Oscillation: Any to-and-fro motion about a fixed point is called oscillation.

Wave: Wave is the disturbance that transfers energy from one position to
another.

Periodic motion: Any motion that repeats itself at regular intervals is

called periodic motion or harmonic motion.

Example: Earth is moving around the sun. After a period of 365 days, it
repeats its motion.

Oscillatory motion: A type of periodic motion in which particles make to

and fro motion about a mean position.

Mean position

A linear SHM is regarded as a projection of a uniform circular motion on the

diameter of a circle. Let 𝑆 be a particle moving on a circumference of a
circle of radius 𝑎 with a uniform velocity 𝑣. 𝜔 be the uniform angular
velocity of the particle.

As the particle 𝑆 moves around the circle, the foot of the perpendicular 𝑃
oscillates along the diameter 𝑌𝑌 ′ . Thus, 𝑃 executes to and fro oscillatory
motion about the point 𝑂. This oscillatory motion of 𝑃 about 𝑂 is called
SHM.
At any instant, the distance of 𝑃 from the center 𝑂 of the circle is called the
displacement.

𝑌 𝑇Τ4
𝜔
𝑃 𝑠
𝑎
𝑦 𝑦
𝜔𝑡 t= 0
𝑇Τ2 𝑂 𝑥 𝑄

𝑌 ′ 3𝑇Τ4

P P

If the particle moved from 𝑋 to 𝑆 in time 𝑡, then ∠𝑆𝑂𝑋 = ∠𝑃𝑆𝑂 = 𝜔𝑡 Since,

𝑂𝑃
s𝑖𝑛𝜃 = 𝑠𝑖𝑛𝜔𝑡 = .
𝑎

∴ 𝑂𝑃 = 𝑦 = 𝑎𝑠𝑖𝑛𝜔𝑡

Where 𝜔 is the angular frequency.

The rate of change of displacement of a vibrating particle is called velocity.

 𝑑𝑦 𝑦2
√ √
= 𝑣 = 𝑎𝜔𝑐𝑜𝑠𝜔𝑡 = 𝑎𝜔 1 − 𝑠𝑖𝑛 𝜔𝑡 = 𝑎𝜔 1 − 2 = 𝜔√𝑎2 − 𝑦 2
2
𝑑𝑡 𝑎

The rate of change of velocity of a vibrating particle is called acceleration.

𝑑2𝑦
2
= 𝛼 = −𝑎𝜔2 𝑠𝑖𝑛𝜔𝑡 = −𝜔2 𝑦
𝑑𝑡
𝑎𝑐𝑐𝑒𝑙𝑒𝑟𝑎𝑡𝑖𝑜𝑛
𝜔2 =
𝑑𝑖𝑠𝑝𝑙𝑎𝑐𝑒𝑚𝑒𝑛𝑡

2𝜋 𝑎𝑐𝑐𝑒𝑙𝑒𝑟𝑎𝑡𝑖𝑜𝑛
𝑜𝑟, 𝜔 = 2𝜋𝑛 = =√
𝑇 𝑑𝑖𝑠𝑝𝑙𝑎𝑐𝑒𝑚𝑒𝑛𝑡

𝑑𝑖𝑠𝑝𝑙𝑎𝑐𝑒𝑚𝑒𝑛𝑡
∴ 𝑇 = 2𝜋√
𝑎𝑐𝑐𝑒𝑙𝑒𝑟𝑎𝑡𝑖𝑜𝑛

Simple Harmonic Motion:

Simple harmonic motion (SHM) is a special type of periodic motion where
the restoring force is proportional to the displacement and acts in the
direction opposite to that of displacement.

Properties of SHM:
❑ Motion is periodic.

❑ It is oscillatory.

❑ Acceleration (force) is proportional to displacement.

❑ Acceleration always acts towards the mean position.

❑ Motion is represented by harmonic functions 𝑠𝑖𝑛𝜃, 𝑐𝑜𝑠𝜃.

Differential equation of SHM
If ‘F’ be the force acting on a particle executing simple harmonic motion and
y be the displacement from equilibrium or mean position, then

𝐹 ∝ −𝑦, … … … (1)

𝑜𝑟, 𝐹 = −𝑘𝑦

𝑑2𝑦
𝑜𝑟, 𝑚 2 = −𝑘𝑦
𝑑𝑡
𝑑2𝑦 𝑘
𝑜𝑟, 2 = − 𝑦
𝑑𝑡 𝑚
𝑑2𝑦 𝑘
𝑜𝑟, 2 + 𝑦 = 0
𝑑𝑡 𝑚

𝑑2𝑦 𝑘
𝑜𝑟, 2
+ 𝜔2 𝑦 = 0, … … … (2), 𝑤ℎ𝑒𝑟𝑒, 𝜔 = √
𝑑𝑡 𝑚

Equation (2) is called the differential equation of motion of a body executing

simple harmonic motion.

Now, to obtain a general solution of the differential equation of SHM,

𝑑𝑦
Multiply both sides of equation (2) by 2 ,
𝑑𝑡

𝑑𝑦 𝑑 2 𝑦 𝑑𝑦 2
2 + 2 𝜔 𝑦=0
𝑑𝑡 𝑑𝑡 2 𝑑𝑡
Integrating with respect to time we get,

𝑑𝑦 2
( ) + 𝜔2 𝑦 2 = 𝐶, … … … (3)
𝑑𝑡
Where, C is a constant of integration.
𝑑𝑦
Boundary condition: 𝐴𝑡 𝑦 = 𝑎, = 0 i.e., velocity is zero at maximum
𝑑𝑡
displacement position.
∴ From equation (3) we get, 𝐶 = 𝜔2 𝑎2

Substituting the value of C in equation (3),

𝑑𝑦 2
( ) + 𝜔2 𝑦 2 = 𝜔2 𝑎2
𝑑𝑡
𝑑𝑦 2
𝑜𝑟, ( ) = 𝜔2 (𝑎2 − 𝑦 2 )
𝑑𝑡

𝑑𝑦
𝑜𝑟, = 𝜔√(𝑎2 − 𝑦 2 )
𝑑𝑡
Integrating with respect to time,
𝑑𝑦
𝑜𝑟, ∫ = ∫ 𝜔 𝑑𝑡
√𝑎2 − 𝑦 2
𝑦
𝑜𝑟, sin−1 = 𝜔𝑡 + 𝜑
𝑎
𝑦
𝑜𝑟, = sin(𝜔𝑡 + 𝜑)
𝑎
𝑜𝑟, 𝑦 = 𝑎 sin(𝜔𝑡 + 𝜑) , … … … (4)

This is the general solution of the differential equation of SHM.

Velocity and acceleration of a body executing SHM

Displacement of a body executing SHM is given by,

𝑦 = 𝑎 sin(𝜔𝑡 + 𝜑)
𝑦
sin(𝜔𝑡 + 𝜑) =
𝑎
Velocity of a body executing SHM is given by,
𝑑𝑦
= 𝜔 acos(𝜔𝑡 + 𝜑) = ±𝜔𝑎√1 − 𝑠𝑖𝑛2 (𝜔𝑡 + 𝜑)
𝑑𝑡
 𝑑𝑦 𝑦2
𝑜𝑟, = ±𝜔𝑎√1 − = ±𝜔√𝑎2 − 𝑦 2
𝑑𝑡 𝑎2

𝑑𝑦
∴𝑣= = ± 𝜔√𝑎2 − 𝑦 2
𝑑𝑡

The acceleration of a body is,

𝑑2𝑦
2
= −𝜔𝑎2 sin(𝜔𝑡 + 𝜑)
𝑑𝑡
𝑑2𝑦 𝑦
𝑜𝑟, = −𝑎𝜔2 ×
𝑑𝑡 2 𝑎

𝑑2𝑦
∴ 𝛼 = 2 = −𝜔2 𝑦
𝑑𝑡
-ve sign indicates that the direction of y and 𝛼 are opposite to each other.

Maximum velocity (at mean position, 𝑦 = 0)

𝑑𝑦
𝑣𝑚𝑎𝑥 = ( ) = ±𝜔√𝑎2 − 02 = ±𝜔𝑎
𝑑𝑡 𝑚𝑎𝑥

Minimum velocity (at extreme position 𝑦 = 𝑎), 𝑣𝑚𝑖𝑛 = 0

Maximum acceleration (at extreme position, 𝑦 =a),

𝑑2𝑦
𝛼𝑚𝑎𝑥 = ( 2) = −𝜔2 𝑦 = −𝜔2 𝑎
𝑑𝑡 𝑚𝑎𝑥

Minimum acceleration (at mean position,𝑦 = 0),

𝛼𝑚𝑖𝑛 = 0
 Examples of SHM:

Rigid support

𝜽
l
T T
𝜽

A m m B
m
mg sin𝜽
mg cos𝜽
Mean Position mg

Simple pendulum:
❑ A simple pendulum consists of a point mass (bob) suspended by an
inextensible weightless string in a uniform gravitational field.

❑ When pulled to one side of its equilibrium position and released, the
bob of the pendulum oscillates about the mean position.

