1

A motion that repeats itself in equal interval of time is called periodic motion. Themotion of the earth about its own axis, the motion of the earth round the sun, themotion of electrons in an atom are few common examples of periodic motion.

When a particle in periodic motion moves to and fro along the same path, themotion is called as oscillatory or vibratory motion. The oscillation of a simplependulum, the oscillation of a mass attached to a spring etc. are the few examples ofoscillatory motion.

It may be said from the above discussion that any oscillatory motion shouldalways be periodic, but a periodic motion may not always be oscillatory in nature.

The time required to complete one oscillation is called time period and the

number of oscillations made per second is called the frequency of the motion. So,

the relation between time period and frequency is given by,

The position at which on an oscillatory particle is called itsequilibrium or mean position. The distance of a particle at any instant from theequilibrium position is known as displacement of the particle. The maximumdisplacement on either side of the equilibrium position is called amplitude ofthe motion.

If a particle moves to

and fro in such a way that theacceleration of the particle isalways directed towards afixed point in its path ofmotion and is proportionalto the displacement of theparticle from the fixed pointat any instant, the motion ofthe particle is called simple harmonic motion (S.H.M.).

The graphical representation of simple harmonic motion is given in Fig. 1.

1

A Textbook of Basic Engineering Physics2

If a particle of mass ‘ ’ executes S.H.M. and be the displacement of it from theequilibrium position, the acceleration of the particle is given by,

where is a proportionality constant.

Hence, the force ‘ ’ acting on the particle is given by,

Here, is called representing of the particle.

The characteristics of S.H.M are,

The motion is linear.

The motion is periodic and oscillatory.

The restoring force acting on the body ( the acceleration of the body) isproportional and oppositely directed to its displacement measured from afixed point ( mean position) in the path of the motion.

If be the mass of the particle executingS.H.M. and be its displacement at any instantfrom the mean position ‘ ’ (as shown in Fig. 2),the force acting on the particle at any point isgiven by,

where is the constant of proportionality (force constant) and negative sign is taken,as the restoring force is opposite to the direction of displacement.

The equation may be written as,

or, , , where = natural angular frequency

or,

This is .

The time period of a S.H.M. is given by,

1.1

1.2

2

21.3

1.3

2

2

2

22 2

2

22 0 1.4

2 1.5

Simple Harmonic Motion 3

Multiplying both sides of equation by , we get,

or,

Integrating , we get,

, where is the constant of integration.

Now, the velocity of the simple harmonic particle is zero at its maximum

displacement at (amplitude of the particle).

Hence, we get from the equation , or,So, the equation takes the form,

or,

Proceeding with the positive sign, we have,

On integration,

or, [where = integration constant]

or,

or,

Here is the phase of the

vibrating particle and is the epoch, or initial phase of the particle ( it is the phase

of the particle at ). Here the natural cyclic frequency of vibration . Here if

we proceed with negative sign, the solution will be of the same form .

1.4 2

22

22 2 0

22 2 0

22 2 1.6

1.6 0 2 2 2 2

1.62

2 2 2 2 2

2 2

2 2

sin 1

sin

sin 1.7

Phase is the state of motion of the particle executing S.H.M. and it indicates about theposition and direction of motion at any time.

0 0 2sin

To express the solution of equation in a more general form, let us write relation as,

=

=

=

This is the general solution of equation of a S.H.M., where and

are two constants such that and

1.71.7

sincoscossin

cos sin cossin

cossin 1

1.5 cos

sin 2 2 tan 1

A Textbook of Basic Engineering Physics4

Now, we get from equation by ignoring negative sign,

or, , [ time period of oscillation]

We know, the displacement of a particle executing S.H.M. at any time is

given by,

Velocity = or,

When the particle is at its mean position, .

at time , so, ,

, where =

Now, when the particle is at the mean position of the path for ,

At the extreme positions of the path for , .

We have, from equation ,

Acceleration

When the particle is at the mean position, .

or, [ at , so, ]

When the particle is at extreme positions ( for ), the acceleration

(neglecting the negative sign)

When the particle is at the mean position ( for ), the acceleration,

1.5

2

21.8

2 accelerationdisplacement

2 displacementacceleration

1.9

sin

cos 1.10

0

cos 12

2

0 0 sin cos 12

2

2 2 2 1.11

0

max

min 0

1.10 cos

2 sin 1.12

0

2 sin

2 0 0 sin 1.13

max2

0

min2 0 0

Simple Harmonic Motion 5

A particle executing a simple harmonic motion possesses both potential andkinetic energy at any instant.

