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SHM (Complete in 5 Lecture)

Introduction

Our hearts beat, our lungs oscillate, we shiver when we are cold, we sometimes snore, we can hearand speak because our eardrums and larynges vibrate. We cannot even say vibration properlywithout the tip of the tongue oscillating.

Periodic motion : Any motion in which all the parameters of motion are repeated after a definite time interval,is called as periodic motion. The time which elapses between successive passages in the same direc-tion through any point is termed the periodic time, or the period, of the motion.

The number of periods comprised in one second is called the frequency ; thus, if T is the period, and nis the frequency of a periodic motion, we have n = I/T.

The motion of the hands of a clock is periodic, the period of the motion of the minute hand being onehour, or 3600 seconds. The bob of a pendulum moves periodically, the period being equal to the timeof one complete (to and fro) oscillation. Periodic motion can be along any path.

Oscillatory motion : If in a case of periodic motion particle moves to and fro on the same path, the motionis said to be oscillatory motion. A body that undergoing oscillatory motion always does so about astable equilibrium position. When it is moved away from this position and released, it experiences anet force or torque to pull it back toward equilibrium position. But by the time it gets there, therestoring force/torque would have done some positive work on it. Thus it must have gained somekinetic energy, so it overshoots, the equilibirium position. Now it stops somewhere on the other side,and is again pulled back toward equilibrium. Imagine a ball rolling back and forth in a round bowl ora pendulum that swings back and forth past its lowest position.

Asking Question :

A particle hanging from a thread is launched horizontally such that it completes the circle. Is the motionperiodic? Is the motion oscillatory?

Sol. Motion is periodic but not oscillatory as the particle is not moving to and fro.

Asking Question :

A particle hanging from a thread is launched horizontally with small speed such that it does even reachto the horizontal level. Is the motion periodic? Is the motion oscillatory?

Sol. Motion is periodic and oscillatory as the particle is moving to and fro repeatedly.

TEACHING NOTES

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Imp. characteristics of oscillatory motion :

1. When a particle in stable equilibrium is disturbed, then it has tendency to return to the position ofequilibrium and this tendency is exhibited as oscillatory motion.

2. The force on the body acts towards the mean position i.e. Force is always opposite to the displace-ment vector of the particle w.r.t. mean position. (This force is known as restoring force)

F = r , =

Where r and are displacements and angular displacement from mean position.

3. Energy is also conserved. If enrgy is not conserved than the particle will not be able to repeat theparameters of the motion.

SHM

Optional Explanation of why we study SHM :

Sinusoidal Vibrations : Our study will be mostly related to sinusoidal vibrations which we will show later onarise when net force experienced by an oscillating body has magnitude proptional to the distance fromthe mean position or torque is directly propotional to the angular displacement from mean position.There are two reasons for this one physical, one mathematical, and both basic to the whole subject.The physical reason is that purely sinusoidal vibrations are common in many type of mechanical sys-tems. Such motion is almost always possible if the displacements are small enough. If, for example, wehave a body attached to a spring, the force exerted on it at a displacement x from equilibrium is actually

F(x) = (k1x + k2x2 + k3x

3 + .........0

where k1, k2, k3, etc., are a set of constants, and we can always find a range of values of x within whichthe sum of the terms in x2, x3, etc., is negligible, compared to the term k1x. If the body is of mass mand the mass of the spring is negligible, the equation of motion of the body then becomes

m 22

dtxd

= k1x

It is easy to verify, that above equation is satisfied by an equation of the form

x = A sin (t + 0)

where = (k1/m)1/2. Thus sinusoidal vibration or simple harmonic motion is likely possibility in small

vibrations. But we should remember that in general it is only an approximation (although perhaps avery close one) to the true motion.

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The second reason is the mathematical one. The actual importance of purely sinusoidal vibrations is isproved a famous theorem given by the French mathematician J.B. Fourier in 1807. According toFouriers theorem, any periodic function with a period T can be considered as sum of of pure sinusoi-dal vibrations of periods T, T/2, T/3, etc., with appropriately chosen amplitudes. A thorough familiaritywith sinusoidal vibrations will be stepping stone for our understanding of every conceivable probleminvolving periodic phenomena.