❑ Let the mass of the bob be ‘𝑚’ and the length of the string be 𝑙.

❑ The path of the bob is not a straight line, but the arc of a circle of
radius 𝑙.
 ❑ At point 𝑄 (at a displacement 𝑥), it will experience a force (𝑚𝑔)
vertically downward.

❑ The weight (𝑚𝑔) will resolve into two components along the radius
(𝑚𝑔𝑐𝑜𝑠𝜃) and along the tangent (𝑚𝑔𝑠𝑖𝑛𝜃) to the circle.

❑ The component 𝑚𝑔𝑐𝑜𝑠𝜃 is balanced by tension 𝑇. ∴ 𝑇 = 𝑚𝑔𝑐𝑜𝑠𝜃

❑ Unbalanced force, 𝑚𝑔𝑠𝑖𝑛𝜃 (known as restoring force) creates

tangential acceleration 𝑎.

∴ 𝐹 = −𝑚𝑔𝑠𝑖𝑛𝜃, … … … (1) Rigid support

𝑜𝑟, 𝑚𝑎 = −𝑚𝑔𝑠𝑖𝑛𝜃 P
𝑜𝑟, 𝑎 = −𝑔𝑠𝑖𝑛𝜃, … … … (2)
𝜽
𝑥 𝐴𝑟𝑐
For small value of 𝜃, 𝑠𝑖𝑛𝜃 ≈ 𝜃 = [𝜃 = l ]
𝑙 𝑟𝑎𝑑𝑖𝑢𝑠

From equation (2), T T

𝜽
𝑥
𝑎 = −𝑔𝜃 = −𝑔 , … … … (3)
𝑙
A m m Q
𝑑2𝑥 𝑔 mg sin𝜽
𝑜𝑟, + 𝑥 = 0, … … … (4) m
𝑑𝑡 2 𝑙
O x
mg
Mean Position S
R
𝑑2𝑦
Equation (4) is similar to the differential equation of SHM, + 𝜔2 𝑦 = 0.
𝑑𝑡 2

𝑥 𝑙
Rearranging equation (3) we can write, = − , … … … (5)
𝑎 𝑔

From (5), 𝑔 is constant at a place and 𝑙 is also fixed. ∴ 𝑎 ∝ −𝑥

Therefore, the motion of a pendulum is SHM provided the amplitude of

oscillation is small.

The time period of oscillation of the simple pendulum is,

 𝑑𝑖𝑠𝑝𝑙𝑎𝑐𝑒𝑚𝑒𝑛𝑡 𝑥 𝑙
𝑇 = 2𝜋√ = 2𝜋√ = 2𝜋√
𝑎𝑐𝑐𝑒𝑙𝑒𝑟𝑎𝑡𝑖𝑜𝑛 𝑎 𝑔

Therefore, for a small oscillation, the period of a pendulum for a given value
of g is determined entirely by its length.

Torsion pendulum:

❑ A disk suspended by a wire. One end of the wire is fixed to a solid

support and the other end is fixed to the center of mass of the disk.

❑ When the disk is in equilibrium position, a radial line is drawn from its
center 𝑂 to a point 𝑃 on its rim.

❑ If the disk is rotated in a horizontal plane so that the reference line 𝑂𝑃

moves to a position 𝑂𝑄, the wire will be twisted. The twisted wire will
exert a restoring torque on the disk tending to return to its mean
position.

❑ If the disk is given a small twist and released, it will execute angular
oscillations about its equilibrium position and it is known as torsion
pendulum.

❑ For small twist, the restoring torque is proportional to the angular

displacement, so that,

𝜏 = −𝑘𝜃, … … … (1)

Where, 𝑘 is a torsional constant. The minus sign indicates that the torque is
directed opposite to the angular displacement. It is a certain condition for
angular SHM.

The equation of motion for such a system is based on the angular form of
Newton’s 2nd law,

𝑑2𝜃
𝜏 = 𝐼𝛼 = 𝐼 2 , … … … (2)
𝑑𝑡
Where, 𝐼 is the rotational inertia of the oscillating disk and 𝛼 is the angular
acceleration.

Combining equation (1) and (2), we obtain,

𝑑2𝜃
−𝑘𝜃 = 𝐼 2
𝑑𝑡
𝑑2𝜃 𝑘
𝑜𝑟, = − ( )𝜃
𝑑𝑡 2 𝐼

𝑅 𝑄

𝑑2𝜃 𝑘
𝑜𝑟, + ( ) 𝜃 = 0, … … … (3)
𝑑𝑡 2 𝐼
𝑑2𝑦
Equation (3) is similar to the differential equation of SHM, + 𝜔2 𝑦 = 0.
𝑑𝑡 2

Hence the angular oscillations of the torsion pendulum is simple harmonic.

𝐼
∴ Time period, 𝑇 = 2𝜋√
𝐾
Spring-mass system:

❑ Consider a spring whose upper end is fixed to a rigid support and

mass (𝑚) is attached to its free end. Let the length of the spring be 𝐿.

❑ When the load is attached, the spring extended by an amount 𝑠.

❑ Under this condition, the upward force 𝐹 exerted by the spring is

equal to the weight of the body 𝑚𝑔.

If the spring obeys Hooke’s law,

𝐹 = 𝑚𝑔 = −𝑘. 𝑠, … … … (1)

Where, 𝑘 is the spring constant. The minus sign indicates that the force and
displacement are oppositely directed.
𝑚𝑔
From equation (1), 𝑘 = , … … … (2)
𝑠

Thus, the spring constant can be defined as the tension per unit
displacement.

𝑦=0
𝑦

https://opentextbc.ca/calculusv3openstax/chapter/applications/
 ❑ If the body is now displaced from its equilibrium position and
released, it will oscillate along the vertical direction.

❑ Let ‘𝑦’ denote the displacement of the mass from equilibrium position.

❑ When the body is at a distance 𝑦 above its equilibrium position, the

extension of the spring is (𝑠 − 𝑦). The restoring force it exerts on the
body is 𝑘(𝑠 − 𝑦) and the gravitational force is 𝑚𝑔. Hence, the
resultant force 𝐹 on the body is,

𝐹 = 𝑘(𝑠 − 𝑦) − 𝑚𝑔 = 𝑘𝑠 − 𝑘𝑦 − 𝑚𝑔 = 𝑚𝑔 − 𝑘𝑦 − 𝑚𝑔 = −𝑘𝑦

∴ 𝐹 = −𝑘𝑦, … … … (3)

Therefore, the resultant force is proportional to the displacement of the

body from its equilibrium position and is oppositely directed.

According to Newton’s second law of motion,

𝐹 = 𝑚𝑎 = −𝑘𝑦

𝑑2𝑦
𝑜𝑟, 𝑚 2 = −𝑘𝑦
𝑑𝑡
𝑑2𝑦 𝑘
𝑜𝑟, + ( ) 𝑦 = 0, … … … (4)
𝑑𝑡 2 𝑚
𝑑2𝑦
This is similar to the differential equation of SHM, + 𝜔2 𝑦 = 0.
𝑑𝑡 2

𝑘 𝑘
∴ 𝜔2 = 𝑜𝑟, 𝜔 = √
𝑚 𝑚

𝑚
∴ 𝑇 = 2𝜋√
𝑘
Thus, the body oscillates simple harmonically with a time period mentioned
above.
Mechanical energy: The particle that having oscillatory motion, has both
types of mechanical energy. i.e., potential energy and kinetic energy.

If no non-conservative forces, such as the force of friction act on the

particle, the sum of its kinetic energy and potential energy remains
constant.

𝐸 = 𝐾 + 𝑈 = 𝐶𝑜𝑛𝑠𝑡𝑎𝑛𝑡

Let the displacement of a particle executing SHM at any instant be 𝑦, the

mass of the particle be 𝑚, and its velocity at that instant be 𝑣.
1
∴ kinetic energy, K = 𝑚𝑣 2
2

The potential energy of the particle at the same instant,

𝑦
Potential energy, U = W = ∫0 𝐹. 𝑑𝑦

Whenever a distance is travelled against a force, the body gathers potential

energy equal to the work done on it.

