1

Periodic Motion: The motion which repeats

itself after a regular interval of time is called

Periodic Motion.

UCM of a body is the simplest type of

Periodic motion.

Oscillatory motion: If a body in periodic

motion moves along the same path to and fro

about a definite point (mean position) then

the motion of the body is oscillatory.

e.g. Simple pendulum, mass attached to a

spring and stretched.

All oscillatory motion are periodic but

all periodic motion are not oscillatory.

All objects that oscillate have one thing

in common, each subjected to restoring

force.

S.H.M. is the simplest and the most

fundamental type of periodic motion S.H.M.

is an oscillatory motion for which the graph

between displacement from equilibrium

position is a sine or cosine curve i.e. its

displacement from mean position varies

sinosidally or cosinosidally with time.

If the motion is taking place under the action

of restoring force in which the magnitude of

force is directly proportional to displacement

then such a motion is called linear simple

harmonic motion.

2) Explanation of the mass oscillating at

the end of the spring:

Consider a spring

with one end fixed

and to the other

end, a mass m is

suspended. The

spring gets

stretched by length

𝑙, so that the

upward force due to the stretched spring is

equal to weight mg.

Mathematically,

𝑚𝑔 = 𝑘𝑙

Where ‘𝑘’ is called force constant of the

spring or spring factor, 𝑙 is the extension

spring due to mass.

As all the net force on mass is zero the

position is called mean position or

equilibrium position. When the mass is

further pulled down and released is performs

to and fro motion about the mean position.

When the mass is pulled down the spring is

stretched and due to elastic property of

spring it exerts a restoring force on the mass.

This force is directly proportional to the

displacement and directed towards mean

position. When the mass is released the

restoring force accelerates the mass towards

the mean position. Due to inertia of motion,

the mass moves beyond the mean position.

Again the restoring force acts on the body

which goes on increasing. The K.E. of the

mass is used in doing work against the

restoring force. When the spring is fully

compressed, the mass has no K.E. and the

restoring force reaches its maximum value.

The mass moves towards the mean position,

Thus, the motion is taking place under the

action of restoring force and the magnitude

of force is directly proportional to the

displacement of mass from mean position.

Hence this to and fro motion is linear

S.H.M.

A body performing linear S.H.M.

satisfies the following conditions:

1) The force acting on the body is always

directed towards mean position.

2) Magnitude of force is directly

proportional to its displacement from

mean position.

Both the conditions can be expressed

mathematically as:

𝐹 − 𝑥

𝐹 = −𝑘𝑥 Where k is the force constant

2

Hence force constant is the restoring

force per unit displacement.

S.I. unit of k N/m.

Graph between restoring force and

displacement is a straight line.

Definition of linear S.H.M. The periodic motion of a body along

a straight line in which the restoring force or

acceleration acting on the body is always

directed towards mean position and its

magnitude is directly proportional to the

displacement of the body from the mean

position.

Definitions: 1) Oscillation: The motion from O to B, B

to O, then O to A and back to O is called

one complete oscillation.

2) Amplitude (a): The maximum

displacement of the particle on either side

of mean position.

3) Time period or periodic time (T): The

time taken by the particle to complete

one oscillation.

4) Frequency (n or f): Number of

oscillations completed by the particle in

one second.

Differential equation of S.H.M. Consider a particle of mass m

performing linear SHM about the mean

position O in a straight line AB. Let at time

‘𝑡’ the particle have displacement 𝑥 from

position O. Let 𝛿𝑥 be small change in

displacement in a short time 𝛿𝑡.

𝑣 =

lim

0 t t

x

𝑣 =dt

dx

Let 𝛿𝑣 represent small change in velocity in

short time ‘𝛿𝑡’

𝑎𝑐𝑐𝑒𝑙𝑒𝑟𝑎𝑡𝑖𝑜𝑛 =lim

0 t t

v

𝑎𝑐𝑐𝑒𝑙𝑒𝑟𝑎𝑡𝑖𝑜𝑛 =dt

dv

𝑎𝑐𝑐𝑒𝑙𝑒𝑟𝑎𝑡𝑖𝑜𝑛 =

dt

dx

dt

d

𝑎𝑐𝑐𝑒𝑙𝑒𝑟𝑎𝑡𝑖𝑜𝑛 =2

2

dt

xd

Thus, 2

2

dt

xd represents acceleration of the

particle

From Newton’s law,

𝐹 = 𝑚𝑎𝑠𝑠. 𝐴𝑐𝑐𝑒𝑙𝑒𝑟𝑎𝑡𝑖𝑜𝑛

= 𝑚.2

2

dt

xd

For a particle performing linear S.H.M.

𝐹 = −𝑘𝑥

Where, k force constant.

2

2

dt

xd = x

m

k.