TYPES OF SHM(a) Linear SHM : When a particle moves to and from about an equilibrium point, along a straight line. A

and B are extreme positions. M is mean position AM = MB = Amplitude.

A BM

(b) Angular SHM : When body / particle is free to rotate about a given axis executing angular oscillations.

EQUATION OF SHM :The necessary and sufficient condition for SHM is

F = kxwhere k = positive constant for a SHM = Force constant

x = displacement from mean position

or m 22

dtxd

= kx

2

2

dtxd

+ mk

x = 0 [differential equation of SHM]

2

2

dtxd

+ 2x = 0 where = mk

Its solution is x = A sin (t + )

DERIVATIONF = kx

a = mk

x

put mk

= 2

a = 2x

v

0vdv =

x

A

2xdx

2v2

= 2

2 [x2 AA2]

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v = 22 xA

dtdx

= 22 xA

22A

dx = dt

sin1 x/A = t + kat t = 0 x = x0x = A sin {t + }

CHARACTERISTICS OF SHMNote : In the figure shows, path of the particle is on a straight line.

(a) Displacement : It is defined as the distance of the particle from the mean position at that instant.Displacement in SHM at time is given by x = A sin (t + )

(b) Amplitude : It is the maximum value of displacement of the particle from its equilibrium position

Amplitude = 1/2 (distance between extreme points/position)It depends on energy of the system.

(c) Angular Frequency () : = T2

= 2f and its units is rad/sec.

(d) Frequency (f) : Number of oscillations completed in unit time interval is called frequency of oscillations,

f = T1

= 2 , its units is sec

1 or Hz.

(e) Time period (T) : Smallest time interval after which the oscillatory motion gets repeated is called time

period, T = 2

= 2km

Ex. For a particle performing SHM, equation of motion is given as 22

dtxd

+ 4x = 0. Find the time period.

Sol. 22

dtxd

= 4x 2 = 4 = 2

Time period : T = 2

=

Objective : Determination of from basic SHM equation.

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(f) Phase : The physical quantity which represents the state of motion of particle (eg. its position anddirection of motion at ay instant.) argument of sine function is called phase.

(g) Phase constant () : Constant in equation of SHM is called phase constant or initial phase. Itdepends on initial position and direction of velocity.

(h) Velocity (v) : It is the rate of change of particles displacement w.r.t. time at that instant.Let the displacement from mean position is given by

x = A sin (t + )

Velocity, v = dtdx

= dtd

[A sin (t + )]

v = A cos (t + )or, v = 22 xA At mean position (x = 0), velocity is maximum

vmax = AAt extreme position (x = A), velocity is minimum.

vmin = zeroGRAPH OF SPEED (v) VS DISPLACEMENT (x) :

v = 22 xA v2 = 2 (A2 x2)

v2 + 2x2 = 2A2 1Ax

Av

2

2

2

2

Graph would be an ellipse

(i) Acceleration : It is the rate of change of particle velocity w.r.t. time at that instant.

Acceleration, a = dtdv

= dtd

[A cos (t + )]

a = 2A sin (t + )a = 2x

NOTE : Negative sign shows that acceleration is always directed towards the mean position.At mean position (x = 0), modulus of acceleration is minimum

amin = zeroAt extreme position (x = A), acceleration is maximum

amax = 2A

GRAPH OF ACCELERATION (A) VS DISPLACEMENT (x) :a = 2x

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GRAPHICAL REPRESENTATION OF DISPLACEMENT, VELOCITY AND ACCELERATIONIN SHM

Displacement, x = A sin t

Velocity, v = A cos t = A sin (t + 2

)

or v = 22 xA Acceleration, a = 2A sin t = 2 A sin (t + )or a = 2x

Note : v = 22 xA a = 2x

These relations are true for any equation of x.

0A0A0a,onAcceleratiA0A0Av,Velocity0A0A0x,ntDisplacemeT4/T32/T4/T0t,Time

22

1. All the three quantities displacement, velocity and acceleration vary harmonically with time, havingsame period.

2. The velocity amplitude is times the displacement amplitude (vmax = A).

3. The acceleration amplitude is 2 times the displacement amplitude (amax = 2A).

4. In SHM, the velocity is ahead of displacement by a phase angle of 2

.