We know, the displacement, 𝑦 = 𝑎 sin(𝜔𝑡 + 𝜑)

𝑑𝑦
𝑜𝑟, 𝑣 = = 𝜔𝑎 𝑐𝑜𝑠(𝜔𝑡 + 𝜑)
𝑑𝑡
𝑑2𝑦
𝑜𝑟, 𝑎 = 2
= −𝜔2 𝑎 𝑠𝑖𝑛(𝜔𝑡 + 𝜑) = −𝜔2 𝑦
𝑑𝑡
𝐹 = 𝑚𝑎𝑠𝑠 × 𝑎𝑐𝑐𝑒𝑙𝑒𝑟𝑎𝑡𝑖𝑜𝑛 = −𝑚𝜔2 𝑦

∴ The potential energy of the particle is,

𝑦 𝑦
U= ∫0 𝐹. 𝑑𝑦 = ∫0 𝑚𝜔2 𝑦. 𝑑𝑦

1
= 𝑚𝜔2 𝑦 2
2
1
= 𝑚𝜔2 𝑎2 𝑠𝑖𝑛2 (𝜔𝑡 + 𝜑)
2
The kinetic energy of the particle is,
1 1 2
𝐾 = 𝑚𝑣 2 = 𝑚(𝜔𝑎 𝑐𝑜𝑠(𝜔𝑡 + 𝜑))
2 2
1
= 𝑚𝜔2 𝑎2 𝑐𝑜𝑠 2 (𝜔𝑡 + 𝜑)
2
Therefore, the total energy,
1
𝐸 = 𝑚𝜔2 𝑎2 [𝑠𝑖𝑛2 (𝜔𝑡 + 𝜑) + 𝑐𝑜𝑠 2 (𝜔𝑡 + 𝜑)]
2
𝟏 𝟏
∴𝑬= 𝒎𝝎𝟐 𝒂𝟐 = 𝒌𝒂𝟐 = 𝟐𝝅𝟐 𝒏𝟐 𝒂𝟐 𝒎
𝟐 𝟐
[∵ 𝜔 = 2𝜋𝑛, 𝑛 is the frequency of oscillation]

At the maximum displacement, kinetic energy is zero, but the potential

1
energy is 𝑘𝑎2 . At equilibrium position, potential energy is zero, but the
2
1
kinetic energy is 𝑘𝑎2 .
2

Thus, the total energy of the system is the same as the maximum
value of any one of the two forms of energy and is independent of
both time and displacement.

Therefore, mechanical energy is always conserved during SHM.

Average kinetic and potential energy:

The potential energy (P.E.) of the particle at a displacement 𝑦 is given by,

1
= 𝑚𝜔2 𝑦 2
2
The average potential energy,
𝑇1
∫0 2 𝑚𝜔2 𝑦 2 𝑑𝑡
̅=
𝑈 𝑇
∫0 𝑑𝑡
 1 𝑇
𝑚𝜔2 ∫0 𝑦 2 𝑑𝑡
=2 𝑇
∫0 𝑑𝑡

1 𝑇
𝑚𝜔2 ∫0 𝑎2 𝑠𝑖𝑛2 (𝜔𝑡 + 𝜑)𝑑𝑡
=2 𝑇
∫0 𝑑𝑡

We know, the displacement, 𝑦 = 𝑎 sin(𝜔𝑡 + 𝜑)

1−𝑐𝑜𝑠2𝐴
From trigonometry, we know that, 𝑠𝑖𝑛2 𝐴 =
2

1 𝑇 1 − 𝑐𝑜𝑠2(𝜔𝑡 + 𝜑)
𝑚𝜔2 𝑎2 ∫0 𝑑𝑡
̅
𝑈= 2 2
𝑇
∫0 𝑑𝑡

1 𝑇
𝑚𝜔2 𝑎2 ∫0 (1 − 𝑐𝑜𝑠2(𝜔𝑡 + 𝜑))𝑑𝑡
=4 𝑇
∫0 𝑑𝑡
𝑇 𝑇
𝑚𝜔2 𝑎2
= [∫ 𝑑𝑡 − ∫ 𝑐𝑜𝑠2(𝜔𝑡 + 𝜑)𝑑𝑡]
4𝑇
0 0

𝑇
But ∫0 𝑐𝑜𝑠2(𝜔𝑡 + 𝜑)𝑑𝑡 =0

[Since the average value of both sine and cosine function for a complete cycle or a
whole time period T is zero]

∴ The average P.E. of the particle,

𝑚𝜔2 𝑎2
= [𝑡]𝑇0 − 0
4𝑇

𝑚𝜔2 𝑎2 1 1 𝑘
= 𝑇 = 𝑚𝜔2 𝑎2 = 𝑘𝑎2 [∵ 𝜔2 = ]
4𝑇 4 4 𝑚

The kinetic energy (K.E.) of the particle at a displacement 𝑦 is given by,

 1 𝑑𝑦 2 1 𝑑 2
= 𝑚 ( ) = 𝑚 [ 𝑎 𝑠𝑖𝑛(𝜔𝑡 + 𝜑)]
2 𝑑𝑡 2 𝑑𝑡
1
= 𝑚𝜔2 𝑎2 𝑐𝑜𝑠 2 (𝜔𝑡 + 𝜑)
2

The average kinetic energy of a particle over a complete cycle is given by,
𝑆𝑢𝑚 𝑜𝑓 𝑎𝑙𝑙 𝑖𝑛𝑠𝑡𝑎𝑛𝑡𝑎𝑛𝑒𝑜𝑢𝑠 𝑣𝑎𝑙𝑢𝑒𝑠 𝑜𝑣𝑒𝑟 𝑜𝑛𝑒 𝑐𝑦𝑐𝑙𝑒
=
𝑇𝑖𝑚𝑒 𝑝𝑒𝑟𝑖𝑜𝑑 𝑇
𝑇1
∫0 2 𝑚𝜔2 𝑎2 𝑐𝑜𝑠 2 (𝜔𝑡 + 𝜑) 𝑑𝑡
= 𝑇
∫0 𝑑𝑡

𝑚𝜔2 𝑎2 𝑇
= ∫ 2 𝑐𝑜𝑠 2 (𝜔𝑡 + 𝜑)𝑑𝑡
4𝑇 0

𝑚𝜔2 𝑎2 𝑇
= ∫ ⌈1 + 𝑐𝑜𝑠2(𝜔𝑡 + 𝜑)⌉𝑑𝑡
4𝑇 0

𝑚𝜔2 𝑎2 𝑇 𝑇
= ∫ 𝑑𝑡 + ∫ 𝑐𝑜𝑠2(𝜔𝑡 + 𝜑)𝑑𝑡
4𝑇 0 0

[Since the average value of both sine and cosine function for a complete cycle or a
whole time period T is zero]

∴ The average K.E. of the particle,

𝑚𝜔2 𝑎2
= [𝑡]𝑇0
4𝑇

𝑚𝜔2 𝑎2
= 𝑇
4𝑇
1
= 𝑚𝜔2 𝑎2
4
1
= 𝑘𝑎2
4
Thus, average value of P.E. of the particle = average value of K.E. of the particle =
1 1
𝑚𝜔2 𝑎2 = 𝑘𝑎2 = half of the total energy.
4 4
Problem:

Which of the following represent simple harmonic motion?

(i) x = A sin ωt + B cos ωt

(ii) x = A sin ωt+ B cos 2ωt

(iii) x = 𝐴 𝑒 𝑖𝜔𝑡

(iv) x = A ln ωt

(i) x = A sin ωt + B cos ωt

𝑑𝑥
𝑜𝑟, = 𝐴𝜔 𝑐𝑜𝑠𝜔𝑡 − 𝐵𝜔 𝑠𝑖𝑛𝜔𝑡
𝑑𝑡

𝑑2𝑥
𝑜𝑟, = −𝐴𝜔2 𝑠𝑖𝑛𝜔𝑡 − 𝐵𝜔2 𝑐𝑜𝑠𝜔𝑡
𝑑𝑡 2

𝑑2𝑥
𝑜𝑟, = −𝜔2 (𝐴 𝑠𝑖𝑛𝜔𝑡 + 𝐵 𝑐𝑜𝑠𝜔𝑡) = −𝜔2 𝑥
𝑑𝑡 2

This differential equation is similar to the differential equation of SHM.

Therefore, x = A sin ωt + B cos ωt represents SHM.

(ii) x = A sin ωt+ B cos 2ωt

𝑑𝑥
𝑜𝑟, = 𝐴𝜔 𝑐𝑜𝑠𝜔𝑡 − 𝐵(2𝜔) 𝑠𝑖𝑛2𝜔𝑡
𝑑𝑡

𝑑2𝑥
𝑜𝑟, 2
= −𝐴𝜔2 𝑠𝑖𝑛𝜔𝑡 − 4𝐵𝜔2 𝑐𝑜𝑠2𝜔𝑡 = −𝜔2 (A sin ωt+ 4B cos2𝜔𝑡) ≠ −𝜔2 𝑥
𝑑𝑡
Therefore, x = A sin ωt + B cos 2ωt does not represent SHM.

(iii) x = A 𝒆𝒊𝝎𝒕

𝑑𝑥
𝑜𝑟, = 𝐴𝑖𝜔 𝑒 𝑖𝜔𝑡
𝑑𝑡
 𝑑2𝑥
𝑜𝑟, = 𝐴 𝑖 2 𝜔2 𝑒 𝑖𝜔𝑡 = −𝐴𝜔2 𝑒 𝑖𝜔𝑡 = −𝜔2 x
𝑑𝑡 2

This differential equation is similar to the differential equation of SHM.

Therefore, x = A 𝒆𝒊𝝎𝒕 represents SHM.

(iv) x = A ln ωt
𝑑𝑥 𝐴 𝐴
𝑜𝑟, = ( )𝜔 =
𝑑𝑡 𝜔𝑡 𝑡
𝑑2𝑥 𝐴
𝑜𝑟, 2
= − 2
≠ −𝜔2 x
𝑑𝑡 𝑡
Therefore, x = A ln ωt does not represent SHM.

