1

Description of SHM’s

SHM spring pendulum, simple pendulum, physical pendulum

SHM Other systems

Syllabus for IIT-JEE

Simple Harmonic Motion: Linear and Angular Simple Harmonic Motions.

Introduction

Periodic Motion

A motion which repeats itself after equal intervals of time is called a periodic motion. In a periodic

motion, force is always directed towards a fixed point which may or may not be on the path of motion.

Examples of periodic motion:

• Circular motion is a periodic motion.

• Rotation of earth about her own axis is periodic.

• Motion of a pendulum is periodic motion.

Any motion which repeats itself after regular interval of time is called periodic or harmonic motion

and the time interval after which the motion is repeated is called time period. A periodic motion can be

either rectilinear or closed or open curvilinear (Fig. 1)

Motion of

hands of a clock

(A)

Motion of

earth around the sun

(B)

Motion of

a piston

in a cylinder

(C)

Motion of

a ball in a bowl

(D)

Motion of

liquid in a

U-tube

(E)

Some Periodic Motions:

Fig. 1

Oscillatory Motion:

A particle is said to be in oscillatory motion or vibratory if it moves periodically back and forth about

a stable equilibrium position. Motion of pendulum is oscillatory. Oscillatory motion is a periodic

motion between two fixed limits. Every oscillatory motion is a periodic motion but every periodic

motion is not always oscillatory. Oscillatory or vibratory motion is a constrained periodic motion

between certain precisely fixed limits. The world is full of oscillatory motions. The motion of the

balance wheel of a watch, the motion of a piston in an automobile engine, the motion of the needle of

sewing machine or the motion of a ball in a bowl are all oscillatory motions. The motion of molecules

in a solid around fixed lattice sites, the perpetual motion of atoms within molecules, vibrations of the

prongs of a tuning fork and the motion of electric and magnetic fields in an electromagnetic wave are

few other examples of vibratory motion.

Note: Every oscillatory motion is periodic but every periodic motion is not oscillatory, e.g., the uniform

circular motion of a conical pendulum or the motion of electrons around the nucleus in an atom are

periodic but not oscillatory while the motion of a liquid in a U- tube or that of the free end of a

clamped strip are oscillatory and hence periodic.

SIMPLE HARMONIC MOTIONSIMPLE HARMONIC MOTIONSIMPLE HARMONIC MOTIONSIMPLE HARMONIC MOTION

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Some oscillatory motions:

Fig. 2

Simple Harmonic Motion:

If in case of oscillatory motion a particle moves back and forth (or up and down) about a fixed point

(called equilibrium position or mean position) through a force or torque (called restoring force or

torque) which is directly proportional to the displacement but opposite in direction, the motion is

called simple harmonic and abbreviated as SHM. Further if the displacement in case of SHM is linear,

the motion is called ‘Linear SHM’ and if the displacement is angular it is known as angular SHM. The

up and down motion of a partially submerged floating body or of a mass attached to a spring are

examples of linear SHM while the motion of a simple pendulum or physical pendulum or torsional

pendulum (twisting and untwisting motion of a body suspended from a wire) are examples of angular

SHM.

� �

cos2

LT

g

��� 2

mT

k�� 2

LT

g�� 2

IT

mgL�� 2

IT

C��

12T

c��

Different types of pendulum

Fig. 3

12.1 SIMPLE HARMONIC MOTION:

When a body repeats its motion after regular time intervals we say that it is in harmonic motion or

periodic motion. The time interval after which the motion is repeated is called the time period. If a

body moves to and fro on the same path, it is said to perform oscillations. Simple harmonic motion

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(SHM) is a special type of oscillation in which the particle oscillates on a straight line, the acceleration

of the particle is always directed towards a fixed point on the line and its magnitude is proportional to

the displacement of the particle from this point. This fixed point is called the centre of oscillation.

Taking this point as the origin and the line of motion as the X-axis, we can write the defining equation

of a simple harmonic motion as

2x� �� � … (1)

Where 2� is a positive constant. If x is positive, a is negative and if x is negative, a is positive. This

means that the acceleration is always directed towards the center of oscillation.

If we are looking at the motion from an inertial frame

a = F/m.

The defining equation (1) may thus be written as

F/m = - 2� x

Or, F = - m 2� x

Or, F = - kx. … (2)

We can use equation (2) as the definition of SHM. A particle moving on a straight line executes simple

harmonic motion if the resultant force acting on it is directed towards a fixed point on the line and is

proportional to the displacement of the particle from this fixed point. The constant 2k m�� is called

the force constant or spring constant. The resultant force on the particle is zero when it is at the centre

of oscillation. The centre of oscillation is therefore, the equilibrium position. A force which takes the

particle back towards the equilibrium position is called a restoring force. Equation (2) represents a

restoring force which is linear. Figure (4) shows the linear restoring force graphically.

Example 12.1

The resultant force acting on a particle executing simple harmonic motion is 4 N when it is 5 cm away

from the centre of oscillation. Find the spring constant.

Solution:

The simple harmonic motion is defined as

F = - k x.

The spring constant is k = F

x = 2

4 480 / .

5 5 10

N NN m

cm m�

� � �� .

12.2 QUALITATIVE NATURE OF SIMPLE HARMONIC MOTION:

Let us consider a small block of mass m in placed on a smooth horizontal surface and attached to a

fixed wall through a spring as shown in figure (12.2). Let the spring constant of the spring be k.

m

k

0

0

0

P

Q

Fig. 12.2

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The block is at a position O when the spring is at its natural length. Suppose the block is taken to a

point P stretching the spring by the distance OP = A and is released from there.

At any point on its path the displacement x of the particle is equal to the extension of the spring from

its natural length. The resultant force on the particle is given by F = - kx and hence by definition the

motion of the block is simple harmonic.

When the block is released from P, the force acts towards the center O. The block is accelerated in that

direction. The force continues to act towards O until the block reaches O. The speed thus increases all

the time from P to O. When the block reaches O, its speed is maximum and it is going towards left. As

it moves towards left from O, the spring becomes compressed. The spring pushes the block towards

right and hence its speed decreases . The block moves to a point Q when its speed becomes zero. The

potential energy of the system (block + spring), when the block is at P, is 21

2k OP and when the

block is at Q, it is 21

2k OQ . Since the block is at rest at P as well as at Q, the kinetic energy is zero

at both these positions. As we have assumed frictionless surface, principle of conservation of energy

gives 2 21 1

2 2k OP k OQ� or OP = OQ.

The spring is now compressed and hence it pushes the block towards right. The block starts moving

towards right, its speed increases up to O and then decreases to zero when it reaches P. Thus, the

particle oscillates between P and Q. As OP = OQ, it moves through equal distances on both sides of the

centre of oscillation. The maximum displacement on either side from the center of oscillation is called

the amplitude.

Example 12.2

A particle of mass 0.50 kg executes a simple harmonic motion under a force F = –(50 N/m)x. If it

crosses the center of oscillation with a speed of 10 m/s, find the amplitude of the motion.

Solution:

The kinetic energy of the particle when it is at the center of oscillation is

21

2E mv�

210.05 10 /

2kg m s� = 25 J.

The potential energy is zero here. At the maximum displacement x = A, the speed is zero and hence the

kinetic energy is zero. The potential energy here is 21

2k A . As there is no loss of energy,,

2125

2kA J� … (i)

The force on the particle is given by F = - (50 N/m)x.

Thus, the spring constant is k = 50 N/m.

Equation (i) gives 21

50 / 25 .2

N m A J� or A = 1 m.

12.3 EQUATON OF MOTION OF A SIMPLE HARMONIC MOTION:

Consider a particle of mass m moving along the X – axis. Suppose, a force F = – kx acts on the particle

where k is a positive constant and x is the displacement of the particle from the assumed origin. The

particle then executes a simple harmonic motion with the center of oscillation at the origin. We shall

calculate the displacement x and the velocity v as a function of time.

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Suppose the position of the particle at t = 0 is x0 and its velocity is v

0. Thus,

At t = 0, x = x0 and v = v

0.

The acceleration of the particle at any instant is 2F k

x xm m

� �� � � � �

Where /k m� � .

Thus, 2dv

xdt

�� � … (12.3)

Or 2dv dx

xdx dt

�� �

Or, 2dv

v xdx

�� �

Or, 2.vdv x dx�� �

The velocity of the particle is v0 when the particle is at x = x

0. It becomes v when the displacement

becomes x. We can integrate the above equation and write

0 0

2.

v x

v x

v dv x dx�� �� � or,0 0

2 22

2 2

v x

v x

v x�

� �� � � �

� � � � � �

or, 2 2 2 2 2

0 0v v x x�� � � � � . or,2 2 2 2 2 2

0 0v v x x� �� � �

or, 2 2 2 2 20 0v v x x� �� � � or,

220

2

vv x�

�

� �� �� �� �

� �

2

2 2002

vx x

�� �

� �� �� �

Writing

22 200

vx A

�� �

� �� �� �

… (12.4)

The above equation becomes 2 2v A x�� � … (12.5)

We can write this equation as 2 2dx

A xdt

�� � or, 2 2.

dxdt

A x��

�

At time t = 0 the displacement is x = x0 and at time t the displacement becomes x. The above equation

can be integrated as 0

2 20

x t

x

dxdt

A x��

�� �

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or, 0

1

0sin

xt

x

xt

A�� �� � �

or,1 1 0sin sin

xxt

A A�� �� �

Writing 1 0sin

x

A�� � , this becomes

1sin

xt

A� �� � �

or, x = A sin t� �� … (12.6)

The velocity at time t is cosdx

v A tdt

� � �� � � . … (12.7)

12.4 TERMS ASSOCIATED WITH SIMPLE HARMONIC MOTION:

(a) Amplitude:

Equation (12.6) gives the displacement of a particle in simple harmonic motion. As sin t� �� can

take values between – 1 and + 1, the displacement x can take values between – A and + A. This gives

the physical significance of the constant A. It is the maximum displacement of the particle from the

center of oscillation, i.e., the amplitude of oscillation.

(b) Time Period:

A particle in simple harmonic motion repeats its motion after a regular time interval. Suppose the

particle is at a position x and its velocity is v at a certain time t. After some time the position of the

particle will again be x and its velocity will again be v in the same direction. This part of the motion is

called one complete oscillation and the time taken in one complete oscillation is called the time period.

T. Thus, in figure (12.4) Q to P and then back to Q is a complete oscillation, R to P to Q to R is a

complete oscillation, O to P to Q to O is a complete oscillation etc. Both the position and the velocity

(magnitude as well as direction) repeat after each complete oscillation.

We have, x = A sin t� �� .

If T be the time period, x should have same value at t and t + T.

Thus, sin sin[ ].t t T� � � �� � � �

Now the velocity is (equation 12.7) v = cosA t� � ��

As the velocity also repeats its value after a time period, cos cos[ ( ) ]t t T� � � �� � � � .

Both sin cost and t� � � �� � will repeat their values if the angle t� �� increases by 2� or its

multiple. As T is the smallest time for repetition.

2t T t� � � � �� � � � �

or, 2T� ��

or,2

T��

� .

Remembering that /k m� � , we can write for the time period.

22

mT

k

��

�� � ... (12.8)

Where k is the force constant and m is the mass of the particle.

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Example 12.3

A particle of mass 200 g executes a simple harmonic motion. The restoring force is provided by a

spring of spring constant 80 N/m. Find the time period.

Solution

The time period is 2m

Tk

��3

200 102

80 /

kg

N m�

���

2 0.05 0.31 .s s� � �

(c) Frequency and Angular Frequency:

The reciprocal of time period is called the frequency. Physically, the frequency represents the number

of oscillations per unit time. It is measured in cycles per second also known as hertz and written in

symbols as Hz. Equation (12.8) shows that the frequency is

1

2v

T

��

� � ... (12.9)

1

2

k

m�� ... (12.10).

The constant � is called the angular frequency..

(d) Phase:

The quantity t� � �� � is called the phase. It determines the status of the particle in simple harmonic

motion. If the phase zero at a certain instant, x = A sin t� �� = 0 and v = A cos t� � �� = A I � .

This means that the particle is crossing the mean position and is going towards the positive direction.

If the phase is / 2� , we get x = A, v = 0 so that the particle is at the positive extreme position. Figure

(12.5) shows the status of the particle at different phases.

We see that as time increases the phase increases. An increases of 2� brings the particle to the same

status in the motion. Thus, a phase t� �� is equivalent to a phase 2t� � �� � . Similarly a phase

change of 4 , 6 , 8� � � ………… etc, are equivalent to no phase change.

Figure (12.6) shows graphically the variation of position and velocity as a function of the phase.

�

��

�

��

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(e) Phase constant:

The constant � appearing in equation (12.6) is called the phase constant. This constant depends on the

choice of the instant t = 0. To describe the motion quantitatively, a particular instant should be called

t = 0 and measurement of time should be made from this instant. This instant may be chosen according

to the convenience of the problem. Suppose we choose t = 0 at an instant when the particle is passing

through its mean position and is going towards the positive direction. The phase t� �� should then

be zero. As t = 0 this means � will be zero. The equation for displacement can then be written as

x = – A sin t� .

If we choose t = 0 at an instant when the particle is at its positive extreme position, the phase is / 2� at

this instant. Thus / 2t� � �� � and hence / 2� �� . The equation for the displacement is x = AA

sin / 2t� �� . or, x = A cos t� .

Any instant can be chosen as t = 0 and hence the phase constant can be chosen arbitrarily. Quite often

we shall choose 0� � and write the equation for displacement as x = A sin t� . Sometimes we may

have to consider two or more simple harmonic motions together. The phase constant of any one can be

chosen as 0� � . The phase constants of the rest will be determined by the actual situation. The

general equation for displacement may be written as

x = A sin t� �� sin '2

A t�

� �� �� � �� �� �

= cosA t� � �� .

Where '� is another constant. The sine form and the cosine form are, therefore, equivalent. The value

of phase constant, however, depends on the form chosen.

Example 12.4:

A particle execute simple harmonic motion of amplitude Along the X –axis. At t = 0, the position of

the particle is x = A/2 and it moves along the positive x – direction. Find the phase constant � if the

equation is written as sinx A t� �� � .

Solution:

We have sinx A t� �� � . At t = 0, x = A/2. Thus, A/2 = A sin � or, sin� =1/2 or, , � = / 6� or 5.

The velocity is cos .dx

v A tdt

� � �� � � At t = 0, v = A cos� � . Now

3 5 3cos cos

6 2 6 2and

� �� � � �

As v is positive at t = 0, � must be equal to / 6�12.5 SIMPLE HARMONIC MOTION AS A PROJECTION OF CIRCULAR MOTION:

Consider a particle P moving on a circle of radius A with a constant angular speed � (figure 12.7). Let

us take the center of the circle as the origin and two perpendicular diameters as the X and Y – axes.

Suppose the particle P is on the X – axis at t = 0. The radius OP will make an angle t� �� with the X

– axis at time t. Drop perpendicular PQ on X – axis and PR on Y – axis. The x and y – coordinates of

the particle at time t are

x = OQ = OP cos t�or, x = A cos t� … (12.11)

And y = OR = OP sin t�or, y = A sin t� . ... (12.12)

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Equation (12.11) shows that the foot of perpendicular Q executes a simple harmonic motion on the X

– axis. The amplitude is A and the angular frequency is � . Similarly, equation (12.12) shows that the

foot of perpendicular R executes a simple harmonic motion on the Y – axis. The amplitude is A and the

angular frequency is � . The phases of the two simple harmonic motions differ by / 2� (remember

cos sin( / 2)).t t� � �� �

Thus, the projection of a uniform circular motion on a diameter of the circle is a simple harmonic

motion.

12.6 ENERGY CONSERVATION IN SIMPLE HARMONIC MOTION:

Simple harmonic motion is defined by the equation F = - kx.

The work done by the force F during a displacement from x to x + dx is dW = F dx = – kx dx the work

done in a displacement from x = 0 to x is

2

0

1.

2

x

W kx dx kx� � � ��Let U (x) be the potential energy of the system when the displacement is x. As the change in potential

energy corresponding to a force is negative of the work done by this force

210

2U x U W kx� � � � .

Let us choose the potential energy to be zero when the particle is at the center of oscillation x = 0.

Then 21

0 02

U and U x k x� � .

This expression for potential energy is same as that for a spring and has been used so far in this

chapter.

As2

,k

k mm

� �� �

We can write2 21

2U x m x�� ... (12.13)

The displacement and the velocity of a particle executing a simple harmonic motion are given by

sinx A t� �� � and cosv A t� � �� � .

The potential energy at time t is, therefore, 2 21

2U m x��

2 2 21sin .

2m A t� � �� �

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And the kinetic energy at time t is 21

2K m v� 2 2 21

cos .2

m A t� � �� �

The total mechanical energy at time t is E = U + K

2 2 2 21sin cos

2m A t t� � � � � � � � �

�

2 21.

2m A�� … (12.14)

We see that the total mechanical energy at time t is independent of t. Thus, the mechanical energy

remains constant as expected.

As an example, consider a small block of mass m placed n a smooth horizontal surface and attached to

a fixed wall through a spring constant k (figure 12.8).

When displaced from the mean position (where the spring has its natural length), the block executes a

simple harmonic motion. The spring is the agency exerting a force F = - kx on the block. The potential

energy of the system is the elastic potential energy stored in the spring.

At the mean position x = 0, the potential energy is zero. The kinetic energy is 2 2 20

1 1.

2 2m v m A�� �

All the mechanical energy is in the form of kinetic energy here. As the particle is displaced away from

the mean position, the kinetic energy decreases and the potential energy increases. At the extreme

positions x = � A, the speed v is zero and the kinetic energy decreases to zero. The potential energy is

increased to its maximum value 2 2 21 1

.2 2

k A m A�� All the mechanical energy is in the form of

potential energy here.

Example: 12.5

A particle of mass 40 g executes a simple harmonic motion of amplitude 2.0 cm. If the time period is

0.20 s, find the total mechanical energy of the system.

Solution:

The total mechanical energy of the system is 2 21

2E m A�� =

2 22

2

1 2 2

2

m Am A

T T

� �� ��� �

� �

=

22 2 2

2

2 40 10 2.0 10

0.20

kg m

s

� � �� �

27.9 10 .J

�� �

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12.7 ANGULAR SIMPLE HARMONIC MOTION:

A body free to rotate about a given axis can make angular oscillations. For example, a hanging umbrella

makes angular oscillations when it is slightly pushed aside and released. The angular oscillations are

called angular simple harmonic motion if.

(a) There is a position of the body where the resultant torque on the body is zero, this position

is the mean position 0� � .

(b) When the body is displaced through an angle from the mean position, a resultant torque

acts which is proportional to the angle displaced, and

(c) This torque has a sense (clockwise or anticlockwise) so as to bring the body towards the

mean position.

If the angular displacement of the body at an instant is� , the resultant torque acting on the body in

angular simple harmonic motion should be T = - k� .