5. In SHM, the acceleration is ahead of velocity by a phase angle of 2

.

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Ex. A particle executing simple harmonic motion has angular frequency s1 and amplitude 1 cm. Find (a)the time period, (b) the maximum speed (c) the maximum acceleration, (d) the speed when thedisplacement is 0.6 cm from the mean position, (e) the speed at t = /6 s assuming that the motionstarts from rest at t = 0.

Objective : Application of basic formula of SHM.

{Home Work : HCV Q.1 to Q.7 }Phasor

One of the best ways of describing simple harmonic motion is obtained by considering it as theprojection of uniform circular motion. Imagine, for example, that a disk of radius A rotates about ahorizontal axis at the rate of rad/sec. Suppose that a peg P is attached to the edge of the disk andthat a horizontal beam of parallel light casts a shadow of the peg on a vertical screen, as shown in Fig.(a). Then this shadow perform simple harmonic motion with period 2/ and amplitude A along avertical line on the screen.

AP

A

xO

A

Screen

Parallel light

(a)

AP

AxOA

Screen

Parallel light

(a)

X

AP

O x

(b)

More abstractly, we can imagine SHM as being the geometrical projection of uniform circular motion.(By geometrical projection we mean simply the process of drawing a perpendicular to a given line fromthe instantaneous position of the point P.) Let the tracing point, P, revolve uniformly at a constantdistance OP from O, the direction of motion being anticlockwise. Through O we draw the rectangularaxes XOX and YOY. From P draw PQ perpendicular to OY. Thus OQ is the component of OP

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resolved parallel to OY. As P revolves about O, the point Q moves up and down along YOY, itsmotion being of the simple harmonic type. When P pases across the axis XOX, Q will pas throughO. When P passes across the axis YOY, Q will be at its position of maximum displacement from itsmean position, O.

The maximum displacement of Q will thus be equal to the radius a of the circular path of P ; it is termedthe amplitude of the s.h.m.

The phase of the s.h.m. at any instant is equal to the angle which has been swept out by the line OP. Ifwe measure the phase from the particular position of OP when Q is moving in a certain direction (sayupwards) through O, the angle XOP = is equal to the phase of the s.h.m.

QOC

b

r

p

oR X

P

X

Y

Y

O

The displacements of Q for various values of the phase angle , are shown by the curve to the right ofFig. ( ). The x coordinate, is equal to the circular measure of the angle XOP, and the distance OQ isplotted on y axis.

Let the amplitude OP = a, while OQ = y. Then y/a = RP/OP = sin . Therefore y = a sin .

plotted to the right of the figure is described by this equation.

Further, if OP completes a rotation (i.e. rotate through an angle 2) in the period T ; then we have =2t/T, and therefore y = a sin (2t/T).

When the tracing point P crosses the axis OX, it is moving, for the instant, parallel to the axis OY.Consequently, at this instant the point Q is moving along OY with a velocity equal to that of the tracingpoint P in its circular path. The velocity of P is equal 2a/T. Hence, the velocity of Q as it passesthrough its mean position O is equal to 2a/T.

When the tracing point P crosses the axis OY, it is moving, for the instant, in a direction perpendicularto OY. At this instant point Q will be stationary. Consequently, the point Q is stationary for an instantat the extreme, on either side of its mean position O.

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Y

Y

XXO

Q

Q P

PS

R R

The velocity corresponding to any phase of a s.h.m. can be determined readily. Let the vector OP(Fig.), of length a, revolve uniformly about O in an anticlockwise direction, its perod being equal to T; and let the projection Q of the point P on the axis YOYis the simple harmonic motion.

Thus OQ = y = a sin (2t/T). Differentiating it with repect to time we get

V = Ta2

cos ( Tt2

).

This gives the instantaneous velocity of the simple harmonic motion at the time t. In terms of the phase

angle , the velocity is equal to Ta2

cos . When = O ; cos = 1, and the velocity is equal to 2a/

T, as already discussed ; when = /2, cos = O, and the velocity is equal to zero ; and so on.