2

2

dt

xd + x

m

k. = 0

Let, 2m

K

2

2

dt

xd + x2 = 0

EXPRESSION FOR DISPLACEMENT,

VELOCITY AND ACCELERATION OF

A PARTIOCLE PERFORMING

LINEAR S.H.M.

Expression for acceleration:

The differential equation of a particle

performing S.H.M. is given by,

2

2

dt

xd + x2 = 0

2

2

dt

xd = - x2 ----------- (1)

The above equation gives the

acceleration of the particle at any instant.

Negative sign indicates that acceleration and

displacement are oppositely directed.

Graph between acceleration and

displacement is straight line.

Expression for velocity

3

We have, 2

2

dt

xd = - x2 --------- (A)

But, 2

2

dt

xd =

dt

dv

=dx

dv

dt

dx

= 𝑣dx

dv

Substituting in (A)

𝑣dx

dv= − x2

𝑣𝑑𝑣 = − x2 𝑑𝑥

Integrating both sides.

dxxdvv .. 2

2

2v =

2

22 x + C ………..(B)

Where C constant of integration.

When the particle is at extreme position

𝑥 = 𝑎, 𝑣 = 0

0 =2

22a + 𝐶

𝐶 =2

22a

Substituting value of C in equation (B)

2

2v =

2

22 x +

2

22a

v2 = - x2 2 + 22a

= 222 xa

v = 22. xa

Graph between velocity and

displacement is an ellipse

Expression for displacement,

v = 22. xa

dt

dx =

22. xa

2

2 xa

dx

= dt

Integrating both sides,

2

2 xa

dx = . dt

Sin-1 (x/a) = t +

a

x = sin ( t + )

𝑥 = a sin ( t + )

Case I: When the particle starts from

mean position,

𝑡 = 0, 𝑥 = 0 0 = 𝑎 𝑠𝑖𝑛 (0 + ) 0 = 𝑠𝑖𝑛 𝑠𝑖𝑛 = 0 = 0 𝒙 = 𝒂 𝒔𝒊𝒏 ( 𝒕)

Case II: When the particle starts from

extreme position,

𝑡 = 0, 𝑥 = 𝑎 𝑎 = 𝑎 𝑠𝑖𝑛 ( . 0 + ) 1 = 𝑠𝑖𝑛 = /2 𝑥 = 𝑎 𝑠𝑖𝑛 ( 𝑡 + /2) 𝒙 = 𝒂 𝒄𝒐𝒔 𝒕

Maximum velocity and maximum

acceleration of a particle performing

S.H.M.

At mean position

𝒙 = 0

2

2

dt

xd = - x2

= − 2 . 0 𝑨𝒄𝒄𝒆𝒍𝒆𝒓𝒂𝒕𝒊𝒐𝒏 = 𝟎

𝑣 =22 xa

= 02 a

=2a

𝒗 = 𝒂 (𝒎𝒂𝒙)

At extreme position

𝑥 = 𝑎

td

xd2

2

= - x2

= − 2 . 𝑎 𝑨𝒄𝒄𝒆𝒍𝒆𝒓𝒂𝒕𝒊𝒐𝒏 =

2 𝑎 (𝑚𝑎𝑥)(𝐼𝑛 𝑚𝑎𝑔𝑛𝑖𝑡𝑢𝑑𝑒)

𝑣 = 22. xa

𝑣 = 22. aa

𝑣 = 0

a

a 2

velocitymaximum

onaccelerati maximum

=

4

TOTAL ENERGY OF A PARTICLE

PERFORMING S.H.M.:

Consider a particle mass ‘m’ performing

SHM about the mean position O along a

straight line AB. Let at time ‘t’ the particle

be at p having displacement ‘𝑥’ from mean

position and ‘a’ be the amplitude of the

particle

Expression for K. E.

𝐾. 𝐸. = 2

2

1mv

𝑣 =22. xa

𝐾. 𝐸. = 222

2

1xam

Thus, K.E. depends on ‘𝑥’ having

max. K.E. at mean position

𝐾. 𝐸. (𝑚𝑎𝑥) = 22

2

1am ------- (1)

As, 𝑘

𝑚= 𝜔2

∴ 𝑘 = 𝑚𝜔2

𝐾. 𝐸. = 22

2

1xak

𝐾. 𝐸. (𝑚𝑎𝑥) = 2

2

1ka

EXPRESSION FOR P.E.

It is the work done by an external agent

against the restoring force in taking the

particle from its mean position to a given

point on its path.

At position p, the restoring force acting

on particle is,

𝐹 = − 𝑘𝑥 This force is directed towards the mean

position. If the particle is displaced through a

small distance ‘𝑑𝑥’ against the direction of

force work done

𝑑𝑤 = −𝐹 . 𝑑𝑥

negative sign indicates that displacement is

opposite to direction of force, substituting

the value of 𝐹 = − 𝑘𝑥.