If the moment of inertia is I, the angular acceleration is T k

I I� �� � � � .

or,

22

2

d

dt

�� �� � … (12.15)

Where, /k I� �Equation (12.15) is indentical to equation (12.3) except for the symbols. The linear displacement x in

(12.3) is replaced here by the angular displacement� . Thus, equation (12.15) may be integrated in the

similar manner and we shall get an equation similar to (12.6), i.e.,

0 sin t� � � �� � … (12.6)

Where 0� is the maximum angular displacement on either side. The angular velocity at time t is given

by, 0 cos .d

td t

�� � � �� � � � … (12.17)

The time period of oscillation is

22

IT

k

��

�� � … (12.18)

And the frequency of oscillation is

1 1

2

kv

T I�� � … (12.19)

The quantity /k I� � is the angular frequency..

Example: 12.6

A body makes angular simple harmonic motion of amplitude /10� rad and time period 0.05 s. If the

body is at a displacement /10� �� rad at t = 0, write the equation giving the angular displacement as

a function of time.

Solution:

Let the required equation be 0 sin t� � � �� � .

Here 0� = amplitude 10

�� rad

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12 240

0.05s

T s

� �� � �� � �

So that 1sin 40

10rad s t

�� � ��� � � �� � �� �

. … (i)

At t = 0, /10� �� rad. Putting in (i)

sin10 10

� ��� �

� � �� �

or, sin 1� � or, / 2� �� .

Thus by (i),

1sin (40 )10 2

rad s t� �

� � �� � � �� � � �� � �

= 1cos (40 )

10rad s t

�� �� � � � �� �

.

Energy:-

The potential energy is 2 2 21 1

2 2U k I� � �� � and the kinetic energy is

21

2K I� � .

The total energy is E = U + K = 2 2 21 1

.2 2

I I� �� � �

Using 0 sin t� � � �� �

2 2 20

1sin

2E I t� � � �� � 2 2 2

0

1cos

2I t� � � �� �

2 20

1.

2I � �� … (12.20).

Composition of two ‘parallel’ SHM’s

We have already seen that, for a particle performing SHM, the restoring force is given by, F = - kx.

Now, if the particle is acted upon by two restoring forces along same line instead of one, then the

resultant motion can be studied using superposition principle. Hence, if F1 and F

2 are the two forces

acting on the particle, then using superposition principle, we first assume that only the force F1 is

acting on the particle and find the corresponding position of the particle as,

x1 = A

1 sin 1t� �� ... (73)

Then we assume that only the force F2 acts on the particle and find the corresponding position of the

particle as,

x2 = A

2 sin 2t� �� ... (74)

Here, we assume the frequency of the oscillation to be same. The resultant position of the particle is

then given by,

x = x1 + x

2... (75)

1 1 2 2sin sinx A t A t� � � �� � � � �

1 1 1 2 2 2{sin .cos cos .sin } {sin .cos cos .sin }x A t t A t t� � � � � � � �� � � � �

1 1 2 2 1 1 2 2sin [ cos cos ] cos [ sin sin ]t A A t A A� � � � � �� � �

Let, 1 1 2 2cos cosA A� �� = X … (76)

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And 1 1 2 2sin sinA A� �� = YY … (77)

� sin cosx X t Y t� �� �Let, X2 + Y2 = A2 … (78)

We can write,

cos & sin

tan

X A Y A

Y

X

� �

�

� � ���

� � � … (79)

sin cos cos sinx A t A t� � � �� � �

sinx A t� �� � � ... (80)

Equation (80) shows that the resultant of two SHM’s along the same path is itself a SHM. The amplitude

of this resultant SHM is A and it’s relation with A1 and A

2 is given by,

2 21 2 1 2 1 22 cosA A A A A � �� � � � ... (81)

This relation can be easily derived using equations (76) to (78).

Also, the initial phase � , is given by,,

1 1 2 2

1 1 2 2

sin sintan

cos cos

A AY

X A A

� ��

� ��

� �� ... (82)

The difference 1 2� �� in equation (81) is called the phase difference between the two motions.

If 1 2� �� , then the two simple harmonic motions are in phase and we get, A = AA1 + A

2 &

1 2tan tan tan� � �� � . In this case, the amplitude is the maximum possible amplitude.

If 1 2� � �� � then the two SHM’s are out of phase and we get, 1 2 2tan tanA A A and � �� � �

If A1 = A

2, then A = 0 and the particle does not oscillate.

For any value of 1 2� �� other than 0 and � the resultant amplitude is between 1 2A A� and AA1 + AA

2.

Vector Combination of Two Simple Harmonic Motion:

Suppose the two individual SHM s are represented by

1 1 sinx A t��

2 2 sinx A t� �� � .

The two SHM s can be represented by vectors with magnitude A1 and A

2 respectively. The resultant A

��

of these two vectors represent the resultant SHM.

2 21 2 1 2| | 2 cosA A A A A �� � �

��and

2

1 2

sintan

cos

A

A A

��

! ��

Case: If the two SHM’s are perpendicular, then

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1 sinx A t�� ... (1)

2 siny A t� �� � ... (2)

By equation (1),

2

21 1

sin, , cos 1x x

tA A

� � �

By equation (2), 2 sin cos cos siny A t t� � � �� �2

2 21 1

cos sin 1x x

AA A

� � � �� � �� � �

2 22

22 1 1

cos 1 siny y x

A A A� �

� �� �� � � �� �� � � �� � � �

2 2 22 2 2

2 2 21 22 1 1

2cos cos sin sin

y x y x x

A AA A A� � � �� � � � �

2 22

2 21 21 2

2 cossin

x y x y

A AA A

��� � � �

The is equation of an ellipse and hence the particle moves in ellipse. Here note that the above cases are

valid only when the frequencies of the two SHMs are equal.

15.3 A Few Graphs Concerning SHM

(a) Graph of Displacement Versus Time

Assuming the displacement equation to be x = a sin t� , clearly the graph will be a sine curve.

(Fig 15.3)

Putting x = 0, we have sin ( t� ) = 0

, n

nt n wheren I t

�� �

�� � ! � �

Putting n = 0, 1, 2, 3 etc, we get

0 1 2 3

2 30, , ,t t t t

� � �� � �

� � � � etc, and putting x = � a (i.e., maximum displacement) we get sin

1t� � � .

2 12

nt n where n I�

�� � � ! .

Putting n = 0, 1, 2 etc, we get.

0 1 2

3 5, ,

2 2 2t t t

� � �� � �

� � � , etc

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(b) Graph of Velocity versus Time.

Assuming x = a sin t� , we get on differentiating

cosdx

v a tdt

� �� � .

Putting v = 0; we get cos t� = 0 2 1 ;2

nt n where n I�

�� � � !

Putting n = 0, 1, 2 etc.

0 1 3

3 5, ,

2 2 2t t t

� � �� � �

� � � etc.

Putting v = a�� (velocity amplitude), we get cos 1t� � � .

nt n where n I� �� � !

,n

nt

��

� � putting n = 0, 1, 2 etc, we get 0 1 2

20 : ; .t t t etc

� �� �

� � � .

2 /O

x

time(t)

3 /2

/

/2

The graph is a cosine curve (Fig. 15.4)

(c) Graph of Acceleration Versus Time:

Assuming x = a sin t� , we have f = - 2 2sinx a t� � �� �

Putting f = 0, the instants at which f = 0, can be found; sin t� = 0.

nt n� �� �

Putting n = 0,1, 2,………..

0 1 2

20, ,t t t

� �� �

� � � � etc.

Putting f = 2�� a we have sin 1t� � � (acceleration amplitude).

2 12

nt n�

�� � � �

Putting n = 0, 1, 2, etc.

0 1 2

3 5, ,

2 2 2t t t

� � �� � �

� � � , etc.

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(d) Graph of Acceleration Versus Displacement

Since acceleration (f) = - 2� (x). The variation is linear (figure. 15.6)

(e) Graph of Velocity Versus Displacement

we have seen earlier that 2 2v a x�� �

2 2 2 2v a x�� � �2 2 2

2 2

2 2 2 21

v x va x

a a� �� � � � � � .

The graph is an ellipse. (Figure. 15.7).

(f) Graph of Variation between Displacement (x) and Phase t� �� .

From the equation x = a sin t� ��

x = 0 at t� �� = n� where n I! . And 2 1 ,2

x a at t n where n I�

� �� � � � � ! .

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(g) Graph of Velocity Versus Phase.

From the equation v = cosa t� � �� .

v = 0 at 2 12

t n where n I�

� �� � � ! . v a at t n where n I� � � �� � � � !

The graph is shown in figure 15.9.

Example:1

A particle executes SHM between two points separated by 10 cm. If the force experienced by it be 2N

at a displacement of 1 cm from the mean position, then find the maximum force experienced by it.

Solution:

Evidently, the maximum displacement 10

52

� � cm.

For a particle executing SHM the force experienced is given by

F = - kx ... (1).

And Fmax

= - k (xmax

) ... (2).

Dividing (2) by (1) max maxF x

F x� max

max

510 .

2 1

FF N� � � �

Example 2

A particle executing SHM oscillates between two fixed points separated by 20 cm. If its maximum

velocity be 30 cm/s, find its velocity when its displacement is 5 cm from its mean position.

Solution:

Evidently, the max, displacement =20

102

cm�

For a particle executing SHM the speed at a displacement ‘x’ is given by.

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2 2v a x�� � ... (1)

and maxv a�� ... (2)

2 2max

v a

v a x�

� 2 2max

10 10 2 3

35 310 5

v

v� � � �

�

2 330 20 3

3v � � � cm/s.

Example 3:

A particle executes SHM with an angular frequency 4� �� rad/sec. If it is at its extreme position

initially, then find the instants, when it is at a distance 3

2 times, its amplitude from the mean position.

Solution:

Let a be the amplitude and � the phase constant for the oscillation. The displacement equation can be

written as

sin 4 4 / secx a t rad� � � �� � ��

Since at t = 0; x = + a.

4 02 2

� �� � �� � � � � .

Now, let tn be the instant, when the particle’s displacement is

3

2

a, nth time (where n I! ).

Evidently3

cos 42

a

aa t�� �

2 14

6 4 24n n

nt n t

�� �� � � � � � .

Taking only the positive values for ‘t’.

1 2 3 4 5

1 5 7 11 13, , , , , .

24 24 24 24 24t s t s t s t s t s etc� � � � �

Example 4

A particle executes SHM with amplitude 8 cm and a frequency 10 sec-1. Assuming the particle to be at

a displacement 4 cm, initially, in the positive direction, determine its displacement equation and the

maximum velocity and acceleration.

Solution

Given a = 8 cm and � = 2 n� = 20 � rad/sec.

Let the phase constant be � .

The displacement equation can be written as x = 8 sin (20� t +� )

Given, at t = 0; x = 4 cm

4 8sin 20 0� �� � �

1sin

2 6

�� �� � � �

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� The displacement equation becomes 8sin 206

x t�

�� �� �� �� �

.

Differentiating the above equation w.r.t time ‘t’

160 cos 206

dxv t

dt

�� �� �

� � �� �� �

max 160 / [ cos 20 / 6) 1 ]v cm s when t� � �� � � � � � .

Differentiating once again w. r. t time t.

22

23200 sin 20 / 6

d xf t

d t� � �� � � �

2max 3200 /f cm s�� � � [when sin (20 / 6) 1t� �� � � ].

Example 5

If the maximum speed and acceleration of a particle executing SHM be 20 cm/s and 100 � cm/s2, find

the time period of oscillation.

Solution

Given max| | 20 /v a cm s�� � … (1).

And 2 2max| | 100 /f a cm s� �� � … (2).

Dividing equation (2) by (1) we get 5� �� rad/sec.

� Time period 2 2

0.45

T� �� �

� � � sec.

Example 6

If a particle executing SHM possesses a speed of 10 cm/s, when at mean position, find its speed when

at a displacement of half the amplitude.

Solution

Let ‘a’ be the amplitude of oscillation.

From velocity 2 2v a x�� �

Squaring, we get 2 2 2 2v a x�� �

Given at x = 0; v = 10

2 2100 a�� � … (1).

And let at x = ; '2

av v�

22 2 3

4

av �

� ��� � � �� �

� �… (2).

Dividing equation (2) by (1), we get

2' 3

10 4

v� ��� �

� �

3' 10 5 3 /

2v cm s� � � � .

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Example 7

Two particles execute SHM with same frequency and amplitude along the same straight line. They

cross each other, at a point midway between the man and the extreme position. Find the phase difference

between them.

Solution

Let ‘a’ be the amplitude and 1 2and� � their respective phases at any instant, when they cross each

other.

Then their displacement equations can be written as x = a sin (phase) for the two particles we have

1 2sin sin2 2

a aa and a� ��

� Taking the least value, 16

�� � . And taking the next possible value 1 2� ��� .

2

5

6

�� �

Moreover, from v a ���

cos (phase)

If 1 2v and v� �

be the respective velocities of the two particles, then 1 2v v�� �

i.e., 1 2cos cosa a� � � �� 2 1cos cos� � �� � �

1 2

5 7,

6 6 6If then or

� � �� �� � �

� Required phase difference 2 2

2

3

�� �� � � rad.

Note: The phase difference can be in general 2

2 ;3

n n I�

� �� !

Example: 8

A particle executes SHM with time period of 2 sec; find the time taken by it to move from one extreme

position to the nearest mid point between mean position and extreme position.

Solution

Let ‘a’ be the amplitude of oscillation then evidently 2

T

�� �� � rad/sec.

The displacement equation can be written as sinx a t� �� �

Let, for simplicity, the time be reckoned from the instant when the particle was as its extreme position.

Thus, at t = 0; x = a

sin2

a a�

� �� � �

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21

, sin2

Thus x a t�

�� �� �� �� �

Let at 12

at t ve x� � �

1sin2 2

aa t

��� �

� � �� �� �

1 1

5

2 6 3t t

� � �� �� �

� � � � �� �� �

1

1

3t� � s.

Example 9

If two SHMs are represented by

1 210 sin 4 5 [sin 2 8 cos 2 ]2

y t and y t t�

� � �� �� � � �� �� �

compare their amplitudes.

Solution

For the equation 1 10 sin 42

y t�

�� �� �� �

� � the amplitude a

1 = 10 units and for the

equation 2 5 [sin 2 8 cos 2 ]y t t� �� � � . Multiplying and dividing by 1 8 3� �

2

1 815 sin 2 cos 2

3 3y t t� �

� � �� �

� � �15sin 2 t� �� � where tan 8� �

Whose amplitude a2 is obviously 15 units.

� The required ratio 1

2

10 2

15 3

a

a� � � .

Example 10

A particle is executing SHM with amplitude ‘a’ and maximum velocity v0. Find the displacement when

its velocity is 3

2

v,

Solution

From the equation for speed

2 2 2 2v a x�� �

2 2 20v a�� ... (1)

And

22 2 203

4

va x�� � ... (2)

Dividing (1) by (2), we get

2 2

2 2 2

4 31

3 4

a xor

a x a� � �

�

2

2

1

4

x

a� �

2

ax� � � .

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Example 11

A small particle moves along a horizontal circle of radius 5 cm with a uniform angular speed of 6 �rad/sec. The circle stands in front of a vertical wall, whereon a normal beam of light casts a sharp

shadow. What is the maximum speed and maximum acceleration of the shadow? Also find the range

between whose limits the shadow moves.

Solution: The situation is similar to the one discussed above. The shadow executes SHM upon the wall. The

maximum speed max 6 5 30 /v a cm s� � �� � � � and the maximum acceleration 2maxf a��

2 2 26 5 180 /cm s� �� � �

The shadow moves along a straight line over the projection of circle upon the wall and its range is the

diameter i.e., 10 cm.

Example 12

Find the position (s) of a particle executing SHM, when its K.E. is 1

4th its T.E.

Solution

From equation (8) K = 2 2 21

2m a x� � and from equation (11) E =

2 21

2m a � from the problem

4

EK � or 4K = E

2 2 2 2 21 14 ( )

2 2m a x m a� � � �� � �

2 2 2 2 24 4 4 3a x a x a� � � � �

3

2

ax� � � .

Example 13

Find the ratio of K.E. to P.E. for a particle executing SHM, when it is midway between the mean

position and extreme position.

Solution

Given 2

ax � �

Now

2 2 22 2

22 2

1

21

2

m a xK a x

U xm x

�

�

� �� �

22

2

4 3

1

4

aa

K

U a

� ��� �� �

� �� � �.

Example 14

What fraction of total energy is carried by a particle executing SHM, in the form of kinetic energy

when the displacement of the particle is 1

4th the maximum displacement, from one of its extreme

position?

Solution

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23

Displacement of the particle from its mean position is 3

4 4

a aa x� � � (say)

Now, fraction of T.E. in the form of K.E.

2 2 22

2 2

1

2 11

2

m a xK x

E am a

�

�

�� �� � � ��� �� �

23 7

14 16

K

E

� �� � � �� �

� �.

Example 15

A particle of mass 200 g is executing SHM with an amplitude 20 cm. If its kinetic energy be 32×10–3

joule initially, determine its displacement equation, assuming the phase constant to be 4

�.

Solution

Let � be the angular frequency, for the SHM, then displacement equation can be written as

x = 20 sin / 4t� �� where t is in sec. and x is in cm.

Now, initially 20sin4

x�� �

� � �� �

[Putting t = 0]

20

2�

2 400200

2x� � �

Now, kinetic energy at a displacement x, will be 2 2 21

2K m a x�� �

3 2 21 20032 10 400 200 10

2 1000�� �� �

� � � � � �� �� �

2

2

32 2

200 200 10�

�

�� �

� �

84 / sec

2rad�� � �

Thus, the displacement equation will be 20 sin 44

x t�� �

� �� �� �

.

Example 16

A particle executes SHM with force constant 2 × 106 N/m and amplitude 1 cm, and has a total mechanical

energy of 200 J. Find the following (a) maximum K.E., (b) Maximum P.E., (c) minimum K.E., (d)

minimum P.E.

Solution

Given 2 6 22 10 ; 1 10k m a m� �� � � � �

(a) Maximum 2 2

max

1. .

2K E m a v a� � � � ��

6 412 10 1 10 100

2J

�� � � � � � .

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24

(b) From E = K + U

maxE U�� [when K = 0; at the extreme position]

� Maximum potential energy Umax

= 200 J.

(c) Minimum kinetic energy 2

min

1

2m v� = 0 [at the extreme portion]

(d) Minimum Potential energy Umin

= E - Kmax

= 200 – 100 = 100 J.

Note: The particle’s potential energy is not zero at mean position

Example 31

A uniform rod of length l and mass m, is suspended by one end and is displaced slightly, to undergo

angular SHM. Find its time period.

Solution

Consider the rod, in a position making an angle ' '� with the vertical at any instant. (Fig. 15.43)

Net torque acting on the rod about the point of suspension.

sin2

lmg" �� � [ sin ]

m g lfor small angle

l

�� �

�� #

Now, angular acceleration

2

2 2

3d

Id t ml

� " "� �

2 2

23 2

ml d m g l

d t

� �"

�� � � or,

2

2

30

2

d g

ld t

��� � .

Comparing with the standard form

22

20

d

d t

�� �� �

We get,3

2

g

l� � .

� The period 2 2

23

lT

g

��

�� � .