Problem: The total energy of simple harmonic motion is E. What will be the
kinetic energy of the particle when displacement is half of the amplitude?

Ans: The total energy of a particle executing SHM,

1
𝐸 = 𝑚𝜔2 𝑎2 , … … … (1)
2
The kinetic energy of a particle executing SHM,
1
= 𝑚𝜔2 𝑎2 𝑐𝑜𝑠 2 (𝜔𝑡 + 𝜑)
2
1
= 𝑚𝜔2 [𝑎2 − 𝑎2 𝑠𝑖𝑛2 (𝜔𝑡 + 𝜑)]
2
1
= 𝑚𝜔2 (𝑎2 − 𝑦 2 )
2
𝑎 1 1 𝑎2
When 𝑦 = , then K.E. = 𝑚𝜔2 (𝑎2 − 𝑦 2 ) = 𝑚𝜔2 (𝑎2 − )
2 2 2 4

1 3 3 1 3
= 𝑚𝜔2 𝑎2 = × 𝑚𝜔2 𝑎2 = 𝐸 (𝐴𝑛𝑠. )
2 4 4 2 4
 Combination of Simple Harmonic Motions

Consider two simple harmonic motions with the same frequency acting on a
particle simultaneously. The state of motion should be described by the addition
of SHM. The resultant displacement of the particle can be obtained by the
algebraic sum of the individual displacements.

We organize our study under two heads:

❑ Composition of SHMs along the same straight line

❑ Composition of SHMs along two mutually perpendicular straight lines

The resultant of two simple harmonic motions either along straight or at right
angles to each other may be obtained by any of the two methods, viz.,

(i) graphical method and

(ii) Analytical method based on finding the vector sum of the individual
motions.

COMPOSITION OF TWO SIMPLE HARMONIC VIBRATIONS ALONG THE SAME LINE

Two SHMs of the same frequency but different amplitude and phase acted on a
particle simultaneously.

𝑦1 = a cos(𝜔𝑡 + 𝜑1 ) … … … … … (1)

𝑦2 = b cos(𝜔𝑡 + 𝜑2 ) … … … … … (2)

According to the principle of superposition, the resultant displacement of the

particle is given by,

𝑦 = 𝑦1 + 𝑦2 = a cos(𝜔𝑡 + 𝜑1 ) + b cos(𝜔𝑡 + 𝜑2 )

𝑦 = 𝑎(𝑐𝑜𝑠𝜔𝑡 𝑐𝑜𝑠𝜑1 − 𝑠𝑖𝑛𝜔𝑡 𝑠𝑖𝑛𝜑1 ) + 𝑏(𝑐𝑜𝑠𝜔𝑡 𝑐𝑜𝑠𝜑2 − 𝑠𝑖𝑛𝜔𝑡 𝑠𝑖𝑛𝜑2 )

= 𝑐𝑜𝑠𝜔𝑡(𝑎 𝑐𝑜𝑠𝜑1 + 𝑏 𝑐𝑜𝑠𝜑2 ) − 𝑠𝑖𝑛𝜔𝑡(𝑎 𝑠𝑖𝑛𝜑1 + 𝑏 𝑠𝑖𝑛𝜑2 ) … … …(3)

Since the amplitude 𝑎 and 𝑏 and the angle of the epoch 𝜑1 and 𝜑2 of two
vibrations are constant.

Let, 𝑎 𝑐𝑜𝑠𝜑1 + 𝑏 𝑐𝑜𝑠𝜑2 = 𝐴 𝑐𝑜𝑠𝜑 … … … (4)

𝑎 𝑠𝑖𝑛𝜑1 + 𝑏 𝑠𝑖𝑛𝜑2 = 𝐴 𝑠𝑖𝑛𝜑 … … … (5)

∴ 𝑦 = 𝐴 𝑐𝑜𝑠𝜑 𝑐𝑜𝑠𝜔𝑡 − 𝐴 𝑠𝑖𝑛𝜑 𝑠𝑖𝑛𝜔𝑡 … … … (6)

𝑜𝑟, 𝑦 = 𝐴 cos⁡(𝜔𝑡 + 𝜑)

Where, 𝐴 = √𝑎2 + 𝑏 2 + 2𝑎𝑏 𝑐𝑜𝑠(𝜑1 − 𝜑2 ) …... (7)

𝑎𝑆𝑖𝑛𝜑1 +𝑏𝑆𝑖𝑛𝜑2
and 𝑡𝑎𝑛𝜑 = ………………………………. (8)
𝑎𝐶𝑜𝑠𝜑1 +𝑏𝐶𝑜𝑠𝜑2

𝑎𝑆𝑖𝑛𝜑1 + 𝑏𝑆𝑖𝑛𝜑2
𝜑 = 𝑡𝑎𝑛−1 [ ]
𝑎𝐶𝑜𝑠𝜑1 + 𝑏𝐶𝑜𝑠𝜑2

Special Cases:

➢ Same phase: Two simple harmonic motions are in the same phase,

Then 𝜑1 = 𝜑2 = 𝜑 𝑜𝑟 𝜑1 − 𝜑2 = 0, 2𝜋, 4𝜋, … … . . = 2𝑛𝜋,

Where, 𝑛 = 0,1,2, …

From equation (7),

𝐴 = √𝑎2 + 𝑏 2 + 2𝑎𝑏 = 𝑎 + 𝑏 (Constructive interference)

➢ Opposite Phase: Two simple harmonic motions are in opposite phase,

When 𝜑1 − 𝜑2 = 𝜋, 3𝜋, 5𝜋, … … . . = (2𝑛 + 1)𝜋, Where, 𝑛 = 0,1,2, …

𝐴 = √𝑎2 + 𝑏 2 − 2𝑎𝑏 = 𝑎 − 𝑏 (Destructive interference)

Problem:

➢ Two SHMs acting simultaneously on a particle are given by, 𝒚𝟏 = 𝟐𝑺𝒊𝒏𝝎𝒕

and
𝝅
𝒚𝟐 = 𝑺𝒊𝒏 (𝝎𝒕 + ) . Find the equation of the resultant vibration.
𝟑

Answer: The component SHMs are similar to

𝑦1 = 𝑎1 𝑆𝑖𝑛(𝜔𝑡 + 𝛼1 )

𝑦2 = 𝑎2 𝑆𝑖𝑛(𝜔𝑡 + 𝛼2 )
𝜋
Here, 𝑎1 = 2, 𝑎2 = 1, 𝛼1 = 0, 𝛼2 = ,
3

The resultant vibration is given by,

𝑦 = 𝐴𝑆𝑖𝑛(𝜔𝑡 + 𝜑)

The resultant amplitude, 𝐴 = √𝑎12 + 𝑎22 + 2𝑎1 𝑎2 𝐶𝑜𝑠(𝛼1 − 𝛼2 )

= √4 + 1 + 2 × 2 × 𝐶𝑜𝑠(− 𝜋Τ3) = √7 = 2.645

𝑎1 𝑆𝑖𝑛𝛼1 + 𝑎2 𝑆𝑖𝑛𝛼2 0.886
𝑡𝑎𝑛𝜑 = =
𝑎1 𝐶𝑜𝑠𝛼1 + 𝑎2 𝐶𝑜𝑠𝛼2 2.5
19.1 × 𝜋
𝑜𝑟, 𝜑 = 19. 1° = 𝑟𝑎𝑑𝑖𝑎𝑛 = 0.1061𝜋
180
Therefore, 𝑦 = 2.654𝑆𝑖𝑛(𝜔𝑡 + 0.1061𝜋) represents the equation of resultant
vibration.

COMPOSITION OF TWO SHMs AT RIGHT ANGLE TO EACH OTHER

Let us consider two simple harmonic vibrations of the same time period but
different amplitudes and phases acting on a particle simultaneously from
a mutually perpendicular direction. The displacements of the particle along the X
and Y axes can be written as:
 𝑥 = a Sin(𝜔𝑡 + 𝜑) … … … … … (1)

y = b Sin 𝜔𝑡 ………………………… (2)

𝑥
From (1) we get, = Sin(𝜔𝑡 + 𝜑)
𝑎

= Sin 𝜔𝑡 𝐶𝑜𝑠𝜑 + 𝐶𝑜𝑠𝜔𝑡 𝑆𝑖𝑛𝜑

= Sin 𝜔𝑡 𝐶𝑜𝑠𝜑 + √1 − 𝑆𝑖𝑛2 𝜔𝑡 𝑆𝑖𝑛𝜑 …………(3)