𝑑𝑤 = − (−𝑘𝑥) 𝑑𝑥 = 𝑘𝑥 𝑑𝑥

Total work done when the particle moves

from O to p

𝑊 = x

dx

0

= x

dxkx0

.

= x

dxxk

0

.

=

x

xk

0

2

2

= 2

2

1kx

This work done is the P. E. of the particle

𝑃. 𝐸. = 2

2

1kx

Substituting 𝑘 = 𝑚 2

𝑃. 𝐸. =2

1 𝑚 2 𝑥2

Thus, P. E. depends on ‘x’ having

maximum P.E. at extreme position

𝑃. 𝐸. (𝑚𝑎𝑥) = 22

2

1am ------- (2)

Expression for total energy:

𝑇. 𝐸. = 𝐾. 𝐸. +𝑃. 𝐸.

= 22222

2

1

2

1xmxam

= ][2

1 2222 xxam

𝑇. 𝐸 = 22

2

1am ------------ (3)

From (1), (2) & (3) it is observed that T.E. of

particle in S.H.M. is Constant.

Substituting = 2 f

𝑇. 𝐸. = 22242

1maf

𝑇. 𝐸. = 2222 amf

From above equation as there is no term

like x, T.E. is independent of the position

of the particle and its value 2π2m f2 a2 is

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const. at any position, Hence T.E. of the

particle in S.H.M. is conserved.

Extra: If 2 π2m is kept constant and if ‘a’

and ‘f’ are allowed to vary

𝑇. 𝐸. 𝛼 𝑓2

𝑇. 𝐸. 𝛼 𝑎2

Conservation of energy in S.H.M.

(graphically) or Graphical representation of

K.E., P.E.,T.E of a particle performing

S.H.M. varying with displacement from

mean position.

We have already seen that T.E. is conserved

in S.H.M. we can arrive at the same result by

considering the variation of K.E., P.E. in

S.H.M. The parabola with its vertex at origin

represents P.E. of the particle. The inverted

parabola represents K.E. of the particle. The

line parallel to the displacement axis

represents the T.E. The Energy of the

particle continuously interchanges between

K.E. and P.E. but their sum at all times

remains constant.

At mean position

P.E. = 0 where as K.E. is maximum.

At extreme position

K.E. = 0 where as P.E. is maximum.

At any other position, it is partly kinetic

and partly potential

At mean position

𝐾. 𝐸 = 22

2

1am

𝑃. 𝐸. = 0

𝑇. 𝐸. = 𝐾. 𝐸. + 𝑃. 𝐸.

= 22

2

1am + 0

= 22

2

1am

At extreme position:

𝐾. 𝐸. = 0 𝑃. 𝐸. = 22

2

1am

𝑇. 𝐸. = 22

2

1am

At any other position

𝐾. 𝐸. = )(2

1 222 xam

𝑃. 𝐸. = 22

2

1xm

𝑇. 𝐸. = 𝐾. 𝐸. + 𝑃. 𝐸.

= )(2

1 2222 xxam

= 22

2

1am .

T.E. at mean position = T.E. at extreme

position = T.E. at any position

Thus, T.E. is conserved.

RELATIONSHIP BETWEEN UCM AND

S.H.M.: Consider a particle performing U.C.M. with

an angular velocity ‘ω’ along a circle of

radius a. Let AB be any diameter of the

circle Suppose that the particle M is the

projection of the particle P on diameter AB.

It is easy to see that as the particle p revolves

along the circle, the particle M moves to and

fro between the ends A and B. Thus the

particle P performs UCM and the particle M

performs SHM with amplitude a. The circle

drawn is called reference circle and the

particle P is called reference particle. When

P complete one revolution, its projection M

completes one oscillation.

Reference particle initially at P0 at time

t = 0. such that OXP0, After time t, at

tOPP 0 . Total angle described is

(ωt + α). At that instant particle performing

S.H.M. will be at M by drawing PM

perpendicular to AB.

In ∆OPM

𝑂𝑀 = 𝑂𝑃 𝑠𝑖𝑛 (𝜔𝑡 + 𝛼) 𝑥 = 𝑎 𝑠𝑖𝑛 (𝜔𝑡 + 𝛼)

6

This is the displacement of the particle M

at time t.

𝑣 = ) cos( tadt

dx

𝑎𝑐𝑐𝑒𝑙𝑒𝑟𝑎𝑡𝑖𝑜𝑛

= ) sin(2

2

2

tadt

xd

𝑎𝑐𝑐𝑒𝑙𝑒𝑟𝑎𝑡𝑖𝑜𝑛 = x 2

xt sin a

As ‘ω’ is constant, above equation

shows that acceleration of the particle is

directly proportional to its displacement and

negative sign implies that acceleration is

directed opposite to displacement i.e.

directed towards mean position. Thus, the

particle M performing S.H.M. can be

considered geometrically linear S.H.M. as a

projection of a particle performing U.C.M.

EXPLANATION OF PHASE:

Phase is that quantity which gives the

position with direction of motion of the

particle.