Example 32

A square lamina of edge ‘a’ is suspended through a hole punched on a diagonal at a distance of 2 2

a

from its center. Find the period of oscillations.

Solution: Consider, the oscillating lamina (of mass m say) at any position making an angle ' '� with the

vertical. Net torque acting on the lamina about the point of suspension O, will be (Figure. 15.44)

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sin2 2 2 2

a am g mg" � �� � # �

Now, the moment of inertia of the lamina about an axis through O, and normal to its plane will be.

22 2

6 2 2

ma aI m

� �� � � �� �

� �

27

24ma�

Since,

2

2

I d

dt

�" �

2 2

2

7

24 2 2

ma d m g a

dt

� �� �� �� �� �

� �

2

2

(12)

7 2

d g

dt

� ���

Comparing with the standard form

22

2

d

dt

�� �� �

We get2 12

7 2

g� � .

� The period 2 7 2 7 2

212 3

Tg g

�� �

�� � �

15.11 Average Values of a Few Parameters:

Average velocity and acceleration of a particle executing SHM.

The average of any time dependent function (Z = f(t)) say) can be obtained, between the instants t = t1

to t = t2.

As

2 2

1 1

2 2

1 1

t t

t t

t t

t t

Zdt f t dt

Z

dt dt

$ %� �� �

� �.

(a) Average Velocity: Since the motion in a SHM, repeats again and again over a period of one time

period, therefore the average velocity over one complete time period is zero. However, the average

velocity between one extreme position to the other will not be zero. This can be calculated as follows.

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26

O

V

t

Assuming the equation for velocity in terms of time as cosv a t� ��

3 / 4

/ 4

3 / 4

/ 4

cosT

T

T

T

a t d t

v

d t

� �� $ %�

��

3 / 4

/

3 / 4

/ 4

3sinsin sin

4 4

3

4 4

T

T A

T

T

T Ttaa

T Tt

� ���

� � � � � �� � � �� �� � � � � � � �� �

� ��� �

� �

2 3sin sin [ 2 ]

2 2

aT

T

� �� �

& �� � � �� � �' �� � � �� � � �(

�

2{( 1) (1)}

a

T� �

24 aav

T

�

�� $ % � �

2

�� times maximum velocity ... (24).

Alternatively: Average velocity over one half oscillation = Average speed from one extreme position

to another. [� there is no variation in direction]

tan cov 2 4

/ 2

total dis ce ered a a

total time taken T T� � � .

(b) Average Acceleration: Arguing in the same way as for velocity, the average acceleration over one

complete oscillation will be zero. However if we are not interested in one complete oscillation, it may

not turn out to be zero.

Let us find out the average acceleration between the instants when the particle crosses its mean position

two consequtive times.

Assuming the equation for acceleration as 2 sinf a t� �� �

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27

/ 22

0

/2

0

sinT

T

a t d tf

d t

� ��$ %�

��

/ 22

0

2 cosT

ta

T

��

��� �

� � � � � �� � �

22cos cos 0

2

a T

T

� ��

� �� �� �� �� � �

22 22

cos 0aa

T

���

� � � � � � � .

22 2, | |

aor f

�

� �$ %� � time maximum acceleration. … (25)

Average Kinetic Energy, Potential Energy and Total energy.

(a) Average kinetic energy over one complete oscillation.

The instantaneous K.E. of a particle executing SHM is given by 2 2 21

cos2

K m a t� ��

2 2 2

0

0

1cos ( )

2

T

T

m a t dt

K

d t

� �� $ %�

��

2 2

0

0

1 cos 21

2 2

T

T

tm a dt

t

��

� ��� �� �� ��

�

2 2

0

sin 21

4 2

rtm a

tT

���

� �� �

� � �

2 22 21 1

04 4

m aT m a

T

��� � � ... (26)

(b) Average potential energy over one complete oscillation.

The instantaneous P.E. of a particle executing SHM is given by 2 2 21

sin2

U m a t� �� .

2 2 2

0

0

1sin

2

T

T

m a t d t

U

d t

� �� $ %�

��

2 2

0

0

1 cos 21

2 2

T

T

tm a d t

t

��

� ��� �� �� ��

�

2 2

0

sin 21

4 2

Ttm a

tT

�� � �� �

� � �

2 22 21 1

04 4

m aT m a

T

��� � � ... (27)

(b) Average total energy over one complete cycle.

Since the total energy is independent of time

2 21

2E m a�� $ %� .

Example 37

What is the ratio between the maximum to average kinetic energy of a particle executing SHM.

Solution

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28

We known that the maximum K.E., is given by 2 2

max

1

2K m a�� and average K.E., is given by

2 21

4averageK m a��

max 2

1average

K

K� � .

Example 38

The average speed over the period of complete oscillation of a particle executing SHM is 2 cm/sec.

The particle attains this speed when it is at P, distant 4 cm from the mean position O. Find (i) the

amplitude (ii) the time period. Also show that the least time taken by the particle to travel from O to P

is

21

2

4 4sin

44

�

�

�� ��� �� �� � �

. [Express your answers in terms of � ].

Solution

Let a cm and � rad/sec be the amplitude and angular frequency of oscillation.

Then average speed <v> is given by <v> = 2 a�

�.

22

a�

�� �

a� �� � . ... (1)

Also, the speed (v) at a displacement (x) is given by

2 2 2 2v a x�� �

2 2 2 2 2v a x� �� � � ... (2)

2 24 16� �� � �

2 22 4 4

16 4

� �� �

� �� �� � � �� �� �

� �

2

41 ,

4From a cm

� �� �

� � �� 2

2 8sec

4and From T

� �� �

� ��

.

Let the displacement equation for the particle be x = a sin � t. [Here for the sake of simplicity phase

constant � is taken zero].

2

2

44

sin44

t��

�

��

�.

Let at t = t1 the particle be at 0

21

2

44

0 sin44

t��

�

�� �

�1 0t� �

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Now, let the particle be at P, at t = t2 for the first time

Then

22

2

44

4 sin44

t��

�

��

�

2224

4sin

4

t� ��

� �� �

21

22

4 4sin

44t

�

�

�� ��� �� �� �� � �

� The required time interval = t2 – t

1

21

2

4 4sin

44

�

�

�� ��� ��� �� � �

Example 39

A particle executes SHM whose displacement changes with time as x = a sin t� . Find the average

speed between the instants when it passes its mean position and its displacement is 2

a for the first

time.

Solution

Since the displacement equation is given as x = a sin t� when x = 0; t = 0 and when x = 2

a;

6t

��

� .

Also, instantaneous velocity v = a cos t� � .

� Required value <v> =

/6

0

/6

0

cosa t d t

d t

� �

� �

� ���

/6

0

/6

0

sinsin sin 0

6

06

ta a

t

� �

� �

� ���

��

� � � �� � �� �� �� �� �� � �� �� �

�� �� �

1 6 3

2

au

� �� �

� �� � �� �

� �.

General Method to show that a given body (or a particle) executes SHM and to find its time period.

Step 1: Establish the equilibrium condition (if any) for the body at its mean position.

Step 2: Imagine the body to be displaced (linear or angular as the case may be) from its mean position

slightly. Find the net restoring force (in case of linear SHM) or torque (in case of angular SHM) acting

on the body in terms of displacement, ending up finally in the form of

F = – kx (for linear SHM)

or k" �� � (for angular SHM).

Where k should be a constant.

Step 3 : Dividing F by M or (" by I, as the case may be) find the net acceleration as :

22

2

d xx

dt�� � or

22

2

d

dt

�� �� �

Knowing the value of � , time period can be calculated as

2T

��

� .

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Alternatively: One can also find the total energy of the system, in step 2 and differentiate the expression

with respect to time t and equate it to zero. [� The T.E. in SHM remains constant]. One should then

eventually end up in equation of the form as obtained in step 3. [See example 25, alternative].

A few examples are solved stepwise to illustrate the idea.

EXAMPLES

Example 1

Write down the equation of a simple harmonic motion with an amplitude of 5 cm if 150 oscillations

are performed in one minute and the initial phase is 450.

Solution

Given: Amplitude (A) = 5 cm

Initial phase 045 / 4 rad� �� � .

Frequency of oscillation = n = 150 per minute

= 150

60 per second

5

2per second�

� Angular frequency = 15

2 2 ( ) 52

n rad s� � � � �� � �

� The displacement of the particle at any instant t is given by

sin 5sin 54

x A t t cm�

� � �� �� � � �� �

� �.

Example 2

In what value of time t (i.e., after t = 0) will a particle, oscillating according to the equation x = 7 sin

0.5 t� , move from the position of equilibrium to the maximum displacement?

Solution

The displacement of the particle is given by x = 7 sin 0.5 t�

Given: xmax

= amplitude = 7 Alternate Method:

7 7 sin 0.5 t�� �2

0.5 4 secTT

��� � �

1 sin 0.5 t�� � At t = 0, x = 7 sin (0) = 0

0.52

t�

�� � Time to reach from x = 0

1 sect ond� � i.e., equilibrium) to x = 7

(Amplitude) that is a quarter

Oscillation t = 1 sec.

Example 3

The equation of an oscillating particle is given by 2 sin ( )2 4

x t cm� �

� � . Find

(i) Period of oscillation (ii) Maximum velocity

(iii) Maximum acceleration and (iv) Initial displacement

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Solution

Given that the displacement of the particle is 2 sin ( )2 4

x t cm� �

� �

Compare it with x = A sin t� ��

� Amplitude (A) = 2 cm

Angular frequency1

2rad s

�� �� .

(i) Period of oscillation 2 2

4/ 2

T s� �� �

� � � �

(ii) Maximum velocity maxv A�� � =1

22

cm s�

� �� �

(iii) Maximum acceleration = 2maxa A�� = 2

2 2/ 2 22

cm s�

� �� �

(iv) Initial displacement = 0 2 sin ( ) 24

x cm�

� � . (i.e., at t = 0).

Example 4

The equation of a simple harmonic motion is given as 6sin10 8 cos 10x t t� �� �

Where x is in cm and t is in seconds. Find the amplitude, period and initial phase.

Solution

Put 6sin10 8 cos 10 sinx t t A t� � � �� � � �

Expanding, we get, cos sin sin cosx A t A t� � � �� �

1, 10We get rad s� � �� �

2 20.2

10The Period T s

� �� �

� � � � � .

Also A cos � = 6 and A sin � = 8

Squaring and adding, we get A2 = 62 + 82 = 100

� Amplitude = A = 10 cm.

sin 8tan 1.333

cos 6

A

A

��

�� � �

� Initial phase = � = 530 8’.

Example 5

A particle performing S. H. M has velocities of 3 cm s-1 and 4 cm s-1 respectively when the displacements

are 4 cm and 3 cm. Find the amplitude, period and maximum velocity of the particle.

Solution

The velocity v of the particle performing SHM when its displacement x is given by

2 2v A x�� �

2 2 2 2v A x�� � �

We have, v = 3 cm s-1 when x = 4 cm.

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2 29 16A�� � � … (I)

Also, v = 4 cm s-1 when x = 3 cm.

� 2 216 ( 9)A�� � … (II)

Dividing (I) by (II),

2

2

9 16

16 9

A

A

��

�Solving, A = 5 cm

Substituting this in (I), we get � = 1 rad s–1

� Period T = 2 / 2 s� � ��

Velocity is maximum, when x = 0

1max 1.0 5v A cm s� �� � � �

Example 6

A particle executing linear SHM had a maximum, velocity of 40 cm s–1 and a maximum acceleration of

50 cm s-2. Find its amplitude and the period of oscillation.

Solution

Let A be the amplitude and � be its angular velocity..

Then maximum velocity 1max 40v A cm s� �� � � ... (I)

Maximum acceleration 2 2max 50a A cm s� �� � � ... (II)

Dividing (II) by (I), we get, 2

150. ., 1.25

40

Ai e rad s

��

��� �

� Period of oscillation = T = 2 2

5.031.25

s� ��

� �

Amplitude = A = 40

321.25

Acm

��

� � .

Example 7

A particle of mass 50 g executes SHM along a path of length 20 cm at the rate of 600 oscillations per

minute. Find its total energy. Also find the potential and kinetic energies when the displacement x =

A/2 where A is the amplitude.

Solution: mass (m) = 50 g = 5 × 10-2 kg

Angular frequency 12 600

2060

rad s�

� � ��� � �

Amplitude (A) 20

10 0.12

cm m� � �

� Total energy = 2 21

2m A� 2 2 21

5 10 400 10 0.9872

J�� �� � � � � � .

When, x = 2

A, Potential energy =

22 2 21 1

0.2472 2 4

Am x m J� �� �

� Kinetic energy = 2 2 21

2m A x� �

22 21

( ) 0.7402 4

Am A J�� � � .

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Example 8

A body is supported by a light spiral spring and causes a stretch of 1.5 cm in the string. If the mass is

now set in vertical oscillation of small amplitude, what is the periodic time of the oscillation?

Solution

Let m be the mass of the body in kg.

Then, since 1.5 cm = 0.015 m, mg = k × 0.015, at equilibrium position ... (I)

Where k is the spring constant in N m–1

K = 1

0.015

m gN m�

If the mass is displaced by x below its equilibrium position, the net restoring force

F = K (0.015 + x) – mg = Kx

i.e. the net force (restoring) is opposite to the displacement

i.e., F K x� ��� �

i.e., the motion is Simple Harmonic

2m

T periodK

�� � � = 0.015 0.015

2 29.81

m

m g� �� = 0.25 s

1. A body of mass 1 kg is executing harmonic motion which is given by x = 6.0 cos 100 / 4t �� cm.

What is the (i) amplitude of displacement, (ii) frequency, (iii) initial phase, (iv) velocity, (v) acceleration,

(vi) maximum kinetic energy?

Solution:

The given equation of SHM is x = 6.0 cos 100 / 4t cm��

Comparing it with the standard equation of SHM, x = A cos t� �� , we have

(i) Amplitude A = 6.0 m

(ii) Frequency � = 100/sec.

(iii) Initial phase / 4� ��

(iv) Velocity 2 2 2

100 36v A x x�� � � �

(v) Acceleration 22 4100 10x x x�� � � � �

(vi) Kinetic energy2 2 2 21 1

2 2mv m A x�� � � .

When x = 0, the kinetic energy is maximum i.e.,

2 2 4

max

1 1 36. 1 10 18 .

2 2 100K E m A meter joules� � �

� � � � � �� �� �

2. A particle is executing SHM with amplitude A and has maximum velocity V0. Its sped at displacement

A/2 will be :

a. 03 / 2V b. 0 / 2V c. V

0d. V

0/4

Solution : (A)

2 2v A x�� �

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For x = A/2

22/ 2v A A�� �

00

33 / 4 [ ]

2

Vv A V A� �� � ��

3. The velocity time graph of a harmonic oscillator is shown in the following figure. The frequency of

oscillation is :

a.25 Hz b. 50 Hz c. 12.25 Hz d. 33.3 Hz.

Solution: (A)

T = 0.04 s1 1

25 .0.04

v HzT

� � � �

4. A body is executing SHM. When its displacement from the mean position is 4 cm and 5 cm, the

corresponding velocity of the body is 10 cm per sec. and 8 cm per sec. respectively. Then the time

period of the body is

a. 2� sec b. 2

� sec c. � sec d.

3

2

� sec

Solution: (C)

Using 2 2v A x�� �

2 210 4A�� � ... (i)

& 28 5A�� � ... (ii)

(i)

2

2 2104A

�� �

� � �� �� �

(ii)

22 28

5A�

� �� � �� �

� �2 2

2 2

100 644 5

� �� � � �

2

36 69

3�

�� � � �

2 2

6 / 3T

v

� ��� � � �

5. A linear harmonic oscillator has a total mechanical energy of 200 J. Potential energy of it at mean

position is 50 J. Find.

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a. The minimum kinetic energy

b. The minimum potential energy

c. The potential energy at extreme positions.

Solution:

At mean position, potential energy is minimum and kinetic energy is maximum. Hence

Umin

= 50 J (at mean position) and Kmax

= E – Umin

= 200 – 50

= 150 J (at mean position).

At extreme positions, kinetic energy is zero and potential eneryg is maximum.

maxU E� � = 200 J (at extreme position).

Single Answer Correct:

1. A particle is executing SHM. Which of the following is maximum when the bob is at the mean position?

a. Time period b. Amplitude c. velocity d. acceleration.

Solution: (C)

At the mean position, velocity is maximum and acceleration is minimum.

2. The displacement of a particle in simple harmonic motion in one time period is

a.A b. 2 A c. 4 A d. zero.

Solution: (D)

At the end of one complete vibration, the particle returns to the initial position.

3. Two bodies A and B of equal masses are suspended from two separate mass less springs of spring

constants k1 and k

2 respectively. If the bodies oscillate vertically such that their maximum velocities

are equal, the ratio of the amplitude of A to that of B is:

a.k1 / k

2b. 1 2/k k c. k

2/k

1d. 2 1/k k .

Solution : (D)

1 1 21 1 2 2

2 2 1

a ka a or

a k

�� �

�� � �

4. The equation of SHM of a particle is a + 4 2� x = 0, where ‘a’ is instantaneous linear acceleration at

displacement x. The frequency of motion is

a.1 Hz b. 4� Hz c. 1

4 Hz d. 4 Hz

Solution: (A)

Comparing with 2 2 20, 4 ; 2 2 , 1 .a x v v H z� � � � � �� � � � � �

5. A particle executes SHM of amplitude 5 cm and period 3 s. The velocity of the particle at a distance 4

cm from the mean position (take � = 3) is

a.8 cm s-1 b. 12 cm s-1 c. 4 cm s-1 d. 6 cm s-1.

Solution:

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2 2 2 22v a x a x

T

��� � � � 12 3

25 16 63

cms��

� � �

6. Figure shows the displacement – time graph of a simple harmonic oscillator. The amplitude, time

period and initial phase of the oscillator, are respectively.

a.1 cm, 2 s, � /2 b. 2 cm, 4 s, � /4 c. 2 cm, 4 s, zero d. 2 cm, 2 s, zero.

Solution: (C)

Graph begins from the origin. So, initial phase is zero.

Multiple Answers:

12. A particle vibrates in SHM along a straight line. Its greatest acceleration is 5� 2 cm s-2 and when its

distance from the equilibrium position is 4 cm, the velocity of the particle is 3� cm s-1. Then:

a. The amplitude i 10 cm b. The period of oscillation 2 sec.

c. The amplitude is 5 cm d. The period of oscillation 2 sec.

Solution: (B), (C)

(Acceleration)max

= 2 25a� �� . ... (1)

Velocity at y = 4 cm is 2 2 2 24 3v a y a� � �� � � � �

or, 216 3a� �� �

Squaring, 2 2 2

16 9a� �� � ... (2)

So,

2 22

2 2

16 9 9

55

a

a

� �

� �

�� �

or,

2 16 9 9 415

5 10

aor a

a

� �� � � � or - 3.2 cm.

� a = 5 cm.

From (1), 2 25 5� �� � .

or, � �� .

� T = 2 sec.

So, choice (B) and (C) are correct and choice (A) and (D) are wrong.