𝑦
From (2) we get, = Sin 𝜔𝑡 …………………(4)
𝑏

Substituting this value of Sin 𝜔𝑡 in equation (3) we get,

𝑥 𝑦 𝑦2
= Cos𝜑 + √1 − . Sin𝜑
𝑎 𝑏 𝑏2

𝑥 𝑦 𝑦2
𝑜𝑟, − Cos𝜑 = √1 − . Sin𝜑 …………..(5)
𝑎 𝑏 𝑏2

Squaring both sides of equation (5) we get,

𝑥2 𝑦2 𝑥𝑦 𝑦2
2
+ 2
𝐶𝑜𝑠 2 𝜑 − 2 Cos𝜑 = (1 − ) 𝑆𝑖𝑛2 𝜑
𝑎 𝑏 𝑎𝑏 𝑏2

𝑥2 𝑦2 𝑦2 𝑥𝑦
or, 2
+ 2
𝐶𝑜𝑠 2 𝜑 + 𝑆𝑖𝑛2 𝜑 − 2 Cos𝜑 = 𝑆𝑖𝑛2 𝜑
𝑎 𝑏 𝑏2 𝑎𝑏

𝑥2 𝑦2 𝑥𝑦
or, 2
+ (𝐶𝑜𝑠 2 𝜑 + 𝑆𝑖𝑛2 𝜑) − 2 Cos𝜑 = 𝑆𝑖𝑛2 𝜑 [𝐶𝑜𝑠 2 𝜑 + 𝑆𝑖𝑛2 𝜑 = 1]
𝑎 𝑏2 𝑎𝑏

𝒙𝟐 𝒚𝟐 𝒙𝒚
𝟐
+ −𝟐 𝐂𝐨𝐬𝝋 = 𝑺𝒊𝒏𝟐 𝝋 …………(6)
𝒂 𝒃𝟐 𝒂𝒃

This is the general equation of the resultant vibration of two simple harmonic

vibrations.

The displacement of the particle will be along a curve whose shape will depend
upon the value of the phase difference 𝜑 between the two vibrations.

Special cases:
Case I: When 𝜑 = 0, 2𝜋, 4𝜋, … … … = 2𝑛𝜋,

where, 𝑛 = 0,1,2, … …

Hence, 𝑆𝑖𝑛𝜑 = 0 and 𝐶𝑜𝑠𝜑 = 1

Substituting these values in equation (6),

𝑥2 𝑦2 2𝑥𝑦 𝑥 𝑦 𝑥 𝑦
2
+ − =0 𝑜𝑟 , ( − )2 = 0 𝑜𝑟, ±( − )=0
𝑎 𝑏2 𝑎𝑏 𝑎 𝑏 𝑎 𝑏

𝑏
𝑜𝑟, 𝑦 = 𝑥………………………………(7)
𝑎

𝜑=0
2b

2a

This is an equation of straight line passing through the origin.

i.e., the particle vibrates simple harmonically along the line.

𝑏
Resultant amplitude= √𝑎2 + 𝑏 2 & Phase angle = 𝑡𝑎𝑛−1
𝑎

𝜋 1
Case II: When 𝜑 = 𝑟𝑎𝑑𝑖𝑎𝑛, 𝑇ℎ𝑒𝑛 𝐶𝑜𝑠𝜑 = 𝑆𝑖𝑛𝜑 =
4 √2

Substituting these values in equation (6),

𝑥2 𝑦2 √2𝑥𝑦 1
2
+ − = ……………………….(8)
𝑎 𝑏2 𝑎𝑏 2

Equation (8) represents the equation of an oblique ellipse.

 𝜋
𝜑=
4
2b

Case III:
2a

𝜋
When 𝜑 = 𝑟𝑎𝑑𝑖𝑎𝑛, 𝑇ℎ𝑒𝑛 𝐶𝑜𝑠𝜑 = 0 𝑎𝑛𝑑 𝑆𝑖𝑛𝜑 = 1
2

Substituting these values in equation (6),

𝑥2 𝑦2
2
+ = 1 ………………..(9)
𝑎 𝑏2

Equation (9) represents a symmetrical ellipse whose center coincides with the
origin.

𝜋
𝜑=
2
2b

2a
If 𝑎 = 𝑏 i.e., the amplitudes of the two vibrations are equal, then equation (9)
reduces to,
𝜋
2 2 2 𝜑= &𝑎=𝑏
𝑥 + 𝑦 = 𝑎 ……………(10) 2

This is the equation of a circle of radius ‘a’.

2b
2a
3𝜋 3𝜋 1 3𝜋 1
Case IV: When 𝜑 = 𝑟𝑎𝑑𝑖𝑎𝑛, 𝑆𝑖𝑛 = and 𝐶𝑜𝑠 =−
4 4 √2 4 √2

Substituting these values in equation (6),

𝑥2 𝑦2 √2𝑥𝑦 1
2
+ + = ……………………….(10)
𝑎 𝑏2 𝑎𝑏 2

Equation (10) represents again the equation of an oblique ellipse with an axis
𝝅
rotated by with respect to that in case II.
𝟐

3𝜋
𝜑=
4
2b

2a

Case V: When 𝜑 = 𝜋 𝑟𝑎𝑑𝑖𝑎𝑛, 𝑇ℎ𝑒𝑛 𝐶𝑜𝑠𝜑 = −1 𝑎𝑛𝑑 𝑆𝑖𝑛𝜑 = 0

Substituting these values in equation (6),

𝑥2 𝑦2 2𝑥𝑦 𝑥 𝑦 𝒙 𝒚
2
+ + =0 𝑜𝑟 , ( + )2 = 0 𝒐𝒓, ± ( + ) = 0
𝑎 𝑏2 𝑎𝑏 𝑎 𝑏 𝒂 𝒃

𝑏
𝑜𝑟, 𝑦 = − 𝑥……………………………….(11)
𝑎

This is an equation of straight line passing through the origin with negative
slope.
𝒃
The line is now inclined with negative x-direction at an angle of 𝒕𝒂𝒏−𝟏 (− ) and
𝒂
𝐀= √𝒂𝟐 + 𝒃𝟐 .

𝜑=𝜋
2b

2a

Lissajous figure: When a particle is influenced simultaneously by two simple

harmonic motions at right angle to each other, the resultant motion of the particle
traces a curve. These curves are called Lissajous figures. It is named after the
French Physicist Jules Antoine Lissajous (1822-1880).

COMPOSITION OF TWO SHM AT RIGHT ANGLE TO EACH OTHER

Let us consider two simple harmonic vibrations acting on a particle simultaneously

from mutually perpendicular direction and having time periods in the ratio 1:2.
Their equations can be written as:

𝑥 = a Sin(2𝜔𝑡 + 𝜑) … … … … … (1)
 y = b Sin 𝜔𝑡 ………………………… (2)
𝑥
From (1) we get, = Sin(2𝜔𝑡 + 𝜑)
𝑎

= Sin 2𝜔𝑡 𝐶𝑜𝑠𝜑 + 𝐶𝑜𝑠2𝜔𝑡 𝑆𝑖𝑛𝜑

= 2 Sin 𝜔𝑡 𝐶𝑜𝑠 𝜔𝑡 𝐶𝑜𝑠𝜑 + (1 − 2𝑆𝑖𝑛2 𝜔𝑡)𝑆𝑖𝑛𝜑 …………(3)

𝑦
From (2) we get, = Sin 𝜔𝑡
𝑏

𝑦2
Cos 𝜔𝑡 = √1 − 𝑆𝑖𝑛2 𝜔𝑡 = √1 −
𝑏2

Substituting this value of Sin 𝜔𝑡 and 𝐶𝑜𝑠 𝜔𝑡 in equation (3) we get,

𝑥 𝑦 𝑦2 𝑦2
= 2. . √1 − 2
Cos𝜑 + (1 − 2. ) Sin𝜑
𝑎 𝑏 𝑏 𝑏2

𝑥 𝑦2 2𝑦 𝑦2
[ − (1 − 2. ) Sin𝜑 ] = Cos𝜑 √1 −
𝑎 𝑏2 𝑏 𝑏2

𝑥 𝑦2 2𝑦𝐶𝑜𝑠𝜑 𝑦2
[( − Sin𝜑) + 2. Sin𝜑] = √1 − …………..(4)
𝑎 𝑏2 𝑏2 𝑏2

Squaring both sides of equation (4) we get,

𝑥 2 4𝑦 4 2
𝑥 2𝑦 2 4𝑦 2 𝐶𝑜𝑠 2 𝜑 𝑦2
( − Sin𝜑) + 4 𝑆𝑖𝑛 𝜑 + 2 ( − Sin𝜑) 2 Sin𝜑 = (1 − 2 )
𝑎 𝑏 𝑎 𝑏 𝑏2 𝑏

𝑥 2 4𝑦 4 4𝑦 2
or, ( − Sin𝜑) + (𝑆𝑖𝑛2 𝜑 + 𝐶𝑜𝑠 2 𝜑) − (𝑆𝑖𝑛2 𝜑 + 𝐶𝑜𝑠 2 𝜑) +
𝑎 𝑏4 𝑏2
4𝑦 2 𝑥
𝑆𝑖𝑛𝜑 = 0
𝑏2 𝑎

𝑥 2 4𝑦 2 𝑦 2 𝑥
∴ ( − Sin𝜑) + (𝑏2 + 𝑎 𝑆𝑖𝑛𝜑 − 1) = 0…………(5)
𝑎 𝑏2

This is the general equation of a curve having two loops.