Consider equation

𝑥 = 𝑎 𝑠𝑖𝑛 (𝜔𝑡 + 𝛼)

The quantity (𝜔𝑡 + 𝛼) is called the

phase. It is the angle which enables us to

determine not only the displacement but also

direction of motion of the particle. The

constant α is called the initial phase (epoch

angle). It shows initially where the particle

starts from. If 𝛼 = 0, particle initially starts

from mean position and 𝛼 = 𝜋 /2, particle

initially starts from extreme position.

FIND ANALYTICALLY RESULTANT

AMPLITUDE AND EPOCH OF TWO

S.H.M. OF SAME PERIOD, DIFFERENT

AMPLITUDE AND DIFFERENT

INITIAL PHASES ALONG THE SAME

LINE.

Two S.H.M.s along x axis having same

period but different amplitude and different

initial phases can be written as,

𝑥1 = 𝑎1 𝑠𝑖𝑛 (𝜔𝑡 + 𝛼1)

𝑥2 = 𝑎2 𝑠𝑖𝑛 (𝜔𝑡 + 𝛼2)

Where a1 and a2 are their amplitudes and

α1 and α2 are their initial phase since the two

S.H.M. are along the same straight line, the

resultant displacement 𝑥 is equal to

algebraic sum of their displacement.

𝑋 = 𝑥1 + 𝑥2 X = 𝑎1 𝑠𝑖𝑛 (𝜔𝑡 + 𝛼1)

+ 𝑎2 𝑠𝑖𝑛 (𝜔𝑡 + 𝛼2)

𝑋 = 1111 sin coscos sin atata

+ 2222 sin coscos sin atata

𝑋 = 2211 coscosa tsin a

2211 sinsin tcos aa

Let,

2211 coscos aa = R cos -------(1)

2211 sinsin aa = R sin --------(2)

𝑋 = 𝑅 t cossin cos t sin R 𝑋 = 𝑅 𝑠𝑖𝑛 t

Above equation shows that resultant

displacement is also S.H.M. having

amplitude 𝑅 and initial phase ‘𝛿’

To calculate Resultant amplitude R

Squaring and adding equation (1) and (2)

𝑅2𝑐𝑜𝑠2𝛿 + 𝑅2𝑠𝑖𝑛2𝛿= (𝑎1𝑐𝑜𝑠𝛼1 + 𝑎2𝑐𝑜𝑠𝛼2)2

+ (𝑎1𝑠𝑖𝑛𝛼1 + 𝑎2𝑠𝑖𝑛𝛼2 )2

= 2

22

21

22

1 coscos aa +

2121 coscos2 aa

+ 2

22

21

22

1 sinsin aa

+ 2121 sinsin2 aa

R2 (sin2 + cos2 ) =

)cos(sin 1

2

1

22

1 a +

)cos(sin 2

2

2

22

2 a +

)sinsincos(cos2 212121 aa

R2 = )cos(2 2121

2

2

2

1 aaaa

R = )cos(2 2121

2

2

2

1 aaaa

To Calculate Resultant Phase 𝛿 . Dividing equation (2) by (1),

2211

2211

coscos

sinsin

cos

sin

aa

aa

R

R

tan =2211

2211

coscos

sinsin

aa

aa

7

=

2211

22111

coscos

sinsintan

aa

aa

Case I: If the phase difference between

them is zero i.e. if the two S.H.M. are

in phase

1 =2 or

1 -2 = 0

𝒄𝑜𝑠(1 − 2 ) = 1

𝑅 = 1.2 21

2

2

2

1 aaaa

𝑅 = 2

21 )( aa

𝑅 = 𝑎1 + 𝑎2 Special case : If a1 = a2 = a

𝑅 = 𝑎 + 𝑎 𝑅 = 2𝑎

Case II: If the phase difference between

them is 𝝅

𝟐

1 - 2 = 𝝅

𝟐

Cos ( 1 - 2 ) = 0

𝑅 = 0..2 21

2

2

2

1 aaaa

𝑅 = 2

2

2

1 aa

Special case : If a1 = a2 = a

𝑅 =22 aa

𝑅 =22a

𝑅 = 𝑎 2 Case III: If the phase difference between

them is π i.e. if the two S.H.M. are

opposite in phase

21

𝑐𝑜𝑠 21 = −1

𝑅 = 12 21

2

2

2

1 aaaa

𝑅 = 21

2

2

2

1 2 aaaa

𝑅 = 2

21 aa

𝑅 = 𝑎1 – 𝑎2 Special case

If a1 = a2 = a

𝑅 = 𝑎 – 𝑎 𝑅 = 0 RELATION BETWEEN TIME PERIOD

AND ANGULAR VELOCITY AND TO

DERIVE THE EXPRESSION FOR

PERIOD AND FREQUENCY OF S.H.M.