13. When the particle is half way to its end point in a S. H. M (where E is the total energy) :

a. 4

EPE � b.

5

4

EPE � c. PE = KE d. KE = 3 PE.

Solution : (A), (D)

Given, 2

ay � and total energy = E

We also know that total of energy a particle in SHM 2 21

2E m a��

Potential energy of particle at some point (PE)

22 2 21 1

.2 2 2

am y m� � � �

� � � �� �

2 21 1

4 2 4

Em a�� �

� �� �� �

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KE2 2 21

2m a y�� �

2 22 2 21 1 3

2 4 2 4

a am a m� �

� �� � �� �� �

� �

3

4E�

So, KE = 3 (PE)

Choice (D) is correct and choices (B) and (C) are wrong.

EXERCISE

Single Answer Correct.

22. If the displacement equation of a particle be represented by y = A sin pt + B cos pt, the particle

executes.

a. a uniform circular motion b. a uniform elliptical motion.

c. a SHM d. a rectilinear motion.

26. What fraction of the total energy is potential when the displacement is one - half of the amplitude?

a. 1

4b.

2

4c.

3

4d.

3.5

4.

27. The displacement of a particle executing simple harmonic motion is given by : y = 10 sin 63

t�� �

�� �� �

.

Here, y is in meter and t is in second. The initial displacement and velocity of the particle are respectively.

a. 5 3 and 30 m s-1 b. 20 3 and 30 m s-1.

c. 15 3 m and 30 m s-1 d. 15 m and 5 3 m s-1.

28. The equation of a SHM of amplitude ‘a’ and angular frequency ' '� in which all distances are measured

from one extreme position and time is taken to be zero at the other extreme position is

a. x = a sin t� b. x = a cos t�

c. x = a – a sin t� d. x = a + a cos t� .

29. The amplitude and the periodic time of SHM are 5 cm and 6 s respectively. At a distance of 2.5 cm

away from the mean position, the phase will be:

a. � /6 b. � /4 c. 5� /12 d. � /3.

30. Two particles execute SHM of same amplitude and frequency along the same straight line. They pass

one another when going in opposite directions, each time their displacement is half of the their

amplitude. The phase difference between them is:

a. 900 b. 300 c. 1200 d. 600.

31. The maximum displacement of a particle executing SHM is 1 cm and the maximum acceleration is

(1.57)2 cm s-2. Then the time period is

a. 0.25 s b. 4.00 s c. 1.57 s d. (1.57)2 s.

32. Which of the following quantities are always positive in a simple harmonic motion?

[Letters have usual meanings].

a. .F a�� �

b. .v r� �

c. .a r� �

d. . sF r�� ��

.

33. The ratio of the KE and PE possessed by a body executing SHM when it is at a distance of 1/n of its

amplitude from the mean position is

a. n2 b. 2

1

nc. n2 + 1 d. n2 – 1.

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34. The velocities of a body executing SHM are v1 and v

2 when the displacements are x

1 and x

2 respectively.

The period is

a.

2 22 1

2 21 2

2x x

v v�

�

�b.

2 22 1

2 21 2

1

2

x x

v v��

�c.

2 21 2

2 22 1

2v v

x x�

�

�d.

2 1

2 1

2x x

v v�

�

� .

Using the following paragraph, solve Q. 50 & 51

A particle is said to possess SHM if (a) acceleration is directly proportional to displacement, (b)

acceleration is directed towards a fixed point. The average of such a function in a complete cycle is

zero. The PE and KE of such SH function will be half the total energy 2 21

2m A�� �

� �� �

, where A is the

amplitude and � is the angular frequency of the mass m. The average of a function of time in a

complete cycle is given by, 0 0

T T

avY Y d t d t� � � .

50. If the average PE of an oscillating system is E, the average KE of a system with double amplitude and

frequency will be:

a. E b. 4

Ec. 4 E d. 16 E.

51. The ratio of average of displacement’s A sin t� and A sin 2 t� over half of cycle is:

a. 0 b. ) c. 1 : 1 d. 1 : 2.

Assertion Reason Type:

In each of the following questions two statements are given as Assertion ‘A’ and Reason ‘R’. Examine

the statements carefully and answer the questions according to the instructions given below.

Mark (A) if both A and R are correct and R is the correct reason of A.

Mark (B) if both A and R are correct and R is not the correct reason of A.

Mark (C) if A is correct and R is wrong.

Mark (D) if A is wrong and R is wrong.

52. Assertion (A): Every oscillatory motion cannot be treated as simple harmonic under small

oscillation

Reason (R): Even under larger oscillation restoring force should be proportional to displacement

for being SHM.

53. Assertion (A): Two SHM’ s along x and y axes with angular frequency ratio 1 2:� � = 1 : 2, with same

amplitude results in a parabola path on super – position.

Reason (R): The x and y displacements are related as y � x2.

Simple Harmonic Motion

22. (C) 23. (B) 24. (B) 25. (D) 26. (A)

27. (A) 28. (D) 29. (A) 30. (C) 31. (B)

32. (A) 33. (D) 34. (A) 35. (C) 36. (B)

37. (B) 38. (B), (C) 39. (A), (C) 40. (A), (B) 41. (A), (B), (C)

42. (B), (C) 43. (A), (D) 44. (A), (B), (C) 45. (A), (B), (C) 46. (A), (B), (D)

47. (D) 48. (D) 49. (B) 50. (D) 51. (B)

52. (B) 53. (A).

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15.7 Spring Pendulum

(a) Horizontal oscillations of mass attached to an ideal spring

C PxF=kx

Consider a block of mass m, attached to one end of a mass less spring, obeying Hooke’s law, with its

other end attached to a rigid support. The block rests on a horizontal frictionless surface. (Fig. 15.29).

Let the block be slightly pulled towards right and left free to oscillate. It executes SHM. Consider the

position of the block, at any instant, when its displacement is ‘x’ from the mean position. (towards

right). The net force acting upon the block is the restoring force due to spring. (towards left).

Thus, F k x� ��� �

[where ‘k’ is the force constant]

or,F k

f xm m

� � �

���� �

.

Comparing with the standard from 2f x�� ��� �

.

We have 2 k

m� �

Now, time period of motion.

22

mT

k

��

�� � .

Energy Conservation in Spring Oscillations:

Let ‘x0’ be the maximum compression (or extension) in the spring. In such situation.

2 20

1 10 0;

2 2K m U k x� � �

And20

1

2E K U k x� � � ... (a).

Now, if v0 be the maximum speed, of the block, when at mean position, then

20

1 1; 0 0

2 2K mv U K� � �

20

1 1; 0 0

2 2K mv U K� � �

20

1

2E K U mv� � � �

But, as the block moves from its mean position, to either of its extreme position, it utilizes, its kinetic

energy 20

1

2m v

� �� �� �

to elongate or compress the spring, which gets stored in the spring in the form of

potential energy 20

1

2k x

� �� �� �

. Thus, from (a) and (b), it is evident at the total energy of the system (i.e.,

spring + block) remains constant.

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Note: Oscillations of a spring pendulum is independent of the value of ‘g’ at that place.

Horizontal oscillations, of a block attached to a spring, on a rough surface:

Consider a block of mass ‘m’, attached to a spring of force constant k, oscillating on a rough horizontal

surface with coefficient of friction ' '* (Fig. 15.30). Let a0 be the initial compression, of the spring

and a1 be the maximum extension, for the first time. From energy considerations.

2 20 1 0 1

1 1

2 2k a k a m g a a*� � �

0 1 0 1 0 12

ka a a a m g a a*� � � � � 0 1

2 m ga a

k

*� � �

This gives the decrease in amplitude, during each a swing.

(b) Vertical oscillations of block attached to an ideal spring.

Consider a block of mass m attached to one end of an ideal spring, with its other end fixed to a rigid

support. Let ' 'y� be the initial extension in the spring. (Fig. 15.31). If ‘k’ be the force constant of the

spring, then for equilibrium (vertical) of the block ky0 = mg (numerically). Now, if the block be

slightly displaced from its mean position it starts executing SHM.

Let ‘y’ be the displacement of the block, at any instant from its mean position (downward). The net

force acting on the block upward will be

F = k (y0 + y) – mg

or, F k y� ��� ��

or, acceleration F k

f ym m

� �� � �� �

� �

���� ��

Comparing the above equation with the standard from 2f y�� ��� ��

, it is evident that the block executes

SHM, with an angular frequency k

m� � .

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� Time period of oscillations 2

2m

Tk

��

�� � .

Note: Even the vertical spring oscillations is also independent of the value of ‘g’ at that place.

(c) Horizontal spring oscillations, taking into consideration, the mass of the spring.

As the block oscillates, so does the spring. However the velocity at each point of the spring is not the

same. In the elongated (or compressed) state, the displacements of each section can be assumed to be

linearly proportional from their respective distances from the fixed end.

Thus, if ‘y’ be the displacement of the block (i.e., the end of the spring connected to the block) at any

instant, then the displacement of a section distant ‘x’ from the fixed end will be y x

L where L is the

length of spring; similarly if v be the velocity of block then d y

vd t

� (Fig. 15.32).

According the velocity v’ of the aforementioned section will be 'x d y x v

vL d t L

� � .

Now, kinetic energy of the entire spring can be obtained as the sum (integral) of the kinetic energies of

each sections.

Thus, total K.E. of the spring = 21 ''

2

M d xv

L

& �� �' �� �

� �( �[Here M’ is the mass of the spring and dx, the length of the elemental section].

2 2 2 3

2 300

1 ' 1 '

2 2 3

LLM v x d x M v x

L L L

� � � �

� � ��

21 '

2 3

M v� �� � �� �

� �

� Total kinetic energy of the spring + block system

221 ' 1

2 3 2

M vM v

� �� �� �� �

� � [where M is the mass of

the block]. Now, the total mechanical energy of the system is given by E = K + U.

22 21 ' 1 1

2 3 2 2

M vM v k y

� �� � �� �� �

� �

Since the total mechanical energy should remain conserved.

0d E

d t� �

' 2 22 0

6 2 2

M v d v M v d v k d yy

d t d t d t

� � � � � �� � � �� � � � � �

� � � � � �

'0

3

M d v d yM k y v

d t d t

� �� � � �� �� �

� � �� .

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42

Again, since

2

2

d v d d y d y

d t d t d t d t

� �� �� �

� �

2

20

'

3

d y k y

M yd tM

� � �� ��� �� �

Comparing with the standard form

22

20

d yy

d t�� �

We find, that the complete system executes SHM, with an angular frequency '

3

k

MM

� �� ��� �� �

� Time period of oscillations

'

32

MM

Tk

�

� ��� �� ��

Note:

(1) The above equation can be rewritten as 2effM

Tk

�� where '

3eff

MM M

� �� �� �� �

.

(2) The oscillations of a block attached to a spring, remains SHM only for small elongation (or

compression) when the restoring forces generated in the spring remains linear. For large

deformations in spring, the restoring forces become non linear.

(3) Unless, otherwise specified, a spring should always be treated as ideal and mass less.

Composite spring oscillations.

(a) When springs are connected in series: Consider three springs with stiffness constants k1, k

2 and

k3 respectively. If x

1, x

2 and x

3 be the respective elongations of the three springs, then, the displacement

of the block x is given by x = x1 + x

2 + x

3. ... (a)

Now, if F be the restoring force acting along each of the spring.

Then F = k1 x

1 = k

2 x

2 = k

3 x

3... (b)

Now, the net force acting on the block will also be F.

From (b), 1 2 3

1 2 3

;F F F

x x and xk k k

� � � .

Substituting them in equation (a). 1 2 3

F F Fx

k k k� � � or

1 2 3

1 1 1F x

k k k

� �� � � �� �

� �

�� �.

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If m be the mass of the block, then acceleration

1 2 3

1 1 1

F xf

mm

k k k

�� �

� �� �� �

� �

�� ���

.

Replacing 1 2 3

1 1 1

k k k� � with

1

effK we have effk x

fm

� �

�

�� which proves the fact that the oscillation

are SHM, with angular frequency effk

m� � .

Time period of oscillation = 2

2 .eff

mT

k

��

�� � where

1 2 3

1 1 1 1

effk k k k� � � .

(b) Where springs are connected in parallel:

It can be shown that the oscillations of the block are SHM with a time period T given by

2eff

mT

k�� where k

efff = k

1 + k

2 + k

3.

(c) Figures 15.35 below shows a special arrangement of two springs virtually appearing to be in

series; however they are not.

If x�

be the displacement of the block at any instant then, the net force acting on it

1 2 1 2F F F k x k x� � � � ��� ��� ��� � �

= 1 2k k x� ��

� The effective spring constant for the above set up is kefff

= k1 + k

2

� Time period 2eff

mT

k�� .

Example 26

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A small block of mass m is attached to three springs with respective stiffness constants k1, k

2 and k

3

(see fig 15.37). Find the period of oscillations, if the block is slightly pressed against the spring with

stiffness constant k3. Will your answer be the same, if the block be pressed against the spring with

stiffness constant k1?

Solution

Figure 15.38 below shows, the respective elongations occurring in the springs of stiffness constant k1

and k2 when the spring of stiffness constant k

3 is compressed by ‘x’.

Net restoring forces acting on the block along the spring of stiffness constant k3 will be

0 01 2 3cos 45 cos 45F F F F� � �

�� �� ��

1 2 3

1 1

2 2k x k x k x

� � � � �� � �

� � �

1 2

2 2

k kF x

� � �� �

�

�� �

Thus, the effective stiffness constant is given by 1 23

2eff

k kk k

�� �� �� �� �

� Time period of oscillations

2eff

mT

k�� � .

If the block be compressed against spring of spring constant k1, the net force will be (See Fig. 15.9)www.N

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2 20 0

1 3 3cos 45 sin 45F F F F� � ���

2 2

3 31

2 2

k x k xk x

� � � �� � �� � � �

� � � �

1/22

3 31

2 2

k kF k x

� � � �� �� � � �� � � �� �� � � � �

�� �

� Effective stiffness constant

1/22 2

3 31

2 2eff

k kk k

� � � �� �� � �� � � �� �� � � � �

.

Example 27

Figure 15.40 shows three springs A, B and C with force constants k1, k

2, and k

3 receptively, connected

to a block of mass m, passing through a light pulley. Assuming all other things to be mass less, find the

period of oscillations.

Solution

Evidently, springs A and B are in series, with effective spring constant 1 2

1 2

'k k

kk k

�� which in turn is in

parallel with the spring C

� Effective spring constant 1 2

2

1 2

eff

k kk k

k k

� �� �� �

�� �

� Time period 2eff

mT

k�� 1 2

1 2 2 3 1 3

2m k k

k K k k k k�

��

� �

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Alternative: (from energy considerations).

Let y0 be the initial compression in the spring C and, for the sake of simplicity, let the spring A and B

in the relaxed position, when the block is at rest. (Fig. 15.41) For vertical equilibrium of the block mg

= k3 y

0. Let ‘y’ be the displacement of the block (below its mean position), which would also be the

additional compression in the spring C.

Let y1 and y

2 be the respective elongation of the springs, A and B respectively. Since total extension in

spring A = Net compression in springs B and C

� y1 = y – y

2

� y = y1 + y

2... (1)

Also, since the tension in the two ends of the string (passing through the pulley) should be same

therefore,

k1 y

1 = k

2 y

2... (2)

From (1) and (2) 2 2

1 2

1 2 1 2

k y k yy and y

k k k k

� � � �� �� � � �

� �� � � �

Total mechanical energy will be 22 2 2

1 1 2 2 3 0

1 1 1 1

2 2 2 2E k y k y k y y mv mgy� � � � � �

Since, total mechanical energy remains constant, during oscillation

0d E

d t� �

i.e.,

2 2 2231 2 2 1

1 2 1 2

02 2 2 2

kk k y k k yd mvy y m g y

d t k k k k

& �� � � �� �� � � � �' �� � � �

� �� � � �� �(

2 21 2 2 1

3 02 2

1 2 1 2

0k k y k k yd y d y d y mv d v m g d y

k y yd t d t d t d t d tk k k k

� � � � � � �� �

2 21 2 2 1

32 2

1 2 1 2

0k k y k k y m d v

k yd tk k k k

� � � � �� �

21 2

3 21 2

0k k m d y

k yk k d t

� �� � � �� �� ��� �

3 0[ ]d y

v and m g k yd t

� ��

21 2

321 2

1, 0

k kd yOr k y

m k kd t

� �� � �� �� ��� �

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Comparing with the standard form

22

20

d yy

d t�� �

we find that 2 1 2 2 3 1 3

1 2

1 k k k k k k

m k k�

� �� �� � �

�� �

Time period 1 2

1 2 2 3 1 3

22

m k kT

k k k k k k

��

�

�� �

� � .

Example 28

A horizontal spring block system (of force constant k) and mass M executes SHM with amplitude A.

When the block is passing through its equilibrium position an object of mass m is gently placed on it

and the two move together. Find the new amplitude.

Solution

Let the initial angular frequency be � and the velocity of the block at its mean position be v..

Then � = k

M

Also v = max. vel. Of the block k

AM

�

Now, if ‘v’ be the combined speed, then, from law of conservation of linear momentum.

(M + m) v’ = Mv.

'Mv

vM m

� ��

And, the new angular velocity will be 'k

M m� �

�.

Again from max ' ' 'v A ��

'M v k

AM m M m

� �� �

[where A’ is the new amplitude]

'M k k

A AM m M M m

� �� � [From eqn. (1)]

'M

A AM m

� ��

Example 29

Two bodies A and B of equal mass are suspended from two separate mass less springs of force constant

k1 and k

2 respectively. If the bodies oscillate vertically such that their maximum velocities are equal,

then find the ratio of the amplitudes of A and B.

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Solution

If 1 2and� � be the respective angular frequencies of the oscillations related to A and B.

Then 1 2

1 2

k kand

m m� �� � [where m is the mass of each body A and B]

Now, if a1 and a

2 be their respective amplitudes then:

max 1 1 2 2v a a� �� �

1 1 2

2 2 1

a k

a k

��

� � � .

Example 30

If a body attached to a spring executes vertical oscillations with time period T, find the new time

period, if the same body be attached to a new arrangement in which the spring is subdivided into four

parts of equal length each and all of them connected in parallel.

Solution

Let k be the stiffness constant of the spring and m be the mass of the body. Then time period ‘T’ of the

oscillations will be given by

2m

Tk

�� ... (1)

Now, since the stiffness constants of a spring varies inversely as its length

� Stiffness constant of each part = 4 k

Further, the effective stiffness constant, when all of them are connected in parallel will be 4 (4K) = 16

k.

Now, the new time period 2

' 216 4

m mT

k k

��� �

'4

TT� � [From (1)]

3. A solid cylinder is attached to a horizontal mass less spring so that it can roll without slipping along a

horizontal surface. The spring constant for the spring is k. If the system is released from rest, by

slightly stretching it, show that the center of mass of the cylinder executes SHM and find its frequency

of oscillation (Fig. 15.55).

Solution:

Step: 1 At equilibrium, no force acts horizontally, so, as such no equilibrium condition exists.