Thus, a point executing a s.h.m. moves with a maximum velocity equal to 2a/T on passing through itsmean position. Its velocity diminishes as it moves away from its mean position, being equal to (2a/T)cos when the phase angle is , and attaining the value zero at the extreme position. Subsequently, asthe point returns towards its mean position, its velocity increases, and once more attains the value 2a/T on moving through the mean position.

The use of a uniform circular motion as a purely geometrical basis for describing SHM embodies morethan we have so far chosen to recognize. This circular motion, once we have set it up, defines SHMof amplitude A and angular frequency along any, straight line in the plane of the circle.

Very Easy Ex.(a)A particle starts from mean position and moves towards positive extreme as shown. Find the equationof the SHM. Amplitude of SHM is A.

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Also write the equation of SHM for the situation shown below.

(b)

(c)

(d)

Ans. (a) x = A cos t ; (b) x = A cos t ; (c) x = A sin (t + 150)

Objective : To calculate initial phase angle & write SHM equation.

Ex. Write equation of S.H.M of angular frequency and A amplitude if the particle is situated at 2A

at

t = 0 and is going towards mean position.

Sol.

At t=0 particle was at 2A

and was going towards mean position as shown in the figure(a) below..

AMAA/ 2

The same situation is described by making reference point on acircle of radius A as shown in figure(b)

below. A/ 2

The reference point can be located at any of the two positions G and H as shown in figure for having displace-

ment 2A

, but to move towards mean position it should be H. Position H Corrospondes to angle

.

Thus = 43

and x = A sin (t + 43

)

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Alternative Mathematical solution Putting t= 0 in equation x = A sin (t + )

We get 2A

= A sin

Thus, = 4

, 43

but at t = 0 ; V < 0 so 4

is rejected

x = A sin (t + 43

)

Objective : the ease of using phasor diagram is quite evident in this solution.

Ex. If initial position of particle performing SHM is +A/2 & moving towards centre having angular frequency find minimum time when it will be at(1) A/2 (2) Mean postion (3) A (4) + A

Ex. x = A cos ( 4

t)

Convert this equation into standard form

Objective : sign of A and should be positive

Ex. Find distance traveled by a particle of time period T & amplitude A in time T/12 starting from rest.

Sol. AMAxt t = 0

l

Eqn of SHM x = A cos t

xt = A cos

12

2 TT

xt = 23A

l = A 2

3A

l = 2

)32( A

Objective : Explain SHM is not linear motion

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Ex. Two particles P and Q are executing SHM across same straight line whose equations are given as yP = A sin (t + 1) and yQ = A cos(t + 2). An observer, at t = 0, observers the particle P at a

distance 2A moving to the right from mean position O while Q at 23

A moving to the left from

mean position O as shown. Find, (2 1)

Sol. yP = A sin (t + 1)at t = 0

2A

= Asin1 sin1= 21

1 = /4, 3/4

& vP = Acos1 since vP = (+)ive so, 1 = /4

yQ = A cos(t + 2).at t = 0

2A3

= Acos2 cos2= 23

2 = /6 = 5/6

+ /6 = 7/6& vQ = Asin2, since vQ = ()ive so, 2 = 5/6

= 5/6 /4 = 12310

= 7/12Ans.

Objective : Application of phasor diagram

NOTE : If mean position is not at the origin, then we can replace x by x x0 and the equation becomesx x0 = A sin t, where x0 is the position co-ordinate of the mean position.

Ex. A particle is moving on Y-axis under a variable force F = ky + c.

(a) Is the motion of the particle can be called SHM ? Hence or otherwise write y position of particle asfunction of time.

Sol.

We can write the force as

F = k (y kC

)

lets define s = y kC

,..........................................eqn()

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Where s is displacement from mean position because at y = kC

, F is zero.

Thus mean position is at y = ( kC

)

Thus force can be written as F = -ks...................................()

which is equation of SHM about y = kC

Differentiating eqn() wrt time twice we get 22

dtyd

= 22

dtsd

Thus F = m 22

dtyd

can also be written as F = m 22

dtsd

Substituting in eqn ()

m 22

dtsd

= ks

Comparing with standard eqn we know that the solution is

s = A sin ( t + )

= mk

Replacing with value of s = y - kC

We get y = A sin (t + ) + kC

(b) If in the previuos question F = 10y + 20, and mass of the particle is m = 2.5 kg and is released fromrest from y = +3, at t = 0. Write the equation of SHM.