Special cases:

Case I: If 𝜑 = 0, 𝜋, 2𝜋, 𝑒𝑡𝑐. , 𝑆𝑖𝑛𝜑 = 0,

Then equation (5) becomes,

𝑥2 4𝑦 2 𝑦 2
2
+ (𝑏2 − 1) = 0 … … … … … … … … (6)
𝑎 𝑏2
𝜋
Case II: If 𝜑 = , 𝑆𝑖𝑛𝜑 = 1,
2

Then equation (5) becomes,

𝑥 2 4𝑦 2 𝑦 2 𝑥
(𝑎 − 1) + 𝑏2
(𝑏2 + 𝑎 − 1) = 0

𝑥 2 4𝑦 2 𝑥 4𝑦 4
𝑜𝑟, ( − 1) + (𝑎 − 1) + =0
𝑎 𝑏2 𝑏4

2
𝑥 2𝑦 2
𝑜𝑟, [( − 1) + 2
] =0
𝑎 𝑏

𝑥 2𝑦 2
𝑜𝑟, ( − 1) + =0
𝑎 𝑏2

2𝑦 2 𝑥
𝑜𝑟, = − ( − 1)
𝑏2 𝑎

𝑏2 𝑥
𝑜𝑟, 𝑦 2 = − (𝑎 − 1)
2

𝑏2
∴ 𝑦2 = − (𝑥 − 𝑎)
2𝑎

This represents the equation of Parabola with vertex at (a,0).

The resulting curves for different values of 𝜑 are as follows:

𝜑=0 𝜑 = 𝜋Τ4 𝜑 = 𝜋Τ2

𝜑 = 3𝜋Τ4 𝜑=𝜋
 DAMPED HARMONIC OSCILLATIONS

A system executing simple harmonic motion is called a harmonic oscillator. A

harmonic oscillator produces oscillations.The oscillations can be of two types:

➢ Undamped Oscillation

➢ Damped oscillation

Question: What are damped oscillations?

➢ Oscillations in the presence of frictional forces are called damped

oscillations.

➢ It results in the dissipation of the energy of an oscillator.

Body is surrounded by air

Vibrating body experiencing some opposing force in the form of air

friction

Vibrating body does work against this opposing force

To do work, energy is needed

Vibrating body uses its energy to do the work against this opposing force

It loses energy
 UNDAMPED OSCILLATION DAMPED OSCILLATION

1. Oscillations whose amplitude remains 1. Oscillations whose amplitude

constant with time are called undamped decreases gradually with time are called
oscillations. damped oscillations.

2. Undamped oscillations will oscillate 2. Damped oscillations do not continue

indefinitely. for a longer time, die out eventually.

3. In undamped oscillation, energy of the 3. Damping is progressive diminution of

vibrating object does not get dissipated. amplitude of oscillations in an oscillatory
Hence, there is no power loss. system, caused by dissipation of stored
energy.

4. If the bob of a pendulum is displaced 4. Example of a damped oscillator is a

in vacuum and then released, the bob swinging pendulum, in which the
will continue to execute SHM of constant vibrations slow down and stops over
amplitude. time.
Amplitude

Amplitude

Time
Time
If an oscillator moves in a resistive medium, its amplitude goes on decreasing. So,
the energy of the oscillator is used in doing work against the resistance of the
medium. The motion of the oscillator is then said to be damped.

The damping force or the resistive force is proportional to the velocity of the
oscillator.

Damping force is given by,

𝑑𝑥
𝑓𝑟 ∝ −𝑣 = −𝑏
𝑑𝑡
Where 𝑏 is the damping constant giving the strength of damping. Its value
depends upon the properties of the medium, such as the density, shape, and
dimension of object.

EQUATION OF MOTION OF DAMPED HARMONIC OSCILLATOR:

In the case of the damped harmonic oscillator, two forces act on it.

(i) Elastic restoring force

(ii) Damping force

A frictional drag force is one which is always directed in the opposite direction to
the instantaneous velocity of the object upon which it acts and is directly
proportional to the magnitude of this velocity.

The equation of motion of damped oscillator is given by,

𝐹 = 𝐹𝑒 + 𝐹𝑟
Differential equation of damped harmonic oscillator is:
𝑑2𝑥 𝑑𝑥
𝑚 = −𝑏 − 𝑘𝑥
𝑑𝑡 2 𝑑𝑡 ➢ Restoring force is
always proportional
𝑑2𝑥 𝑏 𝑑𝑥 𝑘 to the displacement
+ + 𝑥=0 of the body.
𝑑𝑡 2 𝑚 𝑑𝑡 𝑚
➢ Damping force is
𝑑2𝑥 𝑑𝑥
+ 2𝜆 + 𝜔2 𝑥 = 0 … … … … … … (1) proportional to the
𝑑𝑡 2 𝑑𝑡
velocity of the body.
𝑏 𝑘
Where, 2𝜆 = and 𝜔2 =
𝑚 𝑚

To solve equation (1) let us take the trial solution,

′
𝑥 = 𝐴𝑒 𝑚 𝑡 … … … … … … … . (2)

Substituting this in equation (1) we get,

𝑑𝑥 ′
𝑚′2 𝑥 + 2𝜆𝑚′ 𝑥 + 𝜔2 𝑥 = 0 Here, = 𝐴𝑚′ 𝑒 𝑚 𝑡 = 𝑚′ 𝑥
𝑑𝑡

𝑑2𝑥 ′
𝑜𝑟, 𝑚´2 + 2𝜆𝑚´ + 𝜔2 = 0 = 𝐴𝑚′2 𝑒 𝑚 𝑡 = 𝑚′2 𝑥
𝑑𝑡 2

−𝑏±√𝑏2 −4𝑎𝑐
∴ 𝑚′ = −𝜆 ± √𝜆2 − 𝜔 2 [∵ ]
2𝑎

𝑚1′ = −𝜆 + √𝜆2 − 𝜔 2

𝑚2′ = −𝜆 − √𝜆2 − 𝜔 2

The two possible solution of equation (1) are,

′
𝑥1 = 𝐴𝑒 𝑚1𝑡
′
𝑥2 = 𝐵𝑒 𝑚2𝑡

The sum of above two solution will satisfy equation (1),

𝑥 = 𝑥1 + 𝑥2
′ ′
𝑥 = 𝐴𝑒 𝑚1𝑡 + 𝐵𝑒 𝑚2𝑡
 2 −𝜔2 ) 𝑡 2 −𝜔2 ) 𝑡
𝑜𝑟, 𝑥 = 𝐴𝑒 (−𝜆+√𝜆 + 𝐵𝑒 (−𝜆−√𝜆
2 −𝜔2 )𝑡 2 −𝜔2 )𝑡
𝑜𝑟, 𝑥 = 𝐴𝑒 −𝜆𝑡 𝑒 (√𝜆 + 𝐵𝑒 −𝜆𝑡 𝑒 (−√𝜆
2 − 𝜔2 )𝑡 2 − 𝜔2 )𝑡
∴ The general solution is, 𝑥 = 𝑒 −𝜆𝑡 [𝐴𝑒 (√𝜆 + 𝐵𝑒 −(√𝜆 ]………..(3)

The term 𝑒 −𝜆𝑡 is an exponentially decreasing term with increasing time. i.e.,
amplitude goes on decreasing with time.

Case I: (Overdamped motion) :

If 𝜆2 > 𝜔2 , the indices of 𝑒 are real and we get,

𝑥 = 𝑒 −𝜆𝑡 [A𝑒 𝛼𝑡 + 𝐵𝑒 −𝛼𝑡 ], … … … (4) Where, 𝛼 = √𝜆2 − 𝜔 2

Now, let us replace 𝐴 and 𝐵 by two other constants 𝐶 and 𝛿 such that,
𝑐 𝑐
𝐴 = 𝑒 𝛿 and 𝐵 = 𝑒 −𝛿
2 2
𝑐 𝑐 𝑐 𝑐
∴ 𝐴 + 𝐵 = 𝑒 𝛿 + 𝑒 −𝛿 = (𝑒 𝛿 + 𝑒 −𝛿 ) = 2𝑐𝑜𝑠ℎ𝛿 = 𝐶𝑐𝑜𝑠ℎ𝛿
2 2 2 2
𝑐 𝛿
𝐴 𝑒
And = 2
𝑐 −𝛿 = 𝑒 2𝛿
𝐵 𝑒
2

Using the new constant equation (4) becomes,

𝑐 𝑐
𝑥 = 𝑒 −𝜆𝑡 [ 𝑒 𝛿 𝑒 𝛼𝑡 + 𝑒 −𝛿 𝑒 −𝛼𝑡 ]
2 2
𝑐
= 𝑒 −𝜆𝑡 [𝑒 (𝛼𝑡+𝛿) + 𝑒 −(𝛼𝑡+𝛿) ]
2
𝑐
= 𝑒 −𝜆𝑡 2cosh(𝛼t + 𝛿)
2

= 𝐶𝑒 −𝜆𝑡 cosh(𝛼t + 𝛿)

𝒙 = 𝑪𝒆−𝝀𝒕 𝒄𝒐𝒔𝒉[(√𝝀𝟐 − 𝝎𝟐 )𝒕 + 𝜹] … … (𝟓)