Step: 2 Let the system be, at any instant, such that, the spring is stretched by an amount x. If v be the

linear velocity of the center of mass of the cylinder then total energy of the system will be

E = translational K.E. of cylinder + Rotational K.E. of cylinder + Elastic P.E. in the spring.

2 2 21 1 1

2 2 2M v I k x�� � �

2 22 2

2

1 1 1

2 2 2 2

M R vM v k x

R

� �� � �� �� �

� �

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[� Moment of inertia I for the cylinder about the axis of rotation is 2

2

M R and for pure rolling

motionv

R� � ].

Now, since no work is done against friction in pure rolling motion, so T.E. remains conserved.

or, 0d E

d t� .

2 23 10

4 2

dM v k x

d t

� � �� � �

2

2

3 20 0

2 3

M v d v k x d x d x kx

d t d t Md t� � � � � �

Comparing the above equation with

22

2,

d xx

d t�� it is evident that, the cylinder executes SHM, with

an angular frequency 2

3

k

M� � .

� Frequency of oscillation

1 2

2 2 3

kn

M

�� �

� �.

4. A uniform rod of length L and mass M is pivoted at its center. It is held in position by a system of

springs as shown in fig 15.56. Find the time period of oscillation, if the rod be slightly turned and

released.

Solution:

Step 1: Evidently, there is no requirement of any equilibrium conditions, for no torque exists or acts

when the rod is in its equilibrium position.

Step 2: Consider the rod, at any position, when displaced by an angle � about point O. (Fig. 15.57)

The net torque acting upon the rod clockwise.

1 22 2

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50

[where F1 and F

2 are the restoring forces generated in the two springs respectively]

2 2

L Lk y k y k L y� � �

Now, tan2 2

L Ly � �� # .

2

2

k L" �

�� �

But,

2 2 2

2 21

12

d M L d

d t d t

� �" � �

2 2 2 2

2 2

6

12 2

M L d K L d k

Md t d t

� � � �� �� � � �

Comparing with the standard from

22

2

d

d t

�� �� � .

It is evident that the rod executes angular SHM with angular frequency 6 k

M� � .

� Time period of oscillation 2

26

MT

k

��

�� �

12. Two blocks of masses m1 and m

2 are connected to the two ends of a mass less spring of spring constant

k, and are placed on a smooth horizontal surface. The stwo blocks are pressed slightly against each

other so as to compress the spring and released. Show that the oscillations are SHM and find the

period of oscillations.

Solution:

Let l be the original length of the spring. If x1 and x

2 be the displacement of the two blocks at any

instant from their respective mean positions, the net compression in the spring will be x1 + x

2 = x (say)

[Fig. 15.66].

The net force on block m1 = F = - k x

21

1 2

d xm k x

d t� � � ... (1)

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21

21

d x k x

md t� � � .

Similarly the net force on block m2 will be F = – kx

22

2 2

d xm k x

d t� � �

22

22

d x Kx

md t� � � ... (2)

Now from x1 + x

2 = x

Differentiating twice w. r. t time t, we get

2 2 21 2

2 2 2

d x d x d x

d t d t d t� � ... (3)

From (1),(2) and (3) we get

2 2

2 21 2 1 2

1 1k x k x d x d xk x

m m m md t d t

� �� � � � � � �� �

� �

Now let 1 1

1 1 1

m m *� � (say)

Then

2

2

d x kx

d t *� � which proves the fact that the oscillations are SHM with an angular

frequencyk

�*

� .

� Time period 2

2TK

� *�

�� �

1 2

1 2

2m m

k m m��

�

Thus, the two blocks of masses m1 and m

2 connectd to the two ends of the spring oscillate a a single

block of mass.

1 2

1 2

m m

m m* �

� which is known as the reduced mass.

14. A solid uniform cylinder of mass m performs small oscillations due to the action of two springs

each of stiffness factor k. Assuming pure rolling find the period of oscillations (Fig. 15.68)

Solution

As the cylinder is slightly disturbed from its rest position, the spring force come into action, leading

the cylinder to perform angular oscillations about the point of contact of the cylinder with the horizontal

surface as the instantaneous center of rotation. If � be the angular displacement of the cylinder about

the instantaneous center of rotation C. (Fig. 15.69).

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Then, the net torque acting on the cylinder about C will be " = F (2R) + F (2R) [where F is the

restoring force on each cylinder]

22 2 2 8k R R k R� �� � � .

Now since I" �� .

2 2 22 2

2 2

3

2 2

m R d dm R m R

d t d t

� �� �� � �� �� �� �

2 22 2

2 2

1638

2 3

kd dm R k R

md t d t

�� ��� � � � � �

Comparing the above equation with the standard form

22

2,

d

d t

�� �� � it is evident that the cylinder

performs angular SHM with an angular frequency 16

3

k

m� �

� Period of oscillation 2 3

2 316 2

m mT

k k

� ��

�� � �

OBJECTIVE QUESTIONS:

18. The period of oscillation of a spring pendulum is T. If the spring is cut into four equal parts, then the

time period corresponding to each part will be :

a. T b. 2

Tc. 2 T d.

4

T.

19. A vertical spring of mass 120 g is fixed to the ceiling at its upper end and a mass of 370 g is suspended

at its lower end. If the spring constant be 810 Nm-1 then its frequency of oscillation is:

a. 145

2s

��

b. 145 10

2 37s

��

c. 11

45s

��

d. 11 37.

180 10s

��

.

30. Three identical springs with spring constants k each are arranged in two different manner (A) and (B)

as shown in figure 15.72. The ratio of frequencies of the mass, in the two cases will be:www.NetB

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a. 1 : 1 b. 2 : 3 c. 3 : 2 d. 2 2

3 3

32. Springs of spring constants k, 2k, 4k, 8k …………. ) are connected in series. A mass m kg attached

to the lower end of the last spring a allowed to vibrate. The period of oscillation is:

a. 2m

k� b. 4

m

k� c.

22

m

k� d. ) .

Answers:

18. (B) 19. (A) 30. (A) 32. (C).

18. From 1

2 ,m

T we have Tk K

� ��

Now, 1 2 3 4

1 1 1 1 1 1 4

'eff

ork k k k k k k

� � � � � [Here keff

= k and k1 = k

2 = k

3 = k

4 = k’]

� k’ = 4k.

� Time period will be twice the original value.

19.

1

2

3

s

kn

Mm

��

� the formula for where m = mass of block Ms = mass of spring.

30.2

2 ,2

eff

eff

K kmT k

k k k�� �

� in either case.

32. For series connection effective spring constant keff

is given by

1 2

1

1 1 1..........

eff

n

k

k k k

�� � �

11 1 1 1

..........2 4 8

effkk k k k

�

� � � � � � )� � �

1

1

2 3

1 1 1 11 ..........

12 2 21

2

k k

�

� � � � �� � � � � ) �� � � �� � � � � �� �� � �

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[Summing up G.P. with a = 1; r = 1

;2 1

aS

r) �

�]

2 2/ 2eff

m mT

k k� �� � �

Only one Option is correct

Answers: 17 (C) 21. (B) 23. (C) 32. (B) 33. (B)

35. (C) 63. (A) 66. (A).

More than one option is correct.

2. A simple pendulum of length 1 m with a bob of mass m swings with an angular amplitude 300.

Then: (g = 9.8 m/s2)

(a) Time period of pendulum is 2s

(b) Tension in the string is greater than mg cos 150 at angular displacement 150.

(c) Rate of change of speed at angular displacement 150 is g sin 150.

(d) Tension in the string is mg cos 150 at angular displacement 150.

Ans: 2, b, c.

Match the column:

9. In case of second’s pendulum, match the following (consider shape of earth also) :

Table – 1 Table – 2

(A) At pole (P) T > 2 s

(B) On a satellite (Q) T < 2 s

(C) At mountain (R) T = 2 s

(D) At center of earth (S) T = 0

(T) T = )

Answers: 9.. a q; b t; c p; d t.

Assertion – reason

22. Assertion: If a pendulum clock is taken to a mountain top, it gains time.

Reason: Value of acceleration due to gravity is less at heights.

(a) A (b) B (c) C (d) D (e) E

24. Assertion: The frequency of a second pendulum in an elevator moving up with an acceleration half

the acceleration due to gravity is 0.612 s-1.

Reason: The frequency of a second pendulum does not depend upon acceleration due to gravity.

(a) A (b) B (c) C (d) D (e) E

Answers: 22. d 24 c.

7. On a smooth inclined plane a body of mass M is attached between two springs. The other ends of the

springs are fixed to firm supports. If each spring has a force constant k, the period of oscillation of the

body Is : (assuming the spring as mass less):www.NetB

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a. 22

M

k� b.

22

M

k� c.

sin2

2

M

k

�� d.

2 sin2

M

k

�� .

8. A block of mass m is suspended by different springs of force constant shown in figure.

Let time period of oscillation in these four positions be T1, T

2, T

3 and T

4. Then:

a. T1 = T

2 = T

4b. T

1 = T

2 and T

3 = T

4

c. T1 = T

2 = T

3d. T

1 = T

3 and T

2 = T

4.

9. An object suspended from a spring exhibits oscillations of period T. Now the spring is cut in two

halves and the same object is suspended with two halves as shown in figure. The new time period of

oscillations will become :

a. 4 : 1 b. 1 : 4.

15. A mass m = 8 kg is attached to a spring as shown in figure and held in position so that the spring

remains Unstretched. The spring constant is 200 N/m. The mass m is then released and begins to

undergo small oscillations. The maximum velocity of the mass will be : (g = 10 m/s2)

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18. Let T1 and T

2 be the time periods of two springs A and B when a mass m is suspended from the

separately. Now both the springs are connected in parallel and same mass m is suspended with them.

Now let T be the time period in this position. Then:

a. T = T1 + T

2b.

1 2

1 2

T TT

T T�

�

c. 2 2 21 2T T T� � d. 2 2 2

1 2

1 1 1

T T T� � .

22. A ball of mass 2 kg hanging from a spring oscillates with a time period 2� seconds. Ball is removed

when it is in equilibrium position, then spring shortens by :

a. 10 m b. 5 m c. 20 m d. 2� m.

24. In the figure, the block of mass m, attached to the spring of stiffness k is in contact with the completely

elastic wall, and the compression in the spring is e. The spring is compressed further by e by displacing

the block towards left and is then released. If the collision between the block and the wall is completely

elastic then the time period of oscillations of the block will be:

a. 2

3

m

k

�b. 2

m

k� c.

3

m

k

�d.

6

m

k

�.

31. Frequency of a particle executing SHM is 10 Hz. The particle is suspended from a vertical spring. At

the highest point of its oscillation the spring is Unstretched. Maximum speed of the particle is: (g = 10

m/s2)

a. 2� m/s b. � m/s c. 1

�m/s d.

1

2� m/s.

34. The two blocks of mass m1 and m

2 are kept on a smooth horizontal table as shown in figure. Block of

mass m1 but not m

2 is fastened to the spring. If now both the blocks are pushed to the left so that the

spring is compressed a distance d. The amplitude of oscillation of block of mass m1, after the system

is released is:

a. 1

1 2

md

m m� b. 2

1 2

md

m m� c. 2

1 2

2 md

m m� d. 1

1 2

2 md

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41. Two masses M and m are suspended together by a mass less spring of force constant k. When the

masses are in equilibrium, M is removed without disturbing the system. The amplitude of oscillation

is:

a. M g

kb.

m g

kc.

M m g

k

�d.

M m g

k

�

48. A ball of mass when dropped from certain height as shown in diagram, strikes a wedge kept on smooth

horizontal surface and move horizontally just after impact. If the ball strikes ground at a distance d

from its initial line off all, then the amplitude of oscillation of wedge after being hit by the ball will be

a. 2

md M g

M k hb.

2

d k M g

m h c.

2

k gd

M hd.

2

k gd

m h

51. The angular frequency of a spring block system is 0� . This system is suspended from the ceiling of an

elevator moving downwards with a constant speed v0. The block is at rest relative to the elevator. Lift

is suddenly stopped. Assuming the downward as a positive direction, choose the wrong statement:

a. The amplitude of the block is 0

0

v

�

b. The initial phase of the block is �

c. The equation of motion for the block is 0

0

sin t

v�

� .

d. The maximum speed of the block is v0.

54. Two blocks with masses m1 = 1 kg and m

2 = 2 kg are connected by a spring of spring constant k = 24

N/m and placed on a frictionless horizontal surface. The block m1 is imparted an initial velocity v

0 =

12 cm/s to the right. The amplitude of oscillation is :

a. 2 cm b. 1 cm c. 3 cm d. 4 cm.

56. Two bodies M and N of equal masses are suspended from two separate mass less springs of spring

constants k1 and k

2 respectively. If the two bodies oscillate vertically such that their maximum velocities

are equal, the ratio of the amplitude of vibration of M to that of N is :

a. k1/k

2b. 2 1/k k c. k

2/k

1d. 1 2/k k .

61. A spring of force constant k is cut into two pieces such that one piece is double the length of the other.

Then the long piece will have a force constant of:

a. 2/3 k b. 3/2 k c. 3 k d. 6 k.

Answers : 7. a 8. b 15. b 18 d, 22. a. 24 a. 31. d 34. a, 41. a 48. a 51. b 54. a 56 b, 61. b.

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7. 2 22

eff

Mk k T

k�� � � . Period (or k

efff) is independent of � .

8. Effective force constant in case (iii) and (iv) is keff

= 2k + 2k = 4k.

Therefore, 1 2 2m

T Tk

�� �

And 3 4 24

m mT T

k k� �� � � .

15. Mean position will be at k, x = mg.

or, 8 10 2

0.4200 5

m gx m m

k

�� � � � .

This is also the amplitude of oscillation

i.e., A = 0.4 m.

now, max

kv A A

m�� �

2000.4 2 /

8m s� � .

18

2

1 1 21 1

42

m mT or k

k T

��� �

2

2 2 22 2

42

m mT or k

k T

��� �

Now,

2

2

42

m mT or k

k T

��� � .

In parallel k = k1 + k

2

Substituting the values of k, k1 and k

2 we get :

2 2 2

1 2

1 1 1

T T T� �

22. 2m

Tk

��

22

2 2

4 242 /

2

mk N m

T

��

�� � � �

Now, mg = k x0

0

2

2

m g gx

k

�� � � = g meter = 10 m.

24.k

m� �

General equation of motion, x = 2e cos t�

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e = 2e cos t�

3t

�� �

3

kt

m

�� .

3

mt

k

�� �

22

3

mT t

k

�� � �

31. Mean position of the particle is m g

k distance below the unstretched position of spring. Therefore,

amplitude of oscillation is m g

Ak

�

2 20k

fm

� � �� � � (f = 10 Hz)

2

1

400

m

k �� � .

Therefore, the maximum speed of particle will be :

max 2

120 /

2400

gv A m s� �

��

� �� � �� �� �

� �.

More than One Option is correct:

3. A constant force F is applied on a spring block system as shown in figure. The mass of the block is m

and spring constant is k. The block is placed over a smooth surface. Initially the spring was Unstretched.

Choose the correct alternative (s) :

a. The block will execute SHM

b. Amplitude of oscillation is 2

F

k.

c. Time period of oscillation is 2m

k�

d. The maximum speed of block is

22 F x k x

m

�. (Here, x = F/k)

12. A block of mass m is attached to a mass less spring of force constant k, the other end of which is fixed

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from the wall of a truck as shown in figure. The block is placed over a smooth surface and initially the

spring is Unstretched. Suddenly the truck starts moving towards right with a constant acceleration a0.

As seen from the truck:

mk

a. The particle will execute SHM

b. The time period of oscillation will be 2m

k�

c. The amplitude of oscillation will be 0m a

k.

d. The energy of oscillation will be

2 20m a

k.

16. In a spring – block system force constant of spring is k = 16 N/m, mass of the block is 1 kg. Maximum

kinetic energy of the block is 8J. Then :

a. Amplitude of oscillations is 1 m.

b. At half the amplitude potential energy stored in the spring is 2 J.

c. At half the amplitude kinetic energy is 6 J.

d. Angular frequency of oscillation is 16 rad/s.

Answers: 3. a, c, d 12. a, b, c 16. a, b, c.

3. The situation is similar as if a block of mass m is suspended from a vertical spring and a constant force

mg acts downwards. Therefore, in this case also block will execute SHM with time period:

2m

Tk

��

At compression x where F = k x ; F

xk

� .

This is also the amplitude of oscillation. Hence F

Ak

�

At mean position speed of the block will be maximum. Applying work energy theorem.

F. x = 2 21 1

2 2k x m v�

or,

22 F x k x

vm

�� .

Potential energy at extreme position.

12. Free body diagram of the truck from non inertial frame of reference will be as follows :

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This is similar to a situation when a block is suspended from a vertical spring.

Therefore, the block will execute simple harmonically with time period

2m

Tk

�� . Amplitude will be given by A = x = 0ma

k (ma

0 = kx)

Energy of oscillation will be

2 2 22 0 01 1

2 2 2

m a m aE k A k

k k

� �� � �� �

� �.

16.16

1

k

m� � � = 4 rad/s.

2max

18

2m v � .

211 8

2A�� � � �

211 16 8

2A� � � � � or A = 1m.

At half the amplitude 0.5 ,2

Ax m� � potential energy stored in the spring will be

221 1 1

16 22 2 2

U K x J� �

� � � � �� �� �

Only when spring block system is horizontal. At half the amplitude 2 2v A x�� �

2 24 1 0.5v � �

2 3 /v m s�

21

2K E m v� � � �

211 2 3 6

2J� � � �

Match the Column

3. In spring – block system match the following:

Table – 1 Table – 2

(A) If k (the spring constant) (P) Speed will become 16 times

is made 4 – times

(B) If m (the mass of block) is (Q) Potential energy will becomes four times

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62

made 4 – times

(C) If k and m both are made (R) Kinetic energy will remain unchanged

4 – times

(T) None.

8. In the two block spring system, force constant of spring is K = 6N/m. Spring is stretched by 12 cm and

then left. Match the following:

Table – 1 Table – 2

(A) Angular frequency of (P) 4.8 × 10-3 SI unit.

oscillation

(B) Maximum kinetic (Q) 3 SI unit

energy of 1 kg

(C) Maximum kinetic (R) 2.4 × 10-3 SI unit

Energy of 2 kg

(S) None.

12. In a spring – block system; match the following:

Table – 1 Table – 2

(A) If mass of the block is (P) Energy of oscillation becomes 4 times

Doubled

(B) If amplitude of oscillations (Q) Speed of particle becomes 2 times

Is doubled

(C) If force constant is doubled (R) Time period becomes 2 times

(D) If angular frequency is doubled (S) Potential energy becomes 2 times.

Answers: 3. a q, b t, c. q 8. a q, b s, c p 12. a r, b p, c s, d q.

Assertion – Reason

17. Assertion: A block of mass m is attached to a spring inside a trolley of mass 2m as shown in figure.

(i) all surfaces are smooth. Spring is stretched by A and released. Both trolley and block oscillate

simple harmonically with same time and same amplitude.

Reason: The problem is similar to a two – body problem as shown in figure. (ii).

a. A b. B c. C d. D e. E.