The general expression for displacement

at time t is given by.

𝑥 = 𝑎 𝑠𝑖𝑛 t

Let 𝑥1 be the displacement of the particle at

time

2t

𝑥1 = 𝑎 𝑠𝑖𝑛

2t

= 𝑎 𝑠𝑖𝑛 2 t

= 𝑎 𝑠𝑖𝑛 t

= 𝑎 𝑠𝑖𝑛 t

𝑥1 = 𝑥 Thus, the displacement of the particle at time

t and

2t are same.

It can be shown that the direction of

displacement are same since, the position

and direction repeats after time /2

𝑇 =

2

𝜔2 =𝑘

𝑚

=m

k

Substituting = mk /

𝑇 =

m

k

2

𝑇 =k

m2

𝑛 =T

1

𝑛 =m

k

2

1

PHYSICAL SIGNIGICANCE OF

CONSTANT :

𝑎𝑐𝑐𝑒𝑙𝑒𝑟𝑎𝑡𝑖𝑜𝑛 = − 𝜔2𝑥 magnitude of acceleration = 2 𝑥

2 =ntdisplaceme

onaccelerati of mag

=ntdisplaceme

onaccelerati of magnitude

= tdisplacemnunit /onaccelerati of mag

T =

2

T=

ntdisplacemeunit on / accelerati of magnitude

2

8

SIMPLE PENDULUM:

Simple pendulum is defined as a heavy

point mass suspended by weightless

inextensible tortionsless string from a rigid

support. In practice, we can’t have a heavy

point mass and nor we have a weightless

inextensible tortionsless string. Hence, in a

practical simple pendulum in place of heavy

point mass a small metallic sphere is used

known as bob in place of weightless,

inextensible string, a light thread is used.

The point from which the pendulum

is suspended is called the point of

suspension. As the string is very fine, the c.g.

of the pendulum coincides with centre of

mass of the bob known as the point of

oscillation. The distance between the point of

suspension and the point of oscillation is

called the length of the simple pendulum,

denoted as ‘l’.

TO SHOW THE MOTION OF THE BOB

FOR SMALL ANGULAR

DISPLACEMENT IS IN S.H.M. AND

HENCE CALCULATE TIMEPERIOD:-

Consider a

simple

pendulum of

length ‘l’ and

‘m’ be the mass

of the bob. The

mean position

is its vertical

position, So. If

it is displaced

through small

angle in a

vertical plane

and released it

begins to oscillate about the mean

position O. Let be the small angular

displacement when the bob is at A at a

distance ‘x’ from O. At position A force

acting on the bob-

1) Weight mg vertically downwards.

2) Tension ‘T’ in the string.

Resolving mg into two rectangular

components mg cos along OA

produced, mg sin perpendicular to OA.

‘mg cos ’ is balanced by part of the

tension T, and the remaining part of the

tension provides the centripetal force is

magnitude mv2/l. mg sin which is

unbalanced tends to restore the bob to the

mean position and hence the bob performs to

and fro motion about the mean position.

Therefore, 𝐹 = − 𝑚𝑔 𝑠𝑖𝑛

Negative sign because F is opposite to

angular displacement.

For small angles 𝑠𝑖𝑛 =

1

sin0

lim

𝐹 = − 𝑚𝑔

𝐹 = − 𝑚𝑔 l

x

m

F= − g

l

x

i.e. acceleration = xl

g.

---------(1)

At a given place g is constant and for a

pendulum of given length l is constant

𝐴𝑐𝑐𝑒𝑙𝑒𝑟𝑎𝑡𝑖𝑜𝑛 − 𝑥

And negative sign indicates that acceleration

and displacement are oppositely directed.

The motion of the bob is in S.H.M.

EXPRESION FOR TIME PERIOD: From equation (1),

Magnitude of acceleration / unit

displacement = l

g

𝑇 =

2

T =

𝑇 =lg /

2

𝑇 =g

l2

Graph between l and T2 is a straight line.

From above equation time period is

1) independent of the mass of the bob

2) amplitude of oscillation

𝑛 =T

1

ntdisplacemeunit on / accelerati of magnitude

2

9

𝑛 = gl /2

1

𝑛 =l

g

2

1

ASSUMPTIONS MADE IN THE

DERIVATION OF THE TIME PERIOD:

1) support perfectly rigid

2) string is weightless & inextensible.

3) Resistance and buoyancy of air is

neglected

4) Bob having small angular displacement

hence the motion of the bob is almost linear

CONDITIONS UNDER WHICH THE

MOTION IS S.H.M.

1) Bob having small angular displacement

2) Oscillation of the simple pendulum is in

the same vertical plane.

3) Negligible air resistance.

DIFFERENCE BETWEEN IDEAL

SIMPLE AND PRACTICAL SIMPLE

PENDULUM

LAW’S OF SIMPLE PENDULUM:

1) Law of length: At a given place, period

of simple pendulum is directly

proportional to the square root of the

length of simple pendulum.