Equation (5) represents a continuous return of 𝑥 from its maximum value to zero
when 𝑡 = ∞ without oscillation. This type of motion is called the dead-beat
motion or overdamped or aperiodic motion.
Example: Dead beat galvanometer, pendulum oscillating in a viscous fluid etc.
2 − 𝜔2 )𝑡 2 − 𝜔2 )𝑡
Case II : (Underdamped motion) : 𝑥 = 𝑒 −𝜆𝑡 [𝐴𝑒 (√𝜆 + 𝐵𝑒 −(√𝜆 ]

If 𝜆2 < 𝜔2 , indices of 𝑒 are imaginary and equation (3) can be written as,

𝑥 = 𝑒 −𝜆𝑡 [𝐴𝑒 𝑖𝜃𝑡 + 𝐵𝑒 −𝑖𝜃𝑡 ], Here, 𝜃 = √(𝜔 2 − 𝜆2 )

𝑜𝑟, 𝑥 = 𝑒 −𝜆𝑡 [𝐴𝑐𝑜𝑠𝜃𝑡 + 𝑖𝐴𝑠𝑖𝑛𝜃𝑡 + 𝐵𝑐𝑜𝑠𝜃𝑡 − 𝑖𝐵𝑠𝑖𝑛𝜃𝑡]

= 𝑒 −𝜆𝑡 [(𝐴 + 𝐵)𝑐𝑜𝑠𝜃𝑡 + 𝑖(𝐴 − 𝐵)𝑠𝑖𝑛𝜃𝑡]

Let (𝐴 + 𝐵) = 𝑎𝑠𝑖𝑛𝛾

i(𝐴 − 𝐵) = 𝑎𝑐𝑜𝑠𝛾

𝑥 = 𝑒 −𝜆𝑡 [𝑎𝑠𝑖𝑛𝛾 𝑐𝑜𝑠𝜃𝑡 + 𝑎𝑐𝑜𝑠𝛾 𝑠𝑖𝑛𝜃𝑡]

= 𝑒 −𝜆𝑡 𝑎[𝑐𝑜𝑠𝛾 𝑠𝑖𝑛𝜃𝑡 + 𝑠𝑖𝑛𝛾 𝑐𝑜𝑠𝜃𝑡]

= 𝑎𝑒 −𝜆𝑡 𝑠𝑖𝑛(𝜃𝑡 + 𝛾)

∴ 𝑥 = 𝑎𝑒 −𝜆𝑡 sin[√(𝜔 2 − 𝜆2 )t + 𝛾] … … … (6)

In this case 𝑥 alternates in sign and we have periodic motion but the amplitude
continuously diminishes due to the factor𝑒 −𝜆𝑡 . This situation is called
underdamping with the amplitude 𝑎𝑒 −𝜆𝑡 and the frequency √𝜔 2 − 𝜆2 .

2 − 𝜔2 )𝑡 2 − 𝜔2 )𝑡
Case III: (Critical damped motion) 𝑥 = 𝑒 −𝜆𝑡 [𝐴𝑒 (√𝜆 + 𝐵𝑒 −(√𝜆 ]……(3)

If 𝜔2 = 𝜆2 , √𝜆2 − 𝜔 2 = 0; 𝑜𝑟, 𝜆 = 𝜔;

From equation (3) can be written as,

𝑥 = 𝑒 −𝜔𝑡 (𝐴𝑒 0 + 𝐵𝑒 0 ) = 𝑒 −𝜔𝑡 (𝐴 + 𝐵)

This implies that oscillation is decaying without any damping factor which is
impossible. So, the solution breaks down.
Now, we have to consider that 𝜆2 is not quite equal to 𝜔2 , but very close to each
other. Thus, √𝜆2 − 𝜔 2 = ℎ ≈ 0.

From equation (3),

𝑥 = 𝑒 −𝜆𝑡 [𝐴𝑒 ℎ𝑡 + 𝐵𝑒 −ℎ𝑡 ]

ℎ2 𝑡 2 ℎ3 𝑡 3 ℎ2 𝑡 2 ℎ3 𝑡 3
= 𝑒 −𝜆𝑡 [𝐴(1 + ℎ𝑡 + + + ⋯ ) + 𝐵(1 − ℎ𝑡 + − + ⋯ )+)]
2! 3! 2! 3!

= 𝑒 −𝜆𝑡 [𝐴(1 + ℎ𝑡) + 𝐵(1 − ℎ𝑡)]

∴ 𝑥 = 𝑒 −𝜆𝑡 [ (𝐴 + 𝐵) + (𝐴 − 𝐵)ℎ𝑡], ……….(7)

Let, 𝐴 + 𝐵 = 𝐴´ and (𝐴 − 𝐵)ℎ = 𝐵´

∴ 𝑥 = 𝑒 −𝜆𝑡 (𝐴´ + 𝐵´ 𝑡)…………………(8)

At amplitude, 𝑥 = 𝑥𝑚𝑎𝑥 = 𝑎 (𝑎𝑡 𝑡 = 0)

𝑑𝑥
And =0 at t = 0
𝑑𝑡

Applying these two conditions in equation (8),

𝑎 = 𝑒 0 (𝐴´ + 𝐵´ × 0)

or, 𝐴´ = 𝑎
𝑑𝑥
= −𝜆𝑒 −𝜆𝑡 (𝐴´ + 𝐵´ 𝑡) + 𝑒 −𝜆𝑡 𝐵´
𝑑𝑡

𝑑𝑥
⃒𝑡=0 = −𝜆𝑒 0 (𝐴´ + 𝐵´ × 0) + 𝑒 0 𝐵´ = 0
𝑑𝑡

Or, −𝜆𝐴´ + 𝐵´ = 0 or, 𝐵´ = 𝜆𝐴´ or, 𝐵´ = 𝜆𝑎

Substituting these values equation (8) can be written as,

𝑥 = 𝑒 −𝜆𝑡 (𝑎 + 𝜆𝑎𝑡)

∴ 𝒙 = 𝒂 𝒆−𝝀𝒕 (𝟏 + 𝝀𝒕),…………………(9)
This solution represents a continuous return of 𝑥 from its maximum amplitude to
zero. The second term in equation (9) decays less rapidly than the first term. As
time increases the exponential factor 𝑒 −𝜆𝑡 dominates and the displacement
decreases rapidly reaching the value zero for a finite value of 𝑡. The oscillator just
ceases to oscillate and its motion becomes aperiodic or non-oscillatory. It is called
critical damping.
Logarithmic decrement of damped oscillatory system:
𝑇
Let us consider decrement of the successive amplitudes, at intervals of time =
2
𝜋
.
𝜔

Time Amplitude

𝑡=0 𝐴1 = 𝑎

𝑇 𝜋 𝐴2 = 𝑎𝑒 −𝜆𝑇/2
𝑡= =
2 𝜔

2𝜋 𝐴3 = 𝑎𝑒 −𝜆𝑇
𝑡=𝑇=
𝜔

3𝑇 3𝜋 𝐴4 = 𝑎𝑒 −3𝜆𝑇/2
𝑡= =
2 𝜔

Let the magnitude of successive amplitudes be 𝐴1 ,𝐴2 ,𝐴3 , etc. Using the relation
𝑎 𝑒 −𝜆𝑡 for amplitude we have,
𝐴1 𝐴2 𝐴3
∴ = = = 𝑒 𝜆𝑇/2 =Constant, since 𝜆 and 𝑇 are constants for a given
𝐴2 𝐴3 𝐴4
motion.

Putting, 𝑒 𝜆𝑇/2 = 𝑑 we have,

𝐴1 𝐴2 𝐴3
= = =⋯=𝑑
𝐴2 𝐴3 𝐴4

𝑑 is a constant for the oscillating system and called the decrement or damping
ratio of the swing.
𝜆𝑡
∴ 𝑙𝑜𝑔𝑒 𝑑 = = 𝜇 𝑜𝑟, 𝑑 = 𝑒 𝜇
2
Here, 𝜇 is the natural logarithm of the decrement is called the logarithmic
decrement for the oscillation.

𝐴1
𝐴3
𝐴5 𝐴7

𝐴6
𝐴4

𝐴2

Quality factor: The quality factor, 𝑄, also referred to as the figure of merit, of a
harmonic oscillator is defined as 2𝜋 times the ratio between the energy stored
and the energy lost per period. It is a dimensionless quantity.
𝑒𝑛𝑒𝑟𝑔𝑦 𝑠𝑡𝑜𝑟𝑒𝑑 𝜔
∴ 𝑄 = 2𝜋 =
𝑒𝑛𝑒𝑟𝑔𝑦 𝑙𝑜𝑠𝑡 𝑝𝑒𝑟 𝑝𝑒𝑟𝑖𝑜𝑑 2𝜆

The quality factor measures the quality of a harmonic oscillator. The less the
damping, the better the quality of the harmonic oscillator.