18. Assertion: If a man with a wrist watch on his hand falls from the top of a tower, its watch gives correct

time during the free fall.

Reason: The working of the wrist watch depends on spring action and it has nothing to do with

gravity.

a. A b. B c. C d. D e. E.

Answers: 17. d. 18 a.

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Comprehension Based questions:

Concept: [6 – 8]: In perfectly inelastic collision between two bodies momentum remains constant and

the bodies stick together. Angular frequency of a spring block system is K

m� � and maximum speed

of particle is SHM is A� , where A is the amplitude.

Question: Two identical blocks P and Q have mass m each. They are attached to two identical springs

initially unstretched. Now the left spring (along with P) is compressed by 2

A and the right spring

(along with Q) is compressed by A. Both the blocks are released simultaneously. They collide perfectly

inelastically. Initially time period of both the block was T.

6. The time period of oscillation of combined mass is :

a. 2

Tb. 2 T c. T d.

2

T.

7. The amplitude of combined mass is:

a. 4

Ab.

2

Ac.

2

3

Ad.

3

4

A.

8. What is energy of oscillation of the combined mass?

a. 21

2k A b.

21

4k A c.

21

8k A d.

21

16k A .

Answers: 6. c 7. a 8. d

Question [9.10] : A block of mass 1

2m � kg is attached with two springs each of force constant k = 10 N/m

as shown. The block is oscillating vertically up and down with amplitude A = 25 cm. When the block

is at its mean position one of the spring breaks without changing momentum of block.

9. What is the new amplitude of oscillation.

a. 10.6 cm b. 43.3 cm c. 25 cm d. 62.2 cm.

10. If instead of breaking one spring 2

m mass would have fallen from the block, then what will be new

amplitude?

a. 21.6 cm b. 26.2 cm c. 15.8 cm d. 16.9 cm.

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64

Statement [11 – 12] : Angular frequency in SHM is given by k

m� � . Maximum acceleration in SHM is

2� A and maximum value of friction between two bodies in contact is * N, where N is the normal

reaction between the bodies.

11. In the figure shown, what can be the maximum amplitude of the system so that there is no slipping

between any of the blocks?

*

*�+,-

a. 2

7m b.

3

4m c.

4

9m d.

10

3m .

12. Now the value of k, the force constant is increased then the maximum amplitude calculated in above

question will :

a. rem ain same b. increase c. decrease d. data is insufficient.

Question [18 – 19] : There is a spring block system in a lift moving upwards with acceleration 2

ga �

18. In mean position (of block’s oscillations) spring is :

a. compressed by 2

m g

kb. elongated by

m g

k

c. elongated by 2

m g

kd. elongated by

3

2

m g

k.

19. If maximum extension in the spring is 5

2

m g

k, then maximum upward acceleration (a

1) and maximum.

Downward acceleration (a2) of the block are:

a. 12

ga � b. 1

3

2

ga � c. 2

2

ga � d. 2

3

2

ga � .

Statement [20 – 21]: In case of pure rolling a = R� , where a is the linear acceleration and � the angular

acceleration.

Question: A disc of mass m and radius R is attached with a spring of force constant k at its center as shown

in figure. At x = 0, spring is unstretched. The disc is moved to x = A and then released. There is no

slipping between disc and ground. Let f be the force of friction on the disc from the ground.

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20. f versus t (time) graph will be as :

21. In the problem if k = 10 N/m, m = 2 kg, R = 1 m and A = 2 m. Find linear speed of the disc at mean

position :

a. 40

/3

m s b. 20 /m s c. 10

/3

m s d. 50

/3

m s

Answers: 9. b, 10. a 11 c. 12. c. 18. d 19. b, c 20. c 21. a.

15.6 Simple Pendulum:

A small heavy spherical bob connected, to one end of a light inextensible and flexible string, with its

other end fixed to a rigid and unyielding support, constitutes a simple pendulum.

Let P be the position of the bob at any instant, when the string makes an angle � , with the vertical.

(Figure. 15.14). If ‘l’ be the length of the string then, the net unbalanced force on the bob is F = mg sin

� . [where m is the mass of the bob].

l

Assuming � to be very small sin � # �Net unbalanced force F = mg (� )

Now, if ‘x’ be the displacement of the bob from the mean position C, then � = x

l.

xF m g

l

� ��� � � �� �

� �

���

Acceleration g xF

fm l

�� �

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66

f x�� � , which proves that the bob executes SHM for small amplitudes (less than 40).

Comparing with 2,f x�� �

We get 2 g g

l l� �� � � .

Alternatively: The torque acting about the point of suspension, is T = mg l sin �Now, moment of inertia I of the system about the point of suspension is I = ml2

From T = I� , we get 2 sin

sing

m g l mll

�� � �� � � .

2

2

gd

ldt

�� �� �

Comparing with

22

2

d

dt

�� �� �

2g l

or Tl g

� �� �

Note:

Here ‘l’ is the “effective length” of the simple pendulum, measured from the point of suspension to the

center of gravity of the bob.

A Few Important Facts Regarding Simple Pendulum

(a) Change of time period T due to a change in effective length.

[Law of lengths] T � l [other things remaining constant]

(i) If a child sitting on a swing stands up, then the effective length l decreases, due to an uplift of the

center of gravity of the system. This reduces the time period consequently.

(ii) If a hollow sphere filled with a liquid, and having a hole at its base, be made the bob of a simple

pendulum, the time period, initially increases, due to an increase in the effective length as the center of

gravity gets lowered due to the outflow of liquid from the sphere. The time period once again, gets

restored, when the whole liquid flows out, due to the fact that the center of gravity once again, comes

to the center of the sphere.

(iii) If the string be elastic with Young’s modulus Y, then the length of pendulum will be l + . l [where l is

the original length and . l, increase in length]

From Stress M g l

Y YStrain A l

� � ��.

[where M is the mass of the bob]

l M g

l AY

.� �

� Net length = 1 1l M g

l ll AY

� �.� �� � �� �� �

� � � �

Now, Time period 2 1l M g

Tg AY

�� �

� �� �� �

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Now, since Mg < < A Y

1M g

AY� $$

2 1l M g

Tg AY

�� �

� �� �� �

. [From Binomial expansion]

Obviously, the time period increases, in this case.

(iv) If the effective length be comparable to the radius of earth (Fig. 15.15)

The net unbalanced force on the bob = mg sin � ��

F m g � �� � � [Assuming � �� to be small]

x xF m g

l R

� �� � � �� �� �

� �

� ��� 1 1

m g xl R

� �� � �� �

� �

�

or acc. 1 1

f g xl R

� �� � �� �

� �

�� �

.

Which is of the form 2f x�� ��� �

, indicating that the bob, still executes SHM and its time period is not

infinity. Comparing the above equation with std. form 2 1 1

gl R

� � �� �� �

� �

� Time period

12

1 1T

gl R

��� ��� �� �

Putting l = R, we can determine, the corresponding time period as 2 12

RT hr

g�� #

If l be too large i.e., l/) .

10

l� /

2 84.6 minR

Tg

�� � � .

If l be too small compared to earth’s radius, then 1 1

R l$$ .

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� Neglecting 1

R in comparison to

1

l.

2l

Tg

�� � as derived earlier..

(b) Change of time period with a change in acceleration due to gravity. (Law of gravity)

1T

g� [other things remaining constant]

(i) If a simple pendulum be taken to higher altitudes, its time period increases, due to a decrease in the

value of g. Consider a place P at height ‘h’ above the earth’s surface. (Fig 15.16). The value of g at that

place g’ (say) is given by :

2

2'

g Rg

R h�

�

Where g = acc. due to gravity on the earth’s surface.

Time period 2'

lT

g�� .

2

22

l R h

g R�

�� 2 2 1

R h l l hT

R g g R� �

�� � � �� � � �� � � �

� � � �

Now, if h < < R or h

R < < 1

Then , 2

12

1

Th

gR

��

�� ��� �� �

12

21

hg

R

��� �� �� �� �

(ii) If a simple pendulum be taken to a place below the earth’s surface, the time period increases as a

consequence of decrease in the value of acceleration due to gravity. Consider a place P, at a depth ‘d’

below the earth’s surface (Fig. 15.17) the acc. due to gravity at P is given by :www.NetB

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69

'R d

g gR

�� �� � �� �

[where g is the acc. due to gravity on the earth’s surface]

Now, time period

2'

lT

g��

2l R

g R d�

� .

(iii) If a simple pendulum at equator, be shifted to poles, the time period decreases on account of an

increase in the value of ‘g’. (acc. due to gravity is more at poles compared to equator).

(iv) The time period of a simple pendulum in a satellite or within a freely falling lift is infinite (or the

pendulum does not oscillate due to the absence of any restoring force tending the bob to bring back to

the mean position), on account of the fact that the value of g is zero.

(v) For a simple pendulum oscillating in a lift moving up with an acceleration a or down with a retardation

a the effective value of g (say g’) will be g’ = g + a

2l

Tg a

�� ��

The time period always decreases.

For a simple pendulum, oscillating in a lift moving down with an acceleration a moving up with a

retardation a, the effective value of g (say g’) will be g’ = ( g – a) down if g > a or (a – g) up if a > g.

(Fig. 15.18).

For a simple pendulum oscillating in a vehicle accelerating with an acceleration ‘a’, the time period

decreases due to an increase in the value of ‘g’.

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Let us first determine the equilibrium position of the string (see figure 15.20). If � be the angle made

by string with vertical. Then : T cos � = mg and T sin � = ma

�

�

Dividing tana

g� �

1tana

g� � � �

� � � �� �

(Opposite to the direction of acceleration of vehicle).

Effective external force = 2 2m a g�

� Effective acceleration 1/2

2 2'g a g� �

� Time period 1/22 2

2 2'

l lT

gg a

� �� �� .

(vi) If the bob of simple pendulum is acted upon by additional forces of electrical, magnetic, buoyant,

gravitational origin, the time period will increase or decrease according as the net acceleration in the

downward direction is decreased or increased (see problem No. 18, 19)

(C) Law of Mass: The time period of a simple pendulum is independent of the mass of the bob, other

things remaining constant. For example, if two pendulums, be constructed of same effective length,

but whose one of the bobs is solid and the other hollow; both oscillate with the same time period.

(d) Law of Isochronism: The time period of a simple pendulum is constant for every oscillation

provided the amplitude remains small.

Example 17

Show that the maximum tension in the string of a simple pendulum, with angular amplitude 0� is given

by mg 201 �� . Hence find the maximum mass of bob that can be used, if the breaking load of the

string be F.

Solution

Evidently, the tension is maximum, at the lowest position of the bob. (Fig. 15.21)

2max

max

mvT m g

l� � … (1)

Where vmax

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maxv a���

2

22 2 2 2max 0 0

gv a l g l

l� � �

� �� � � �� �� �

� �

[Where a = linear amplitude and � = ang. Freq.]

Substituting for 2maxv in eqn. (1)

20 2

max 01m g l

T m g m gl

��� � � �

If M be the maximum mass, that can be employed for the bob. Then 201F M g �� �

201

FM

g �� �

�

Example 18

A simple pendulum of length l, is made to oscillate with a solid sphere (density D), in a non viscous

fluid of density 0 ; find the period of oscillation (D > 0 )

Solution

Net force on the bob will be (D - 0 ) Vg (down). Net Acceleration downward (See Fig. 15.22)

01

1D V g

f gDV D

0 0� � �� � �� �

� �

� Time period

2

1

lT

gD

�0

�� �

�� �� �

.

� The time period increases, due to a decrease in effective value of g.

Example 19

A simple pendulum consists of a small sphere of mass m, suspended by a thread of length L. The

sphere carries a negative charge ‘q’. The pendulum is placed in a uniform electric field of strength E

directed (i) vertically upwards (ii) vertically downwards, and (iii) horizontal. Find the time period in

each case.

Solution

(a) The net force downward F = mg + qE

Net acceleration downward F q E

f gm m

� �� � �� �� �

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Time period

2L

Tq E

gm

��� �� � �� �

.

Note: The electric force experienced by a body carrying a negative charge is opposite to the direction

of electric field.

(b) The net force downward = mg – qE [Assuming mg > qE] [Fig. 15.24 (a)]

� Acceleration qE

f gm

� �� � � �� �

Time period 2L

Tq E

gm

��� ��� �� �

The net force upward = qE – mg [Assuming mg < qE] [Fig. 15.24 (b)]

Acceleration q E

f gm

� �� �� �� �

�Time period

2L

Tq E

gm

��� � �� �� �

(c) If ' '� be the angle made by the string with the vertical (Fig. 15.25)

Then T cos � = mg

T sin � = qE

1tan tanq E q E

mg m g� � � � �

� � � � � �� �

Net force on the bob = 2 2

mg qE� .

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� Net acceleration

22 q E

f gm

� �� � � �� �

� Time period 1/ 2

22

2L

T

q Eg

m

��& �� �� ��' �� �

� �� �(

.

Example 20

A simple pendulum is suspended from the ceiling of a carriage, moving along a horizontal curve of

radius of curvature r, with a speed v. If ‘l’ be effective length of the pendulum find (i) the tension in the

string, while oscillating in the mean position and (ii) the period of oscillation (Assume 0� as the

angular amplitude).

Solution

Let us adopt the carriage as the frame of reference. Therefore, in addition to the tension and weight

acting on the bob, a pseudo (centrifugal) force should also be applied on it. (Fig. 15.26)

At the mean position, the pendulum is stable.

� For its equilibrium net vertical and horizontal forces should be zero.

� T cos � = mg

And

2

sinv

T ma where ar

�

� �� �� � �

.

tana

g�� �

or, g sin � - a cos � = 0. … (1)

Now, let � be the displacement (from mean position). at any instant

Then, the net unbalanced force tangential to the motion of the bob.

F = mg sin � �� - ma cos � �� .

= m [ sin .cos cos .sin ) (cos .cos sin .sin )]g a� � � � � � � �� � �

Since � is too small

cos 0 sinx

andl

� �� # #

Where ‘x’ is the displacement of the bob from the mean position

sin cos . cos sinx x

F m g al l

� � � � � �� �� � � �� �� �� �� � � � �� �� � � �� � � �� �� �� �� � � �� �� � �

� ���

.

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74

sin cos / cos sinm g a x l g a� � � � � � � � �

� Acceleration cos sing a x

fl

� �� �� [By virtue of 1]

� Comparing with 2f �� � x, the standard form of a simple harmonic motion, we get

1/ 22

2

sincos

cos sin

vg

g a r

l l

��� �

� �

� ��� ��� � � �� � �� � � �� �

� �� �

Now, maximum velocity (vel. at the mean position) max 0v l� ��

Or,

22 2 2max 0

cos singr vv l

r l

� ��

� ��� � �� �

� �.

Now, from the FBD of the pendulum in its mean position, it is evident that, for equilibrium along the

string.

2maxcos sin

mvT m g m a

l� �� � �

2 22 20 0

sincos sin cos

v vm g g

r r

�� � � � �

� � � �� �

� � �

220

sin1 cos

vm g

r

�� �

� �� � �� �� �

� �

Now, from tana

g� � we have

2 2 2 2sin cos

a gand

a g a g� �� �

� �

2 220

2 2 2 21

g aT m

a g a g�

� �� � � �� �� � �

1/24

2 20 2

1v

m gr

�

� � �� �� � �

(ii) Time period = 1/24

2

2

22

l

vg

r

��

��

� ��� �� �

� �

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Example 21

A simple pendulum, of length l, installed between two inclined walls, each making an angle � with

the vertical, oscillates with a maximum speed of u, colliding elastically with the walls. Find the time

period of oscillation. For the above problem, how is � related to u? (Fig. 15.28)

Solution

Had there been no walls, the linear amplitude of oscillation would be

max[ ]u

a From v a��

� �

But since,g

l� �

la u

g�

Now, let the equation of motion for the pendulum be 0 sin t� � �� [where 0� = angular amplitude]

The time taken by the pendulum to move from O (mean position) to A (extreme position), can be

obtained as sinu g

tlg l

�� �

� � �� �� �

sing lg

tl u

�� �

1sing lg

tl u

��� �

� � � �� �� �

1sing ll

tg u

��� �

� � � �� �� �

Evidently, the period of oscillation T = 4t

14 sin

g llT

g u

��� �

� � � �� �� �

Now, since, the bob collides with walls, the angular amplitude should exceed the value � .

i.e.,u

or u g lg l

� �% %

This is the required relation.

Note: If the bob just misses the collision with wall. Then, in the limiting case, angular amplitude = �

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Then ,14 sin (1)

lT

g

�� 42

lT

g

�� �� � � �

� �2

l

g�� .

This proves the validity of above derivation.

Second’s Pendulum: It is a pendulum with period of oscillation 2 sec or period of swing 1 sec. Length

of a second’s pendulum at mean sea level, may be calculated as follows.

From

2

22

4

l T gT we have l

g�

�

�� � .

Putting T = 2 and taking 2� # g

� 1l m# .

Note: A second’s pendulum always bears a time period of 2 sec. However, the length changes with a

change in acceleration due to gravity.

Loss or Gain of Time by Pendulum Clocks:

Pendulum clocks gain time, if the time period of the pendulum decreases (due to a decrease in length

in winter or due to increase in acc. due to gravity, as when taken from equator to poles). For example,

let the time period of a second’s pendulum (originally) decrease to 1.99 sec. Now, each time. It oscillates,

it counts 2 sec rather than 1.99 s. Thus, after a considerable no. of oscillations, the time counted or

shown by the clock is more than the actual time, and consequently, is goes fast.

Similarly, Pendulum clocks loose time and go slow, when their time period increases.

Gain or loss of time due to a change in surrounding temperature:

Let T be the time period of clock pendulum of effective length l, showing the correct time.

Then, 2l

Tg

�� … (1)

Now, let �. be the change in temperature of the surrounding, the new length will be

' 1l l � �� � . [where � is the coefficient of linear expansion of the material of the pendulum]

� The changed time period

'' 2

lT

g��

12

l

g

� ��

� .� … (2)

Dividing equation (2) by (1).

1/2'1 1

T

T� � � �� � . � � .

11

2� �� �

� � .� �� �

[Expanding by binomial expansion]

11

2T T � �� ��� � � .� �

� �

Also; ' 1

2

T T

T� �

�� �� .� �

� �

i.e., Relative change in time period1

2� �� .

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77

or, time lost or gained each second is 1

2� �� . .

Note: Unless otherwise stated, adopt the time period of clock pendulum as 2 sec.

Example 22

A clock with an iron pendulum keeps correct time at 200 C, how much time will it loose or gain, in a

day, if the atmospheric temperature changes to 400 C. (� for iron = 0.000012/0 C)

Solution

The relative change in time (i.e., time lost or gained per swing i.e., in 1 sec is given by)

6 51 112 10 40 20 12 10

2 2

Ts

T� � � �.

� �. � � � � � �

� Total time lost per day = 5 312 10 86400 12 864 10 10.37 sec

� �� � � � � �

Example 23

The pendulum of a certain clock has a time period 2.02 sec. How fast or slow does the clock run in an

interval of one week.