𝑇 l when, 𝑔 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡.

2) Law of acceleration: For a given length

of simple pendulum, the period of simple

pendulum is inversely proportional to the

square root of acceleration due to gravity.

𝑇g

1 when, 𝑙 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡.

3) Law of mass: The period of the

simple pendulum is independent of mass

of the bob.

4) Law of isochronisms: The period of

simple pendulum is independent of the

amplitude of the bob. This is called

isochronisms and the oscillations of

simple pendulum are said to be

isochronous.

SECOND’S PENDULUM:

A second’s pendulum is a simple

pendulum whose time period is 2 sec

i.e. T = 2 sec.

Expression for length of second’s

pendulum:

We have,

𝑇 =g

l2

For a second’s pendulum T = 2 sec.

2 =g

l2

1 =g

l

Squaring both sides,

1 =g

l2

𝑙 =2

g

GRAPHICAL REPRESENTATION OF

DISPLACEMENT, VELOCITY AND

ACCELERATION OF A BODY

STARTING FROM MEAN POSITION:

For a particle starting from mean

position,

The equation for displacement, velocity

and acceleration are given by

𝑥 = 𝑎 𝑠𝑖𝑛 t

𝑣 = tdt

dx cos a

𝐴𝑐𝑐𝑒𝑙𝑒𝑟𝑎𝑡𝑖𝑜𝑛 =dt

dv

= −𝑎 2 𝑠𝑖𝑛 𝑡

Ideal simple

pendulum

Practical simple

pendulum

1) Heavy point

mass is used

2) String

weightless,

3) Resistance of air

is neglected and

hence amplitude is

constant

1) Heavy metallic

sphere known as

bob is used

2) Light thread is

used

3) Due to resistance

amplitude goes on,

decreasing

exponentially with

time.

10

1) Displacement graph:

Following table gives the values of

displacement for certain values of time

having these value displacement graph is

plotted which represents the variation of

displacement x with variation of time t.

2) Velocity graph:

Following table gives the values of velocity

for certain values of time having these values

velocity graph is plotted which represents

variation of velocity with variation of time t.

3) Acceleration graph:

Following table gives value of acceleration

for certain values of time having these values

acceleration graph is plotted which gives, the

variation of acceleration with variation of

time.

Conclusions from graph.

1) They vary harmonically with time.

2) When displacement is maximum velocity

is zero hence there is a phase difference

of 𝜋

2 between them.

3) When displacement has maximum

positive value of acceleration has

maximum negative value. Hence there is

a phase difference of 𝜋 between them.

For a particle starting from extreme

position, the equation of displacement,

velocity and acceleration are given by

𝑥 = 𝑎 𝑐𝑜𝑠 𝑡

𝑣 = dt

dx= −𝑎 𝑠𝑖𝑛 𝑡.

𝑎𝑐𝑐𝑒𝑙𝑒𝑟𝑎𝑡𝑖𝑜𝑛 =2

2

dt

xd

𝑎𝑐𝑐𝑒𝑙𝑒𝑟𝑎𝑡𝑖𝑜𝑛 = −𝑎𝜔2𝑐𝑜𝑠𝜔𝑡

1) Displacement graph:

Following table gives the values of

displacement for certain values of time using

these values, displacement graph is plotted

which represents the variation of displacement

𝑥 with variation of time 𝑡.

t t x

0 0 0

T/4 /2 a

T/2 0

3T/4 3 /2 -a

T 2 0

t t v

0 0 a

T/4 /2 0

T/2 -a

3T/4 3 /2 0

T 2 a

T t accn

0 0 0

T/4 /2 -a 2

T/2 0

3T/4 3 /2 a 2

T 2 0

11

2) Velocity graph:

Following table gives the values of velocity

for certain values of time. Using these values

velocity graph is plotted which represented

the variation of velocity v with variation of

time t.

3) Acceleration graph:

Following table gives values of acceleration

for certain values of time, using these values

acceleration graph is plotted which

represents, the variation of displacement x

with variation of time t.

DAMPED SIMPLE HARMONIC

MOTION

We know that the motion of a simple

pendulum, swinging in air, dies out

eventually. This is because the air drag and

the friction at the support oppose the motion

of the pendulum and dissipate its energy

gradually. The pendulum is said to execute

damped oscillations. In damped oscillations

although the energy of the system is

continuously dissipated, the oscillations

remain apparently periodic. The dissipating

forces are generally the frictional forces. To

understand the effect of such external forces

on the motion of an oscillator, let us consider

a system as shown in Fig.

Here a block of mass m oscillates vertically

on a spring with spring constant k. The block

is connected to a vane through a rod (the

vane and the rod are considered to be

massless). The vane is submerged in a liquid.

As the block oscillates up and down, the

vane also moves along with it in the liquid.