The kinetic energy of a particle of mass and movingwith velocity executing S.H.M. is given by,

But,

The potential energy of a vibrating particle of mass atany instant is calculated from the total amount of work done in overcoming therestoring force through a distance . Here, force = . The negative sign is takenas the force is opposite to the direction of motion of the vibrating particle. So, thepotential energy,

We know that the displacement of a particle executing S.H.M. at any instant of time is

velocity

Comparing equations and we get that, the volocity of the particle executing

S.H.M. at any instant leads the displacement by a phase difference radian ( two are

in quadrature)

Now differentiating eqn. we get acceleration

Comparing equations and we find that the acceleration at any instant leads the

velocity by radians in phase.

Similarly comparing equations and we find that acceleration leads thedisplacement by radian. [ antiphase]

sin 1

cos2

sin 2

1 2

2

2

cos

2 sin 2 sin 32 3

21 3

Kinetic energy ( )

12

2

sin

cos 12

2 cos 1

2

2

2 2

12

2 12

2 2 2 1.14

Potential energy ( )

2

2

0

12

2 2 1.15

A Textbook of Basic Engineering Physics6

Now, the total energy, =

or, = , where

or, constant

So, the total energy is constant in S.H.M. and is proportional to the square of theamplitude.

So, we may conclude from equation and that, for ,

, but

Similarly, for , , but

So, K.E. is maximum when P.E. is minimum and vice versa.

1.14 1.15

min0 max

12

2 2

0max

12

2 2min 0

Total energy

12

2 2 2 12

2 2 2

12

2 2 2 2 2 2 1.16

In case of free vibration, the restoringforce is proportional to its displacementand it is the only internal force actingon the body. Hence, we can say fromequation and that themaximum amplitude of free vibrationand the total energy of the (free) simpleharmonic motion remain constant withtime [Fig. 4].

1.7 1.16

1

The displacement of a particle of mass executing S.H.M. is indicated by

m. Calculate

the amplitude the angular velocity the time period

the maximum velocity and maximum acceleration

the epoch the phase difference after

the displacement, velocity and acceleration of the particle after one second

0.2 kg

103 12

sin

i ii iii

iv

v vi 1 second

vii

Simple Harmonic Motion 7

The general equation of a simple harmonic motion is given by,

Here,

Comparing equations and , we get

,

executing S.H.M. is,

for

The ,

for

Now, is the initial phase for .

We get from equation ,

The phase after second

when , phase

So, = = = = = =

second,

The

The

the kinetic energy and potential energy of the particle at displacement.

the total energy of the particle.

viii0.04 m

ix

sin 1

103 12

sin 2

1 2

i 10 m

ii3

iii 2 2

3

6 s

iv 2 2

max 0

103

103.14

3 10.46 m s 1

2

max2 10

3.143

210

3

10.95 m s 2

v 3

0

212

vi

13

112

412

312 4

1 s4 12

312

212 6

1806

30

vii 1

103

112

sin 104

sin10

25 2 m

1 second

2 23

102 5 2 2

350 7.4 m s 1

1 second 23

25 2 7.75 m s 2

A Textbook of Basic Engineering Physics8

K.E. and P.E.

So, at displacement

K.E.

P.E. =

According to the problem, or,

or, or,

or, .

Let the density of the liquid be and the area of cross-section of

the glass tube be .

The volume of the liquid displaced by the glass tube

Mass of the tube = = mass of the displaced liquid =

Now, if we consider the tube is pushed by a further length of cm, the additionalupthrust acting on the glass tube,

or,

or, ,

This equation implies that the motion is simple harmonic.

The time period, = = .

viii 12

2 2 2 12

2 2

0.04 m

12

0.23.14

3

2102 0.04 2 10.95 joule

12

0.23.14

3

20.04 2 0.0001 joule

ix 12

2 2 12

0.23.14

3

210 2 10.955 joule

2

The differential equation of motion of a freely oscillating body is given by

. Calculate the natural frequency of the body. 22

218 2 0

22

218 2 0

2

29 2

2 9 2 3 2 3

32

1.5Hz

3

A weight glass tube is floating in liquid with cm of its length immersed. It is

pushed down a certain distance and then released. Compute the time period ofits vibration.

20

g cm 3

20 cm3

20 g

2

2

2

22

2 2 2 20 2 20980

0.89s