Solution

Since the correct time period for a clock pendulum is 2 sec. Therefore, for every 2.02 sec. (one

oscillation period) of actual time elapsed, it would count only 2 sec. In other words, as successive

oscillations are repeated it goes on loosing time, and would run slow.

Loss of time per oscillation ( = 2 sec). = 2.02 – 2 = 0.02 sec.

� Loss of time per sec = 0.01 sec.

� Time lost in 1 week = 0.01 × 86400 × 7 = 6048 sec = 108 min.

Example 24

The linear displacement ‘x’ of a simple pendulum’s bob from its mean position varies as x = a sin (0.

707 � t) where a is a constant and t is in sec. Find its length (take g = � 2 m/s).

Solution

Since the simple pendulum executes SHM. Therefore comparing the displacement equation with

standard form x = a sin t� , we get 2

T� � rad/sec.

Also from, 2

2l

Tg

��

�� � .

We have, 2 g

l� �

2

22

gl m

l

�� � � � .

Example 25

A simple pendulum of 1 meter, suspended from the ceiling of a lift executes oscillation with a period

of 6.25 sec. If the lift is moving up with an acceleration ‘a’ find a (take g = 2� m/s2).

Solution

If ‘a’ be the acceleration of the lift upwards then, net acceleration on the bob downwards, relative to

the lift is g + a

2l

Tg a

�� ��

2

2

16.25 4

a�

�� �

� � � ��� �

2 2 216 9

25 25a a� � �

�� � � � �

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78

Thus, the lift is retarding with a retardation of 2 29

/25

m s�

Objective Questions:

1. The time period of a simple pendulum is independent of :

Ans: (d).

a. The length of the string

b. The radius of the bob

c. Amplitude of oscillation

d. None of these.

2. The time period of a simple pendulum of infinite length suspended on earth’s surface is :Ans: (c)

a. Inifinte b. 59.8 minutes c. 84.6 minutes d. 24. hrs.

3. The ampliutude of oscillation of a simple pendulum decays exponentially with time due to air resistance

and friction at the point of suspension. The total energy of the pendulum. Ans: (A)

a. Decays exponentially with time

b. Decays linearly with time

c. Decays at a steady rate

d. Decays at a rate less than the amplitude.

4. A hollow sphere is filled with water. It is hung by a long thread. As the water flows out of a hole at the

bottom of the sphere, the frequency of oscillation will : Ans: (D)

a. Go on increasing

b. Go on decreasing

c. First increases and then decreases

d. First decreases and then increases.

4. The formula 2l

Tg

�� holds good only for small angular amplitude0

0 4� $ . For large amplitudes

2l

Tg

�2 .

5. See text

564 10

2 29.8

RT s

g� �

�� � .

6. Let the amplitude A be given by A = k1 e- k

2

t .

Then T.E. 2 22

22 2 21 1, . .

2 2

k t k tm A m e or T E e� � �� �� �

7. Freq. of a simple pendulum 1 1

2

gf Evidenty f

l l�

�� at any place.

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79

The effective length l is the distance of the point of suspension O’ form the center of gravity (C.G) of

the bob. As, water flows out, of the hole at the bottom, the C.G descends from center towards the

bottom, increasing the effective length and consequently, f decreases.

However, when all the water has flown out, the C.G. of a hollow sphere is once again at its center and

hence the effective length would decrease, thereby increasing the frequency.

5. A simple pendulum oscillates slightly above a large horizontal metal plate. The bob is given a charge.

The time period: Ans : B

a. Has no effect, whatever be the nature of charge.

b. Always decreases, whatever be the nature of charge.

c. Always increases, whatever be the nature of charge.

d. May increase or decrease depending upon the nature of charge

11. When any charge is given to the bob of the pendulum, it induces opposite charge on the metal plate,

and hence a net force of attraction acts on the bob. Thus, the effective value of g increases.

6. The time period of a simple pendulum is T. In which of the following situations will the period increase?

Ans : B

a. The pendulum is suspendend in lift, going with an acceleration (< g)

b. The bob is given a negative charge, while the point of suspension a positive charge.

c. The pendulum is suspended in a cabin, moving along a circle with uniform speed”

d. The pendulum is suspended in a cabin, moving along a straight path, with a uniform acceleration.

15. The electrostatic force of attraction acting on the bob is up, which decreases the net force of gravity

downwards.

effg g� $ .

So, since 1

eff

Tg

� a decrease in g

increases the time period.

In the rest of all the cases, the effective value of g increases thereby decreasing

the value of T.

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80

35. The bob of a simple pendulum of mass m executes oscillations with a total energy E. At any instant is

at one of the extreme positions. Its linear momentum, after a phase change of6

� rad takes place, will

be:

a. 2 m E b. 2

m Ec. m E d.

3

4

m E.

36. A simple pendulum of length l has a time period T for small oscillations. A fixed obstacle is placed

directly below the point of suspension, so that only the lower quarter of the string continues oscillations

(see figure 15.74). The pendulum is released from rest at a certain point O. The period of oscillation,

assuming small angles will be:

3 /4

O

a. 3

4

Tb.

4

Tc. 1 3

2

T� d. T.

38. Two simple pendulums of time periods 3s and 5s respectively start oscillating simultaneously from

their mean positions. What is the minimum time elapsed, before which they oscillate with the same

phase?

a. 15 s b. 2 s c. 7.5 s d. 8 s.

39. In the above problem no. 38. if the bobs of the pendulums start their oscillations, from two opposite

extreme positions, then, the answer would be :

a. 3.75 s b. 7.5 s c. 4 s d. 1.5 s.

40. Figure 15.75 shows three physical pendulums consisting of identical spheres rigidly connected by a

mass less stiff rod. The pendulums are vertical and are made to oscillate about the point O. If their

respective frequencies be f1, f

2 and f

3 then which of the following is correct :

a. 1 2 3f f f� $ b. 1 2 3f f f� % c. 1 2 3f f f$ $ d. 1 2 3f f f% % .

Answers: 35. a 36. a, 38. c 39. a 40. a.

Only one option is correct

17. A simple pendulum has time period T = 2 s in air. If the whole arrangement is placed in a nonviscous

liquid whose density is 1

2 times the density of bob. The time period in the liquid will be:

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81

a. 2

2s b. 4 s c. 2 2 d. 4 2 s.

21. An accurate pendulum clock is mounted on the ground floor of a high building. How much time will

it lose or gain in one day if it is transferred to top storey of a building which is h = 200 m higher than

the ground floor. Radius of earth is 6.4 × 106 m.

a. it will lose 6.2 s b. it will lose 2.7 s.

c. it will gain 5.2 s d. it will gain 1.6 s.

23. A simple pendulum 4 m long swings with an amplitude of 0.2 m. What is its acceleration at the ends of

its path? (g = 10 m/s2)

a. zero b. 10 m/s2 c. 0.5 m/s2 d. 2.5 m/s2.

32. The maximum tension in the string of a pendulum is two times the minimum tension. Let 0� be the

angular amplitude. Then cos 0� is :

a. 1

2b.

3

4c.

2

3d.

3

5.

33. A pendulum has time period T for small oscillations. An obstacle P is situated below the point of

suspension O at a distance 3

4

l. The pendulum is released from rest. Throughout the motion the moving

string makes small angle with vertical. Time after which the pendulum returns back to its initial

position is :

a. T b. 3

4

Tc.

3

5

Td.

4

5

T.

35. Time period of a simple pendulum of length L is T1 and time period of a uniform rod of the same

length L pivoted about one end and oscillating in a vertical plane is T2. Amplitude of oscillations in

both the cases is small. Then T1/T

2 is:

a. 4

3b. 1 c.

3

2d.

1

3.

63. The period of oscillation of simple pendulum of length L suspended from the roof of the vehicle which

moves without friction, down an inclined plane of inclination � , is given by :

a. 2cos

L

g�

� b. 2sin

L

g�

� c. 2L

g� d. 2

tan

L

g�

�

66. A simple pendulum has time period T1. The point of suspension is now moved upward according to

the relation y = Kt2, (K = 1 m/s2) where y is the vertical displacement. The time period now becomes

T2. The ratio of

21

22

T

T is: (g = 10 m/s2)

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a. 6

5b.

5

6c. 1 d.

4

5.

15.9 Physical Pendulum or Compound Pendulum:

Any rigid body, suspended about a fixed point and capable of oscillating about a horizontal axis

passing through the point, constitutes a physical or a compound pendulum. Figure 15.45 along side

shows a rigid body (of any shape) suspended about a fixed point O (the center of suspension) capable

of rotating about a horizontal axis through O. Let G be the center of gravity at a distance l from the

point of suspension O. If ‘R’ be the vertical reaction upwards at the support O, then for vertical

equilibrium, we have.R = mg

l

When at rest, the line of action of R and mg is the same and hence, there is no net torque on the body.

If the body be given a slight angular displacement about O, it starts oscillating, the oscillations being

angular SHM.

Consider the position of the body; at any instant (during oscillatory motion) when the line joining O

and G makes an angle ' '� with the vertical .

The net torque on the body about the point O will be given by sinm g l" ��

If I be the moment of inertia of the body about O, then

2

2

dI

dt

�" � [where

2

2

d

dt

� = angular acceleration]

2

2sin

dI m g l

dt

��� � �

[The – ve sign, here indicates that the direction (or sense] of � and " are opposite]

[� for small angles sin� = � ]

2

20

d m g l

Idt

� �� � � .

Which reveals the fact that, the body performs angular SHM with angular frequency

m g l

I� �

Now, if T be the time period for each oscillation, then

22

IT

m g l

��

�� � ... (20).

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83

If K be the radius of gyration of the body about a parallel axis through the center of gravity

then, I = mK2 + ml2

2 2

2K l

Tg l

��

� . … (21)

If L be the length of an equivalent simple pendulum (i.e., a pendulum whose time period is same as

that of the compound pendulum)

Then , 2L

Tg

��

Comparing the above equation with equation (21) we get

2KL l

l� � .

Thus, the length of equivalent simple pendulum is

2K

l more than the distance between the point of

suspension O and the center of gravity G. If we choose a point O’, on the line OG produced such that

GO’ =

2K

l, then O’ is known as the center of oscillation.

Interchangeability of Center of Suspension and Center of Oscillation:

Suppose we shift the point of suspension from O to O’. Then time period

2 2

' 2'

K lT

gl�

�� . (from

eq. 21)

Where l’ is the distance of O’ (new point of suspension) from center of gravity G.

But,

2

'K

GOl

� .

2

'K

ll

� �

22 2

2 2

2

2 2

1

' 2 2

K KK K

l lT

K Kg g

l l

� �

� � � �� � �� � � �� � � �� � � �� � �� � � �� � � �� � � �� � � �

.

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2 2

22

l K l

g l�

��

2 2

2K l

g l�

�� which is same as equation (19).

'T T� �

Thus, it is evident that, the time period remains the same for both the center of suspension and center

of oscillation i.e., both are interchangeable.

Minimum Period of Oscillation:

From equation (21), the time period T for a compound pendulum is given by

2

2

Kl

lT

g�

� ��� �� �

� ��

Which can be rewritten as

221

2 2 2K

T l K Kg l

�� �� �� �� � � �� �� �� �� �

21

2 2K

l Kg l

� � �� �� � �� �� �� � �

.

� 2K is constant, therefore T will be min, when

2K

ll

� ��� �

� � is also min. Since the min value of a

square can be zero.

� K = l

Thus, the time period of oscillations is minimum when the distance of the center of suspension from

the center of gravity is same as the radius of gyration of the body about a parallel axis through C.G.

Also, min

22

kT

g�� ... (22)

Example 33

A uniform circular disc of radius R, is suspended about a point (not its center) such that it can oscillate

about a horizontal axis through that point. What is the distance of this point from the center, for the

time period to be minimum? What is the minimum time period?

Solution

The radius of gyration for a disc about its center of gravity (i.e., geometrical center) is given by

2

22

R I M RK k

M M

� �� � �� � �� .

� This is the required distance

Also, min

22

2

RT

g�� (See equation 22).

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22

R

g�� .

Example 34

Find the frequency of oscillations of a uniform thin ring of radius ‘R’ suspended about a point on its

circumference.

Solution

From equation (20) 2I

Tm g l

�� .

Here 22 ,I m R l R� �

� Frequency 2

1 1

2 2 22

m g R gn

Rm R� �� � .

15.10 Torsional Pendulum:

A rigid body suspended by means of a wire, with its one end fixed rigidly to the ceiling, and about

which the body can perform horizontal angular oscillation, is known as a torsional pendulum.

Figure 15.48 shows a body attached to a wire OO’ whose one end O, is fixed to the ceiling. When the

body is at rest, there is no torque on the body. If the body be given a slight “twist”, and left free, it starts

performing angular SHM.

Let O’ P be a horizontal line chosen on the body at rest. Further let O ‘P’ be its position at any instant,

during oscillations. Since the upper end O of the wire is fixed, and the lower end O’ is rotated by an

angle ' '� we say that the wire has a twist ' '� . Due to this twist, a torque (restoring) is developed in the

wire due to elastic forces, which tend to bring the wire and the body back to a position � = 0 (i.e., the

mean position). This restoring torque is directly proportional to the twist, but opposite in sense.

Then, " � �� Or, k" �� � .

Where k is a positive constant of proportionality known as torsional constant (for the wire).

If I be moment of inertia of the body about the wire, then

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2

2

I d

dt

�" �

2

2

I dk

dt

��� � �

2

20

d k

Idt

��� � � .

Which proves the fact that, the body executes angular SHM, with an angular frequency.

k

I� � .

� Time period 2

2I

Tk

��

�� � ... (23)

Example 35

A circular disc of mass m = 500 gm and radius r = 5 cm is suspended by means of a wire of torsional

constant k = 0.36 N – m/radian. Find the time period of oscillations.

Solution

Moment of inertia of the disc about the wire 2

2

m r� .

� From equation (23), time period of the torsional pendulum

2I

Tk

��2

22

m r

k��

Substituting the values of m, r and k.

We get

22

2

5 1012

2 2 36 10T �

�

�

�� �

� �

225 10 5

2 24 36 20 36

� ���

� � �� �

sec.72

T�

� �

Example 36

A torsional pendulum consists of a square lamina, suspended by means of a wire with torsional constant

k = 0.25 N.m/rad. If the time period of oscillation be 2.5 sec, find the moment of inertia I of the square

lamina, about the wire.

Solution:

The time period for oscillations of a torsional pendulum is given by 2I

Tk

�� .

or,

222

2 2

2.5 0.250.04

4 4 3.14

T kI kg m

�

�� � � �

�.

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PRACTICE PROBLEMS:

PROPERTIES OF MATTER:

116.

In a Searle’s experiment, the diameter of the wire as measured by a screw gauge of least count 0.001

cm is 0.050 cm. The length, measured by a scale of least count 0.1 cm, is 110.0 cm. When a weight of

50 N is suspended from the wire, the extension is measured to be 0.125 cm by a micrometer of least

count 0.001 cm. Find the maximum error in the measurement of Young’s modulus of the material of

the wire from these data.

117.

A 5 m long cylindrical steel wire with radius 2 × 10-3 m is suspended vertically from a rigid support

and carries a bob of mass 100 kg at the other end. If the bob gets snapped, calculate the change in

temperature of the wire ignoring radiation losses.

(For the steel wire: Young’s Modulus = 2.1 × 1011 Pa; Density = 7860 kg/m3; Specific heat = 420 J/kg

– K).

118.

A light rod of length 200 cm is suspended 200 cm is suspended form the ceiling horizontally by means

of two vertical wires of equal length tied to its ends. One of the wires is made of steel and is of cross

– section 0.1 sq cm. Find out the position along the rod at which a weight may be hung to produce

(i) equal stress in both wires.

(ii) equal strains in bob wires.12 2 12 21 10 / , 2 10 /brassY dyne cm dyne cm� � �

119.

A sphere of radius 10 cm and mas 25 kg is attached to the lower end of steel wire which is suspended

from the ceiling of a room, the point of support is 521 cm above the floor. When the sphere is set

swinging as a simple pendulum, its lowest point just grates the floor. Calculate the velocity of the ball

at its lowest position.

(Young’s modulus of steel = 20 × 1011 dyne/cm2. Unstretdched length of wire = 500 cm, Radius of

steel wire = 0.05 cm) g = 980 cm/s2.

A uniform cylindrical pulley of mass M and radius R can freely rotte about the horizontal axis O. The

free end of a thread tightly wound on the pulley carries a mass m0. At a cetain angle � it counter

balances a point mass m fixed to the rim of the pulley. Find the frequency of small oscillations of the

arrangement. (Fig. 15.25)

Solution:

Step 1 : Under equilibrium of the system, the total

The net unbalanced torque on the cylinder about its center = r] 0m g R m g R� �� �

0[ (sin cos sin ) ]g R m m� � �� � �

= 0[ (sin cos ) ]g R m m� �� � [ cos 1 sin ]is small and� � � �� ��

cosm R" ��� � .

Now, it I be the moment of inertia of the system about the horizontal axis of rotation then

22 2

02

M RI m R m R� � �

Now, from

2

2

I d

dt

�" �

2 22 2

0 2cos

2

M R dm R m R m g R

dt

�� �

� �� � � � �� �� �

� �

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Step 3 :

2

2

0

cos

2

d m g

mdtR M m

� ����

� �� �� �� �

Comparing with the standard form

22

2

d

dt

�� �� � it is evident that the system executes angular SHM,

with an angular frequency 0

cos

2

m g

mR M m

�� �

� �� �� �

� �.

� Frequency 2

n��

�0

1 2 cos

2 2

m g

R M m m

��

�� �

Note: From equation (1) m0 = m sin �

1 2 cos

2 [ 2 (1 sin )]

m gn

R M m

��

��

� �

Clockwise moment of due to mass m0 should be equal to the anti – clockwise moment due to mass m

about the center of the pulley.

0 sing m R g m R �� � … (1)

Step 2: Let the cylinder be given a slight angular displacement, the cylinder starts executing SHM.

(which is to be proved). Consider, the position, in which the cylinder has a displacement � (in the

clockwise direction). (Fig. 15.53)

[ 2 (1 sin )]2

n v R M m�

�� � � .

2. A point mass ‘m’ is suspended at the end of mass less wire of length l and corss – sectional area ‘A’.

If ‘y’ be the Young’s modulus of elasticity for the wire, find the time period for small vertical oscillations.

Solution:

Step 1: Let . l be the initial extension in the wire, then

since strain = stress / Y

l m g

l AY

.� � ... (1)

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89

Step 2: Let the mass be displaced slightly and left free to oscillate. Consider the position, at any

instant, when the mas is at distance ‘y’ below the mean position.

Net unbalanced force (upwards) acting on the block.

F = Restoring force – weight f block.

l yAY m g

l

. �� � [� Restoring force = strain × Y × A]

AY y

l� [By virtue of eqn. (1)]

AYF y

l

� �� � �� �

� �

�� ��

Which proves the fact that the mass executes SHM.