Sol. From previuos discussion we can write

y = A sin (t + ) + kC

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where, k =10 and C = 20, thus = mk

, gives us = 2

Mean position is 2 as force at y = 2 is zero F = 10 x 2 + 20 = 0

y = A sin (2t + ) + 2

at t = 0 the particle is at y = 3 and is at rest thus its extreme position is y = 3. the distancebetween the mean position and extreme position is 1m. Thus amplitude A = 1.

at t = 0

1 sin + 2 = 3

` sin = 1 = 2

Ans. cos 2t + 2

(Home Work : H.C.V. Q. 8 to 11)

ENERGY OF SHM :Kinetic Energy (KE)

21

mv2 = 21

m2 (A2 x2) = 21

k (A2 x2) (as a function of x)

= 21

mA22 cos2 (t + ) = 21

KA2 cos2 (t + ) (as a function of t)

KEmax = 21

kA2 ; 2T0 kA41KE ;

2A0 kA3

1KE

Frequency of KE = 2 (frequency of SHM)

Potential Energy (PE)

21

Kx2 (as a function of x) = 21

kA2 sin2 (t + ) (as a function of time)

Total Mechanical Energy (TME)Total mechanical energy = Kinetic energy + Potential energy

= 21

k(A2 x2) + 21

Kx2 = 21

KA2

Hence total mechanical energy is constant in SHM.

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Graphical Variation of energy of SHM

Ex. A particle of mass 0.50 kg executes a simple harmonic motion under a force F = (50 N/m)x. If itcrosses the centre of oscillation with a speed of 10 m/s, find the amplitude of the motion.

SPRING-MASS SYSTEM

T = 2km

Ex. A particle of mass 200 g executes a simple harmonic motion. The restoring force is provided by aspring of spring constant 80 N/m. Find the time period.

(1) (2)

Sol. The time period is

T = km2 =

m/N80kg102002

3 = 2 0.05 s = 0.31s.

Explain that in this case along with variable force constant force of mg is acting but still the motion isSHM and for this SHM equation is F = Ax + B for above figure A = k and B = mg

Objective : Application of basic formula.

Ex. A block of mass 100gm attached to a spring of spring constant 100N/m is lyingon a frcitionless floor as shown. In natural length the block is moved to compressthe spring by 10cm and then released. If the collisions with the wall in front areelastic then find the time period of the motion.

Ans 0.133 sec

Objective : Example of incomplete SHM, application of phasor diagram.

Qus. Block A of mass is performing SHM of amplitude a. Another block B of mass m is gently placed on Awhen it passes through mean position and B sticks to A. Find the time period and amplitude of newSHM.

Ans. T = 2 Km2

amplitude = 2a

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Qus. Repeat the above problem assuming B is placed on A at distance 2a

from mean position.

Ans. T = 2 Km2

amplitude = a85

Ex. The left block in figure collides inelastically with the right block and sticks to it. Find the amplitude ofthe resulting simple harmonic motion.

Ex. Two blocks of mass m1 and m2 are connected with a spring of natural length and spring constant k.The system is lying on a smooth horizontal surface. Initially spring is compressed by x0 as shown infigure.Show that the two blocks will perform SHM about their equilibrium position. Also (a) find the timeperiod, (b) find amplitude of each block and (c) length of spring as a function of time.

Objective : To explain SHM in non inertial frame.

OptionalEx. A string of normal length l & force constant k is released from ceiling. Find its time period?

Ml/2

[Assume force constant k is not varying with extension is string]

{Home Work : HCV Q.11 to Q.28 }

METHODS TO DETERMINE TIME PERIOD, ANGULAR FREQUENCY IN SHM(a) Force / torque method(b) Energy method

Ex. The string, the spring and the pulley shown in figure are light. Find thetime period of the mass m.

Sol. (a) Force MethodLet in equilibrium position of the block, extension in spring is x0. kx0 = mg ......... (1)

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Now if we displace the block by x in the downward direction, net force on the block towards meanposition is.

