
2πℓ
2g
= π2ℓ
2g.
2(T1 + T2) = 2 π2ℓg+ π
2ℓ
2g
⎛
⎝⎜
⎞
⎠⎟ = π 1+ 1
2
⎛⎝⎜
⎞⎠⎟ℓg
= 1.71π 0.78 m
9.81 m/s2= 1.51 s.
Question 14:8: Consider two pendulums, a simple
pendulum of length 1 m and a 1 m ruler pivoted at one
end. Which one, if either, has the smaller periodic time?
The periodic time of a simple pendulum is
For a ruler of length pivoted at one end, and
So, the ruler pendulum has the shorter periodic time.
Ts = 2π ℓ
g.
∴Tr = 2π IMgD
== 2π Mℓ2
3Mg2ℓ= 2π 2ℓ
3g,
i.e., Tr = Ts23= 0.816Ts. ∴Tr < Ts.
Ts = 2π 1 m
9.81 m/s2= 2.00 s, so Tr = 1.64 s.
ℓ I = Mℓ2
3
D = ℓ
2.
Torsional oscillator
A torsional oscillator, or torsional pendulum, consists of a
disk-like mass suspended from a wire or thin rod. If the
mass is twisted about the axis of the
wire, the wire exerts a restoring
torque. If twisted and released, the
mass will oscillate back and forth
with simple harmonic motion. If the
angle is sufficiently small so the
wire is not plastically deformed, the restoring torque is
where is the torsion constant. If the moment of inertia of
the mass is I, the angular acceleration of the mass is
which defines simple harmonic motion. The solution is
θθ
τ = −κθ,κ
α = τ
I, i.e., d2θ
dt2= −κ
Iθ,
θ(t) = θmax cos(ωt + δ),
where is the angular amplitude and is the phase
angle, which are both determined by the initial conditions,
and
where T is the periodic time of the oscillation. Therefore, ω = κ
I= 2π
T,
θmax δ
T = 2π I
κ.
Question 14.9: What is the periodic time of torsional
pendulum consisting of a uniform solid sphere of mass
400 g and radius 8.0 cm, which is suspended from a torsion
wire 10 m long. The wire twists when a torque of
is applied to it.
90!
0.04 N ⋅m
First we need to determine the torsion constant.
The moment of inertia of the sphere is
Note that it does not depend on the length of the wire.
τ = −κθ, i.e., κ = τθ= 0.40 N ⋅m
π2
= 0.25 N ⋅m/rad.
I = 25
mr2 = 25
(0.40 kg)(0.08 m)2 = 1.02 ×10−3 kg ⋅m2.
∴T = 2π Iκ= 2π 1.02 ×10−3 kg ⋅m2
0.25 N ⋅m/rad= 0.40 s.
Question 14.10: A uniform bar is suspended in a horizontal
position by a vertical wire attached to its center. When a
torque of is applied to the bar, the bar rotates
through an angle of When released, the bar oscillates
as a torsion pendulum with a period of 0.50 s. Find the
moment of inertia of the bar.
5.0 N ⋅m
12!.
12! F
F
For a torsion oscillator
In this case
If we know the length (or mass) of the bar we can find its