The up and down motion of the vane

displaces the liquid, which in turn, exerts an

inhibiting drag force (viscous drag) on it and

thus on the entire oscillating system. With

time, the mechanical energy of the block

spring system decreases, as energy is

transferred to the thermal energy of the

liquid and vane.

Let the damping force exerted by the

liquid on the system be Fd. Its magnitude is

t t x

0 0 a

T/4 /2 0

T/2 -a

3T/4 3 /2 0

T 2 a

T t v

0 0 0

T/4 /2 -a

T/2 0

3T/4 3 /2 a

T 2 0

T t accn

0 0 - a 2

T/4 /2 0

T/2 a

3T/4 3 /2 0

T 2 - a 2

12

proportional to the velocity v of the vane or

the block. The force acts in a direction

opposite to the direction of v. This

assumption is valid only when the vane

moves slowly. Then for the motion along the

x-axis (vertical direction as shown in Fig).

we have

𝐹𝑑 = – 𝑏 𝑣 where b is a damping constant that depends

on the characteristics of the liquid and the

vane. The negative sign makes it clear that

the force is opposite to the velocity at every

moment. When the mass m is attached to the

spring and released, the spring will elongate

a little and the mass will settle at some

height. This position, shown by O in Fig , is

the equilibrium position of the mass. If the

mass is pulled down or pushed up a little, the

restoring force on the block due to the spring

is

𝐹𝑠 = – 𝑘𝑥

where 𝑥 is the displacement of the mass

from its equilibrium position. Thus the total

force acting on the mass at any time t is

𝐹 = 𝐹𝑑 + 𝐹𝑠

𝑚𝑎 = – 𝑘 𝑥 – 𝑏 𝑣

𝑚𝑎 + 𝑘 𝑥 + 𝑏 𝑣 = 0

𝑚𝑑2𝑥

𝑑𝑡2+ 𝑏

𝑑𝑥

𝑑𝑡+ 𝑘𝑥 = 0

The solution of above equation describes the

motion of the block under the influence of a

damping force which is proportional to

velocity.

The solution is found to be of the form

𝑥 = 𝐴𝑒−𝑏𝑡 2𝑚⁄ cos (𝜔′𝑡 + 𝜑)

amplitude, which is 𝐴𝑒−𝑏𝑡 2𝑚⁄ gradually

decreases with time as shown in

displacement – time graph.

The expression cos (𝜔′𝑡 + 𝜑) shows

that motion is still periodic and S.H.M.

The period of oscillation is given as,

𝑇 =2𝜋

√ 𝑘𝑚 − (

𝑏2𝑚)

2

The period of oscillation increases due to

presence of term − (𝑏

2𝑚)

2

in the denominator.

Thus damping increases the period and

decreases amplitude.

NUMERICALS

Type I : Composition of two simple

harmonic motion of same period and

different amplitude and initial phase.

1. Two parallel simple harmonic motions

are given by

X1 = 20 sin 8 t ,X2 = 10 sin (8 t + 6

)

Find resultant amplitude and initial phase

of resultant simple harmonic motion.

2. Equation of simple harmonic motion is

x = 3 sin ( 5 t) + 4 cos (5 t)

Find amplitude and phase constant of the

motion.

3. Equation of simple harmonic motion is

x = 5sin (10 t + /2)

Find initial phase angle, period,

amplitude and frequency.

4. The displacement of particle in simple

harmonic motion is given by

x = a sin (ωt) + b cos (ωt) in metre,

where, a, b and ω are constant show that

motion is simple harmonic

If a = 3 cm b = 4 cm ω = 2 rad/sec.

Find period, maximum velocity and

maximum acceleration

5. Displacement of particle performing

SHM along the x-axis is given by

x = 4 sin ( t + /3) Find :

i) displacement ii) velocity

iii) Acceleration iv) Frequency

at t = 1 sec.

6. Displacement of particle performing

SHM is given by

x = 8 sin (5 t + /6)

Where time is in sec. and displacement in

metre. Calculate

i) Displacement

ii)Velocity

iii) Acceleration of a

13

Particle at t = 2 sec.

Type II :

7. A particle performs SHM of amplitude

10 cm at what displacement kinetic

energy of particle performing SHM is

two times its potential energy.

8. A body performing linear SHM has

amplitude 5 cm. Find the ratio of its P.E.

to K.E. When its displacement is 3 cm

from mean position.

9. Total energy of particle of mass 0.5 kg

performing SHM is 25 J. What is its

speed when crossing the centre of path.

10. Particle performing linear simple

harmonic motion of amplitude a. What

fraction of total energy is kinetic, when

displacement of half of amplitude.

Type III :

11. Find the change in length of seconds

pendulum if acceleration due to gravity

changes from 9.65 m/s2 to 9.8 m/s2.

12. Period of simple pendulum is double

when its length is increased by 1.2 m.

Calculate the original period.