F = k(x + x0) mg = kx using (1) Hence the net force is acting towards mean position and is alsoproportional to x. So, the particle will perform S.H.M. and its time period would be.

T = 2km

(b) Energy MethodLet gravitational potential energy to be zero at the level of the block when spring is in its natural length.Now at a distance x below that level, let speed of the block be v.Since total mechanical energy is conserved in SHM

mgh + 21

kx2 + 21

mv2 = constant

Differentiating w.r.t. time, we getmgv + kxv + mva = 0where a is acceleration

F = ma = kx + mg or F = k

k

mgx

This shows that for the motion, force constant is k and equilibrium position is x = kmg

.

So, the particle will perform S.H.M. and its time period would be T = km2 .

Objective : Write SHM equation & determine .

Qus. Solve the above problem if the pulley has a moment of inertia I about its axis and the string does not slipover it.

Ans.k

)r/Im(22

SIMPLE PENDULUMIf a heavy point-mass is suspended by a weightless, inextensible and perfectly flexible string from arigid support, then this arrangement is called a simple pendulum.

Time period of a simple pendulum T = 2 g

(some times we can take g = 2 for making calculation simple)

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Note : (a) If angular amplitude of simple pendulum is more, then time period

T = 2 g

161

20

(For other exams)

where 0 is in radians.(b) General formula for time period of simple pendulum.

T = 2

1

R1g

1

(c) On increasing length of simple pendulum, time period increases, but time period of simple

pendulum of infinite length is 84.6 min which is maximum and is equal to T = 2 gR

.

(Where R is radius of earth)(d) Time period of seconds pendulum is 2 sec and = 0.993 m.(e) Simple pendulum performs angular S.H.M. but due to small angular displacement, it is considered

as linear SHM.(f) If time period of clock based on simple pendulum increases then clock will be slow but if time

period decrease then clock will be fast.

(g) If g remains constant and is change in length, then TT

100 =

21

100

(h) If remains constant and g is change in acceleration then, TT

100 = 21

gg

100

(i) If is change in length and g is change in acceleration due to gravity then,

100gg

21

21100

TT

Time Period of simple Pendulum in accelerating Reference Frame :

T = .effg

2 where

geff. = Effective acceleration due to gravity in reference system = |ag|

a = acceleration of the point of suspension w.r.t. ground.Condition for applying this formula: |ag| = constant.

Ex. A block of mass m is suspended from the ceiling of a stationary elevator through a spring of springconstant k it is in equilibrium. Suddenly, the cable breaks and the elevator starts falling freely. Showthat block now executes a simple harmonic motion of amplitude mg/k in the elevator.

Objective : Explain SHM in moving frame & discuss motion of particle from cabin frame.

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Ex. A simple pendulum is suspended from the ceiling of a car accelerating uniformly on a horizontal road.If the acceleration is a0 and the length of the pendulum is , find the time period of small oscillationsabout the mean position.

Sol. We shall work in the car frame. As it is accelerated with respect to the road, we shall have to apply apsuedo force ma0 on the bob of mass m.For mean position, the acceleration of the bob with respect to the car should be zero. If be the anglemade by the string with the vertical, the tension, weight and the psuedo force will add to zero in thisposition.

Hence, resultant of mg and ma0 (say F = m20

2 ag ) has to be along the string.

tan 0 = mgma 0 = g

a0

Now, suppose the string is further deflected by an angle as shown in figure.Now, restoring torque can be given by

(F sin ) = m2Substituting F and using sin = , for small .

2202 m)agm(

or, =

20

2 agso, 2 =

20

2 ag

This is an equation of simple harmonic motion with time period

T = 2

= 2 4/120

2 )ag(

Compound pendulum / Physical pendulumWhen a rigid body is suspended from an axis and made to oscillateabout that then it is called compound pendulum.C = Initial position of centre of massC = Position of centre of mass after time tS = Point of suspensionl = Distance between point of suspension and centre of mass (it remains constant during motion)for small angular displacement from mean positionThe restoring torque is given by

= mglor, I = mgl where, I = Moment of inertia about point of suspension.

or, = Imgl

or, 2 = Imgl

Time period, T = 2 lmgI

I = ICM + ml2

where ICM = momentum of inertia relative to the axis which passes from the centre of mass & parallelto the axis of oscillation.