Step 3: The net acceleration (up) can be obtained as

2

2

d y F AYy

m mld t

� �� � �� �

� �

Comparing with

22

2,

d yy

d t�� � the angular frequency for the oscillations is

AY

ml� �

� The period 2

2ml

TAY

��

�� �

5 An ideal gas is enclosed in a vertical cylindrical container and supports a freely moving pistorn of

mass M. The piston and the cylinder have equal cross sectional area A. Atmosphere pressure is P0 and

when the piston is in equiblibrium the volume of the gas is V0. Assuming the system to be completely

isolated form its surroundings, show that the piston executes SHM, when the piston is displaced slightly

form its equilibrium position, and find the time period (ratio of sp. heat capacity for the gass = 3 ).

Solution

Step 1: When the piston was in equilibrium net force (acting vertically) on it must be zero.

or, 0 1

M gP P

A� � ... (1).

Step 2 : Let the piston, be at any instant, displaced (up) by a distance ‘y’ from its mean position.

If P2 be its new pressure, then for an adiabatic process.

2 1[( ) ] [( )]P l y A P l A3 3� � [where lA = V0]

2 0 1 0[ ] [ ]P V A y P V3 3� � �

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2 1 2 10 00 0

1 1A y A y

P V P V P PV V

33 3 3

� � � � � �� � � � � �

[Expanding by binomial theorem and neglecting the terms containing higher powers of y]

12

0

1

PP

A y

V

� � �� �

� ... (2).

Now, the net unbalanced force acting on the piston (downwards) 0 2P A M g P A� � � .

0 2

M gA P P

A

� �� � �� �� �� � �

.

11

0

1

PA P

A y

V

3

� �� �� �� �� �� �� �� �� �� �� ��� �� �� �� �� � �

[From equations (1) & (2)]

1

0

AA P y

V

3 � � �

�[Neglecting

0

A y

V

3 in comparison with 1]

2

00

A M gF P y

V A

3 � �� � �� �

� �[From equation (1)]

� Now comparing with the standard form of equation for SHM: F = – K (y), it is evident that the

piston executes SHM.

Step 3 : The net acceleration on the piston

2 2

020

d y F A M gP y

M M V Ad t

3 � �� � � �� �

� �.

Compring with

22

2

d yy

d t�� � .

The angular frequency � can be obtained as

2

0

0

A M gP

M V A

3� � �

� �� �� �

� The period 0

20

2M V

TM g

A PA

�3

�� ��� �� �

6. A particle of mass m is attached to the mid point of a wire of length L and stretched between two fixed

points. If T be the tension in the wrie, find the frequency of lateral oscillations.

Solution:

Let the mass m be displaced by an amount ‘x’ from its mean position (while oscillating) laterally. The

net unbalanced force (towards the mean position) will be :

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91

2 sin 22

x

LF T T�

� �� �

� � � �� �� �� �

or 4T x

FL

� �

�

��

Which proves the fact that the mass executes SHM. The net acceleration of the mass m is F

m

��

i.e.,

2

2

4 ( )d x T x

m Ld t

�� .

Comparing the above equation with

22

2

d xx

d t�� � .

It is evident that the angular frequency 4T

m L� � .

� Frequency of oscillations 1 4

2 2

Tn

m L

�� �

� �

7. An elastic string of natural length 2a can support a certain weight extending to a total length of 3a. One

end of the string is now attached to a point on a smooth horizontal table and the same weight is

attached to the other end and can move on the table. Show that, if the weight is pulled to any distance

and let go, the string slacks after a time 2

a

g

� � �� �� �

.

Solution

If ‘m’ be the mass of the weight, and assuming the string to obey Hooke’s law.www.NetB

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92

i.e., F = Kx ... (1).

We have mg = k ( 3a – 2a) = ka ... (1)

[where K is the force constant for the wire]

Let the wire be stretched by an amount ‘a0’. The net restoring force F

0 = K (a

0) acting on the weight.

The weight starts moving under the action of this restoring force and becomes slack when the extension

in it vanishes. If ‘x’ be the extension in the wire, then restoring forces acting on it will be F = - Kx.

( )m g

xa

� � [By virtue of eqn. (1)]

The acceleration of the weight

2

2( )

d x gx

ad t� � ………….. (2).

Which shows that the weight executes SHM.

Let the displacement equation for the weight be given as 0 sinx a t� �� � .

[where � is the phase constant]

Now, initially i.e., at t = 0; x = a0

0 0 sin (0)a a � �� � �

1sin (1)

2

�� �� � �

� Displacement equation becomes

0 0sin cos2

x a t a t�

� �� �� � �� �

� �

If t1 be the time taken by the weight to reach the point where extension becomes 0.

Then 0 10 cos ( )a t��

11 cos (0)

2t

�� �� � � [least value]

12

Or t��

�

� Required time 2 2

a

g

� ��

� � . [� from eqn (2) g

a� � ].

8. A small body is dropped in a tunnel dug along the diameter of earth from a height h (< < R) above the

earth’s surface. Find the time period of motion. Is the motion simple harmonic?

Solution

Let AB be the tunnel dug along the diameter of earth with O as its center. Let P be the pt. (at a height

h above the surface of earth) from which the body (of mass m, say) is dropped.

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93

Let the body be at a distance of ‘y’ from the earth’s center at any instant, (with in the earth).

The net force (gravitational) exerted on ‘m’ due to earth will be2'

10G M m

F ydiwb

� .

Where M � is mass of the spherical portion of earth, with center at O and radius y..

Now, 34

'3

M y� 0� . [where 0 is the average density of earth]

4

3F G m y� 0� ��� ��

The net acceleration of the body will be

2

2

4( )

3

d y FG y

md t� 0� � �

��

Comparing the above equation, with the standard from

22

2,

d yy

d t�� � it is evident that, the body

ocillates simple barmonically within the earth, with an angular frequency 4

3G� � 0� .

Now, since 2

G Mg

R� .

[where M is the total mass of earth and g the acc. due to gravity on the earth’s surface].

3

2

4 4

3 3

Gg R G R

R� 0 � 0� �

� � �� �� �

.

4

3

gG

R� 0� �

g

R�� � . ... (1)

Now, let us consider the body at a distance y from the center (during its downward motion) i.e., at C.

Kinetic energy possessed by the body there will be the sum of the kinetic energy possessed by it at A

and the work done by the gravitational force (restoring force) on it.

i.e., if v be the velocity of the body at C, then

2 2 2 2 21 1 1

2 2 2Amv mv m R y�� � �

2 2 22

gv g h R y

R� � � �

(2 )zg

v h R R yr

� � � �

[ ](2 ) z

d y g d yd t v

R d th R R y

�� � � �

� ��

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� If ‘t’, be the time taken by the body to travel from point A to point O, then

10

2 0(2 )

t

R

d y gd t

Rh R R y

��

� �� � .

11

0

sin2

R

y gt

Rh R R

� � �� �� �� �

1/21

1 sin2

R Rt

g h R

� � � � �� �

.

If ‘t2’ be the time taken by the body to fall from point P to earth’s surface (i.e, point A), then

22

2 1[ , ]

2

ht From s u t at

g� � � .

Now, we know that the total time period of oscillation of a particle executing SHM is equal to four

times the min, time taken by it to travel from its mean position to extreme position or vice versa.

Moreover, the time taken by the body to rise to the maximum height from the earth’s surface, or vice

versa, at both points A, as well as B is the same. So, period of oscillation T = 4 (t1 + t

2).

14sin 2

2

RR h

R hg

�& �� �� �

� �� �' �� ��� �� �( .

Note: If the body be dropped into the tunnel at point A (or B) with an initial velocity u, then in the

above problem 2

ut

g� [From v = u + at]

Alternatively: t1 can be evaluated as follows: Force acting on the body at C.

( )m g

F yR

� � .

So, work done by the gravitational force, on the body, during the time it comes from point A to C,

during its downward motion.

R

R

m g yW d y W F d y

R � � �� � �� ��

2 2

2

m gR y

R� �

Now, total K.E of the body at C will be equal to the sum of K.E. at A and work done by gravitational

force on the body.

i.e., 2 2 2 21 1

2 2 2A

m gmv mv R y

R� � �

22 2

A

g yv v g R

R� � � � .

Now, the velocity at point O can be obtained by putting y = 0

i.e, 20 Av v g R� � … (3)

Also, 2 2A

Ry v g R v

g� � � .

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95

Now, from equation (1) d v g

yd t R

� � .

i.e., 22A

d v gd t

Rv g R v

�� � . [� The negative sign has been dropped, since v increases with t]

At : t = 0; v = vA and at t = t

1 ; v = v

0.

0 1

02 2A

v t

vA

d v gd t

Rv g R v

� �� �

� �0

11 01 0

sin

A

v

t

Av

v gt

Rv g R

� � �� �� �� �

2 21 10

1sin sin A

A A

v v gt

Rv g R v g R

� � � � � �� � �� � � �� � � �

.

1/21

1

2sin

2 2

R ht

g h R

� �& � � �

� � �' �� �� �� �( . [from eqn. (3)]

Notice that the two expressions for ‘t1’ deduced by the two methods are the same

1 1sin cos2 2 2

R R

R h R h

�� � � �� � � �

� �� � � � � �1 1[ sin cos / 2]� � �� �� �� .

1 2sin

2

h

R h

� � � � �

�� � �.

9. A particle of mass m is located in a unidimensional potential field where the potential energy of the

particle depends on the co – ordinate x as U (x) = U0 (1 – cos ax) where U

0 and a are positive constants.

Find the period of small oscillations,w hich the particle performs about the equilibrium position.

Solution

Evidently, the equilibrium position of the particle will be the position corresponding to the least potential

energy.

� U will be minimum when cos ax is maximum

i.e., max0 0 [ (cos ) 1]ax x ax� � � �� .

Thus, the particle oscillates about the point x = 0.

Now, net force acting on it will be given by

0[ (1 cos )]d U d

F U a xd x d x

� � � � � = – U0 a sin ax.

� Oscillations are small

20sin ( )ax ax F U a x� # � � � .

or acceleration

220

2( )

U ad x Fx

m md t� � �

Comparing the above equation with

22

2.

d xx

d t�� � .

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96

It is evident that, the particle executes SHM.

with an agular frequency

22 0a U

m� � .

� Time period 20

22

mT

a U

��

�� � .

10. A uniform rod of mass m suspended by two identical threads each of length 1 is turned through a small

angle about the vertical axis passing through the middle point C (See fig 15.62). Then the rod is

released to perform small oscillations. Find the time period.

Solution

The threads, here are ssumed to be mass less, and inextensible. As, the threads are inextensible, the

rod of mass m, slightly gets elevated, when it is turned through a small angle (� , say).

Let the threads deviate in this process by an angle 4 .

Total mechanical energy of the system (i.e., rod) will be E = rotational K.E. + potential energy.

Rotational 21

. .2

K E I��

221

2 12

M L d

d t

�� � � �� � � � �� � � �� �

.

Where L is the length of the rod. And potential energy U = mgl (1 - cos 4 )

[Taking the equilibrium postion of the rod as datum level]

2211 cos

2 12

M L dE m g l

d t

�4

� � � �� � � �� � � �� � � �� �

.

Since mechanical energy remains conserved in SHM.

0d E

d t� �

221

1 cos 02 12

d M L dmgl

d t d t

�4

& �� � � �� �� � � �� �' �� �� � � �� �� �(

2 2

sin12

m L d d dmgl

d t d t d t

� � 44

� � � �� � � �� �� � � �� � � �� � � �� � � �� � � �

Now, linear displacement of either end of rod.

sin sin2

Ll 4 �� �

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[ & ]2

Ll are small4 � 4 �� � � ... (2)

Also,2

ld L d

d t d t

4 �� �� � �

� �... (3).

Making use of equations (2) and (3), equation (1) can be rewritten as

2 2

012 2

L d d L dg

d t d t d t

� � �4

� �� � � �� �� �� � � �� � � �� �� �

2 2

2 2

30 0

6 2 2

L d L dg

l ld t d t

� � � �� � � � � � � �� � � �� � �� �

Comparing the above equation with the standard from

22

20

d

d t

�� �� � .

It is evident that, the rod executes angular SHM with angular frequency 3 g

l� � .

� Time period 2

23

lT

g

��

�� � .

11. A ring of radius R carries a uniformly distributed charge + Q. A point charge – q of mass m placed at

the center of the ring and is given a small displacement a (< < R), along the axis and released. Does the

particle execute SHM? If so what is the time period? (Fig. 15.63)

Solution

When the charge – q is placed at the center of the ring, it is attracted with equal force (or intensity) in

all directions by the positive charge uniformly distributed along the ring, and hence remains at rest.

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98

However if the charge – q be slightly displaced from the center (along the axis) it starts executing

oscillatory motion due to the electrostatic force of attraction always directed towards the center. Let us

find out whether the motion is SHM or not. Consider the instant, when the charge (- q) is at a distance

of x from the center (x d•a). Consider two elemental lengths on the ring (A and B say). (Fig. 15.64).

They exert forces of attraction F1 and F

2, on the charge (- q) at P, making angles � , with the axis (See

Fig 15.64). Their components normal to the axis F1 sin � and F

2 sin � being equal and opposite

cancel each other. However, their components along the axis F1 cos � and F

2 cos � add up being in

the same direction. Thus, we find that, the component of force normal to the axis due to any elemental

length, is neutralized (or cancelled) by the corresponding component of force, due to an elemental

length diametrically opposite to the former (of the type A and B). Further, their component along the

axis add up.

� The net force on charge (- q) along the axis is 2 20

( ) ( )cos

4

dq qF

R x

�

� 5

�� 6

� .

[Where (dq) is the charge on the elemental length]

2

2 200

( )cos

2 4

Q qdl

R R x

� �� � 5

� � �� � �

�� ��

2

02 20

cos

24

Rq Ql

RR x

���� 5

� ��� �� �

� � �

1/22 2 2 204

q Q x

R x R x� 5

�� �

� � .

Now, since x < < R

� x2 is negligible small compared to R2

� 30

( )4

q QF x

R� 5

�� .

The linear acceleration of the point charge (-q) will be

2

2 30

( )4

d x F qQx

mdt m R� 5� � �

Which porves the fact that the particle executes SHM, with an angular frequency 304

q Q

m R�

� 5� .

� Time period of oscillation

3042

2m R

Tq Q

� 5��

�� � .

13. A solid cylinder of radius r rolls without sliding along the inside surface of a cylinder of radius R1

performing small oscillations. Find the period of oscillation.

Solution

Consider any position of the solid cylinder of radius r, when the line joining the centres of the two

cylinders make an angle � with the vertical. (Fig. 15.67)

Let v be the velocity of the center of mass of the solid cylinder and � its angular velocity..

The total mechanical energy of the solid cylinder will be given by E = translatory K. E. + rotational

K.E. + potential energy.

2 21 11 cos

2 2E mv I m g R r� �� � � � � �

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99

[Here I is the moment of inertia of the solid cylinder about the axis through its center].

Differentiating the expression for E w. r. t time t and equating it to zero.

[Since M.E. remains conserved]

sin . 0d v d d

mv I mg R rd t d t d t

� �� �� � � � … (1)

Now, d

v r R rd t

��� � � and

2

2

d v d dr R r

d t d t dt

� �� � � .

Making use of these in equation (1) we get sin 0d v d

m r I m g rd t d t

�� � � �� � � .

2

sin 02

d v m r dm r m g r

d t d t

��� � � �

2 2

2 2

1( ) ( ) ( ) [ sin ]

2

d dR r R r g

d t d t

� �� �� � � � � � � #�

2

2

2( )

3( )

d g

R rd t

��� � �

�

Comparing with the standard form

22

2( )

d

d t

�� �� � it is evident that the solid cylinder performs SHM

with angular frequency 2

3( )

g

R r� �

� and time period 3( )

22

R rT

g�

�� .

15. A uniform rod of mass m length l performs small oscillations about the horizontal axis passing through

its upper end. Find the mean kinetic energy of the rod averaged over one oscillation period if, at the

initial moment, it was deflected from the vertical by an angle 0� and then imparted an angular velocity

0� .

Solution

Consider the rod, at any instant making an angle � with the vertical. (Fig. 15.70)www.NetB

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100

The net torque acting on the rod, about O will be ( )

sin2 2

l m g lm g

�" �

�� #

But

2 2

23

ml dI

d t

�" �� �

2 2 2

2 2

3

3 2 2

ml d m g l d g

ld t d t

� � ��

� � � �� � � � � �� � � �� � � �� �

Which shows that the rod makes angular oscillations with angular frequency 3

2

g

l� � .

Now if a0 be the angular amplitude, the equation for angular displacement for the rod can be written as

0 sina t� � �� � . [where � is the phase constant]

Since at t = 0; 0� ��

0 0 sin ( )a� �� �

0

0

sina

��� � ... (1)

Also, 0 cosd

a td t

�� � �� �

Again at 00;d

td t

��� �

0 0 cosa� � �� �

0

0

cosa

��

�� � ... (2)

Squaring and adding equations (1) & (2) we get

2 2 22 20 0 00 02 2 2

0 0

1 aa a

� � ��

� �� � � � � � … (3)

Now, the instantaneous K.E. (rotational) of the rod will be

21

12

dK

d t

�� �� � �

� �

2 2 222 20

0

1[ cos ( )] cos ( )

2 3 6

ml amla t t

�� � � � �

� �� � � �� �� �

� �

� Average K.E. over one complete time period will be

2 2 2 2 2 220 0{ cos ( )}

6 12

ml a ml aK t

� �� �$ %� $ � �

[� Average of 2cos ( )t� �� over one complete oscillation is 1

2]

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22 22 2 200 0 0

3 2

12 2 3 12

mglml g l ml

l g g

�� � �� �� �

� � � �� �� �� �� �

16. If a particle moving with SHM has velocities u, v and w at the distances a, b ad c respectively from an

arbitrary point on the straight line, prove that its period T is given by the equation:

2 2 2

2 2

2 2 2

1 1 1

4

1 1 1

u v w

T a b c a b c

a b c

��

Solution

Let A and � be the amplitude and angular frequency of the SHM.

Then from 2 2 2 2

v A x�� � we have

2 2 2 2( )u A a d�� � � ... (1)

2 2 2 2( )v A b d�� � � ... (2)

and2 2 2 2( )w A c d�� � � ... (3)

Now the three equations can be rewritten as

22 2 2

22

uA a d ab

�� � � �

22 2 2

22 0

ua ad d A

�� � � � � � ... (4)

Similarly,

22 2 2

22 0

vb bd d A

�� � � � � ... (5)

And

22 2 2

22 0

wc cd d A

�� � � � � ... (6)

Eliminating d and A from equations (4), (5) and (6) we get

22

2

22

2

22

2

1

1 0

1

ua a

vb b

wc c

�

�

�

� ��� �� �

� �

� �� �� �� �

� �

� ��� �� �

� �

2 2

2 2

2

2 2

1 11

1 1 0

1 1

u a a a

v b b b

w c c c�

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102

[Resolving the original determinant into two determinats]

2 2 2 2 2 2

2

241 1 1 1 1 1

u v w a b cT

a b c a b c�

� � � or

2 2 2

2 2

2 2 2

1 1 1

4

1 1 1

u v w

T a b c a b c

a b c

��

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