13. A period of simple pendulum is found to

change by 40%, when the length of

simple pendulum increases by 0.5 m.

Calculate initial length, initial period of

oscillation at a place where g = 9.8

m/sec2.

14. Period of simple pendulum is found to

increase by 50% when length of the

simple pendulum increases

by 0.6 m. Calculate initial length and

initial period of oscillation.

15. Find maximum velocity of second’s

pendulum if the amplitude of oscillation of

pendulum is 0.15 m.

16. A clock regulated by second’s pendulum

keeps correct time. During summer, length

of the pendulum increases to 1.01m. How

much will the clock gain or lose in one day.

Given g = 9.8 m/sec2

17. A simple pendulum performs 50

oscillations in 3 min at a place A. It requires

same time for 49 oscillation at another place

B compare the value of gravitational

acceleration at A and B.

Type IV:

18. Bar magnet is vibrating in uniform

magnetic field of induction B about an

axis passing through its centre when the

value of B is reduced by 3 10-5 wb/m2.

The period of vibrating magnet is

doubled. Find the value of B.

19. A bar magnet of magnetic moment

7.9Am2 having moment of inertia 10-6

Kgm2. About the transverse axis passing

through its centre performing SHM in the

magnetic induction 3.6 10-5 wb/m2.

How many Oscillation does its performs

in one min.

20. A bar magnet freely suspended in

magnetic induction H makes 60

oscillation in one min. When another

magnet X is brought near to it no. of

oscillation are found to be 30\ min. What

is the induction of magnet X in term of

H, where H > X.

21. Two bar magnets are tied together side

by side and suspended so as to oscillates

about their centre of gravity in horizontal

plane compare their magnetic dipole

moment when one magnet is reversed.

Period of oscillation are 4 sec and 12 sec.

Type V :

22. Particle performing SHM has maximum

velocity 0.21 m/sec. and maximum

acceleration is 0.84 m/sec2. Find

amplitude and period of oscillation.

23. Maximum velocity of particle performing

SHM is 0.16 m/sec. If its maximum

acceleration is 0.64 m/sec2. Calculate

period and amplitude.

24. Calculate period and amplitude of a

particle performing SHM if its maximum

velocity is 32 cm/sec and maximum

acceleration 256 cm/sec2.

25. Maximum velocity of particle performing

SHM is 6.28 m/sec if length of its path is

8 cm. Calculate period.

26. Find the value of displacement of a

particle in SHM when its velocity is half

of the maximum velocity.

27. Particle is performing linear simple

harmonic motion. It’s velocity is 4

cm/sec. when it is at a distance of 3 cm

from mean position and 3 cm/sec when it

14

is at a distance of 4 cm from mean

position. Find the amplitude and period

of simple harmonic motion.

28. Particle executing simple harmonic

motion has velocities V1 and V2 when at

a distances X1 and X2 from the centre of

path. Show that its time period T is given

by

2

22

1

21

222

VV

xxT

29. Particle performing simple harmonic

motion has acceleration of 100 cm/sec2.

When it is at a distance of 25 cm from

mean position. Find its period?

30. Verticle light spring is stretched by 5 cm,

when a body of mass 5 kg is attached to

its free end. The body is further pull

down by 3 cm and released. Find

i) Period ii) Amplitude iii) Frequency

iv)Total energy

v) Velocity at mean position.

31. A body of mass 100 kg is suspended

from rigid support and performs simple

harmonic motion in the vertical direction

if force constant of the spring is 4.9

103 dynes/cm. Find frequency of SHM.

32. Period time of vibration of mass m1

Suspended from light spring is T sec.

When mass m2 is added to first mass and

system is made to vibrate, periodic time

is 2T sec. Compare masses m1 and m2.

33. A particle in SHM has velocity 0.4 m/sec

When it crosses the mean position. Find

its velocity when it is midway between

mean and extreme position.

34. A needle of swing machine moves in a

path of amplitude 2 cm and has

frequency 10 Hz with simple harmonic

motion. Find its acceleration at 40

1th

second after it has crossed mean position.

35. A sewing machine needle moves in a

path of length 0.06 m makes 20

oscillation in one sec. Find its

Acceleration after 120

1 sec after it has

cross mean position.

36. Particle of mass 10 gm performs linear

simple harmonic motion of amplitude 5

cm with period 2 sec. Find its potential

energy, kinetic energy at

6

1th second after it has crossed mean

position.

37. Differential equation for linear simple

harmonic motion of a particle of mass 2 g

is

0162

2

xdt

xd Find force constant

38. A body of mass 2 kg performs linear

simple harmonic motion. The restoring

force acting on it when displacement is

0.6 m from mean position in 3 N. write

down differential equation of its motion.

39. For a particle performing linear simple

harmonic motion show that average

peed over one oscillation is

Aw2 where

A is amplitude of simple harmonic

motion.

*********************************

SPACE FOR EXTRA POINTS