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T = 2 ll

mgmI 2CM

where ICM = mk2

k = gyration radius (about axis passing from centre of mass)

Torsional pendulumIn torsional pendulum, an extended object is suspended atthe centre by a light torsion wire. A torsion wire is essentiallyinextensible, but is free to twist about its axis. When thelower end of the wire is rotated by a slight amount the wireapplies a restoring torque causing the body to oscillaterotationally

when released. The restoring torque produced is given by

= C where, C = Torsional constantor, I = C where, I = Moment of inertia about the vertical axis.

or, = IC

Time period, T = 2 CI

Ex. A uniform disc of radius 5.0 cm and mass 200g is fixed at its centre to a metal wire, the other end ofwhich is fixed to a ceiling. The hanging disc is rotated about the wire through an angle and is released.If the disc makes torsional oscillations with time period 0.20s, find the torsional constant of the wire.

Sol. The situation is shown in figure. The moment of inertia of the disc about the wire is

I = 2

mr2 =

2)m100.5()kg200.0( 22

= 2.5 104 kg-m2.

The time period is given by

T = 2 CI

or, C = 22

TI4

= 2242

)s20.0()mkg105.2(4

= 0.25 22

smkg

Objective : Basic application of torsional pendulum.

{Home Work : HCV Q.28 to Q.54 }

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Superposition of two SHMsIn same direction and of same frequency

x1 = A1 sin tx2 = A2 sin (t + ), then resultant displacementx = x1 + x2 = A1 sin t + A2 sin (t + ) = A sin (t + )

where A = cosAA2AA 2122

21 & = tan

1

cosAAsinA

21

2

If = 0, both SHMs are in phase and A = A1 + A2If = , both SHMs are out of phase and A = |A1 A2|The resultant amplitude due to superposition of two or more than two SHMs of this case can also befound by phasor diagram also.

In same direction but are of different frequencies.x1 = A1 sin 1tx2 = A2 sin 2t

then resultant displacement x = x1 + x2 = A1 sin 1t + A2 sin 2tthis resultant motion is not SHM.

In two perpendicular directionsx = A sin ty = B sin (t + )

Case (i) : If = 0 or then y = (B/A) x. So path will be straight line & resultant displacement will be r = (x2+ y2)1/2 = (A2 + B2)1/2 sin t

which is equation of SHM having amplitude 22 BA

Case (ii) : If = 2

then x = A sin t

y = B sin (t + /2) = B cos t

so, resultant will be 22

Ax

+ 22

By

= 1 i.e. equation of an ellipse and if A = B, then superposition will be

an equation of circle.

Superposition of SHMs along the same direction (using phasor diagram)If two or more SHMs are along the same line, their resultant can be obtained by vector addition bymaking phasor diagram.

1. Amplitude of SHM is taken as length (magnitude) of vector2. Phase difference between the vectors is taken as the angle between these vectors. The magnitude of

resultant of vectors give resultant amplitude of SHM and angle of resultant vector gives phase constantof resultant SHM.

Page-22

For example :x1 = A1 sin tx2 = A2 sin (t + )

If equation of resultant SHM is taken as x = A sin (t + )

A = cosAA2A 2121

tan =

cosAAsinA

21

1

Ex. x1 = 3 sin tx2 = 4 cos tFind (i) amplitude of resultant SHM. (ii) equation of the resultant SHM.

Sol. First write all SHMs in terms of sine functions with positive amplitude. x1 = 3 sin t

x2 = 4 sin (t + /2)

A = 2

cos43243 22 = 169 = 25 = 5

tan =

2cos43

2sin4

= 53

equation x = 5 sin (t + 53)

Ex. x1 = 5 sin (t + 30)x2 = 10 cos (t)Find amplitude of resultant SHM.

Sol. x1 = 5 sin (t + 30)

x2 = 10 sin

2

t

Phasor diagram

A = 60cos1052105 22

= 5010025 = 175 = 5 7

{Home Work : HCV Q.55 to Q.58 }