Problem: A massless spring, suspended from a rigid support, carries a mass of 500
gm at its lower end and the system oscillates with a frequency of 5/sec. If the
amplitude is reduced to half its undamped value in 20 sec, calculate (i) the force
constant of the spring, (ii) the relaxation time of the system and (iii) its quality
factor.
Solution:

(i) The force constant 𝑘 of the mass-spring system is given by,

𝑘
𝜔=√ Where, 𝜔 = angular frequency= 2𝜋𝑛 = 2𝜋. 5 = 10𝜋 rad/sec
𝑚

𝑘
Then, 𝜔2 =
𝑚

Or, k= 𝑚. 𝜔2 = 500 × (10𝜋)2 = 4.934 × 105 dynes/cm.

(ii) The amplitude is reduced to half its undamped value in 20 sec,

𝑎 = 𝑎0 𝑒 −𝜆𝑡
𝑎0 1
𝑜𝑟, 𝑎 = = 𝑎0 𝑒 −20𝜆 𝑜𝑟, 𝑒 −20𝜆 = 𝑜𝑟, 20𝜆 = 𝑙𝑜𝑔𝑒 2 = 0.693
2 2

0.693
𝑜𝑟, 𝜆 =
20

1
The relaxation time, 𝜏 =
2𝜆

20
∴𝜏= = 14.44 𝑠𝑒𝑐 (Ans.)
2×0.693

(iii) The quality factor,

𝜔 10𝜋×20
𝑄= = = 453.7 (Ans.)
2𝜆 2×0.693

Problem: For the damped oscillatory system as shown in figure, the block has a
𝑑𝑥
mass of 1.5 kg and spring constant is 8 N/m. The damping force is given by −𝑏 ,
𝑑𝑡
where b=230 g/s. The block was pulled down 12 cm and released.

(a) Calculate the time required for the amplitude of the resulting oscillations
falls to one-third of its initial value.

(b) How many oscillations are made by the block in this time?
 Solution: (a)

𝑎 = 𝑎0 𝑒 −𝜆𝑡
𝑎0
𝑜𝑟, = 𝑎0 𝑒 −𝜆𝑡
3
1 −𝜆𝑡 −
𝑏𝑡
𝑜𝑟, = 𝑒 =𝑒 2𝑚
3
𝑏𝑡
𝑜𝑟, − = −𝑙𝑛3
2𝑚
2𝑚 × 𝑙𝑛3 2 × 1.5𝑘𝑔
𝑜𝑟, 𝑡 = = 𝑙𝑛3 = 14.3 𝑠 (𝐴𝑛𝑠. )
𝑏 0.230 𝑘𝑔Τ𝑠

(b) The angular frequency is,

2
𝑘 𝑏2 8 𝑁Τ𝑚 (.230 𝑘𝑔Τ𝑠)
𝜔 =√ −
′
=√ − = 2.31 rad/s
𝑚 4𝑚2 1.5𝑘𝑔 4(1.5𝑘𝑔)2

2𝜋
The time period is, 𝑇 = = 2.72 𝑠 and the number of oscillations is,
𝜔′
𝑡 14.3 𝑠
= = 5.27
𝑇 2.72 𝑠
∴ 5 complete oscillations are made by the block in this time.

FORCED OSCILLATIONS
Free Vibrations: Vibrations that occur in the absence of friction and external
forces after initial release of body. Example: Simple pendulum.

Forced Oscillations: When repeated force continuously acts on the system, the
vibrations are said to be forced vibrations. Example: Pendulum in a clock.
 UNDAMPED OSCILLATION DAMPED OSCILLATION

1. Free oscillations are oscillations in1. Forced oscillations are oscillations in

which a body or system oscillates with which the body oscillates with a
its own natural frequency without being frequency other than its natural
acted upon by an external force. frequency under the influence of an
external periodic force.
2. They occur due to the elastic forces 2. They occur due to the action of a
and inertia of the system. periodic force applied externally.

3. Free oscillations diminish gradually 3. Forced oscillations persist as long as

due to the resisting forces called the body is acted upon by an external
damping forces. force.
4. The frequency depends on the mass 4. The frequency is equal to the
and elasticity of the body. frequency of periodic force applied to
the body.

5. When you push a swing just once, it 5. If you push the swing each time it
oscillates at its own natural frequency, so slows down, it will continue to swing
it acts as a free oscillator. because external force is applied to it, so
it becomes a force oscillator.
 FORCED OSCILLATIONS

The amplitude of oscillation for a damped oscillatory motion of a body goes on

decreasing with time due to loss of energy to overcome the resistive forces. But if
an external periodic force is applied to the system to recover the losses, the
amplitude of oscillation does not decay with time and the body will oscillate
regularly under such periodic force. This type of vibration is known as forced
vibration.

If an external force 𝐹(𝑡) = 𝐹0 𝑒 𝑖𝑝𝑡 is applied to a damped oscillator, the motion of

a particle under the combined action of a linear restoring force, damping force
and time-dependent driving force becomes,
𝑑2𝑥 𝑑𝑥
𝑚 + 𝑘𝑥 + 𝑏 = 𝐹0 𝑒 𝑖𝑝𝑡
𝑑𝑡 2 𝑑𝑡

𝑑2𝑥 𝑏 𝑑𝑥 𝑘 𝐹0
𝑜𝑟, + + 𝑥= 𝑒 𝑖𝑝𝑡 ,
𝑑𝑡 2 𝑚 𝑑𝑡 𝑚 𝑚

𝑑2𝑥 𝑑𝑥
or, + 2𝜆 + 𝜔2 𝑥 = 𝑓𝑒 𝑖𝑝𝑡 , … … … … (1)
𝑑𝑡 2 𝑑𝑡

This equation (1) is a second order first degree differential equation for forced
vibration.
𝑏
Where, 2𝜆 = , damping constant
𝑚

𝑘
𝜔 = √ , Natural angular frequency
𝑚

𝐹0
𝑓= , amplitude of driving force per unit mass
𝑚

Let us consider a trial solution,

𝑥 = 𝐴𝑒 𝑖𝑝𝑡 … … … … … (2)
𝑑𝑥
or, = 𝑖𝑝𝐴𝑒 𝑖𝑝𝑡
𝑑𝑡

𝑑2𝑥
or, = −𝑝2 𝐴𝑒 𝑖𝑝𝑡
𝑑𝑡 2
Substituting these values in equation (1) we get,

−𝑝2 A𝑒 𝑖𝑝𝑡 + 2𝜆𝐴𝑖𝑝𝑒 𝑖𝑝𝑡 + 𝜔2 𝐴𝑒 𝑖𝑝𝑡 = 𝑓𝑒 𝑖𝑝𝑡

or, 𝐴[−𝑃2 + 2𝑖𝜆𝑝 + 𝜔2 ] = 𝑓

𝒇
or, 𝑨 = (𝝎𝟐 … … … … … …(3)
−𝒑𝟐 )+𝒊𝟐𝝀𝒑

Let, 𝜔2 − 𝑝2 = 𝐵𝑐𝑜𝑠𝜑 and, 2𝜆𝑝 = 𝐵𝑠𝑖𝑛𝜑

𝐵2 = 4𝜆2 𝑃2 + (𝜔2 − 𝑃2 )2
2𝜆𝑝
∴ 𝐵 = √4𝜆2 𝑃2 + (𝜔 2 − 𝑃2 )2 and 𝑡𝑎𝑛𝜑 =
𝜔2 −𝑝2

Substituting these values in equation (3),

𝑓 𝑓 𝑓𝑒 −𝑖𝜑
𝐴= = 𝑖𝜑
=
𝐵{𝑐𝑜𝑠𝜑+𝑖𝑠𝑖𝑛𝜑} 𝐵𝑒 √4𝜆2 𝑝2 +(𝜔2 −𝑝2 )2

Substituting these values in equation (2),

𝒇
𝒙= 𝒆𝒊(𝒑𝒕−𝝋) , … … … … … … (𝟒)
√𝟒𝝀𝟐 𝒑𝟐 +(𝝎𝟐 −𝒑𝟐 )𝟐

𝑝
Equation (4) represents a SHM of frequency , i.e., same as that of the driving
2𝜋
force, but lagging behind it in phase by 𝜑.

Since, Equation (1) is an inhomogeneous differential equation.

Hence, 𝑥 = 𝐴𝑒 𝑖𝑝𝑡 is not really a complete solution.

The solution will be complete if we add a complementary function which is a

solution of the related homogeneous equation,
𝑑2𝑥 𝑑𝑥
+ 2𝜆 + 𝜔2 𝑥 = 0
𝑑𝑡 2 𝑑𝑡

Applying the boundary condition, another solution is obtained when 𝐹 = 0. This

corresponds to free vibrations.

In the case of free vibrations, the solution is,

𝑥 = 𝐶𝑒 −𝜆𝑡 𝑆𝑖𝑛(𝜔′ 𝑡 + 𝛿), where 𝐶 and 𝛿 are two constants depending on the
initial condition.

So, the general solution of equation (1) is,

𝒇
𝒙 = 𝑪𝒆−𝝀𝒕 𝑺𝒊𝒏(𝝎′ 𝒕 + 𝜹) + 𝒆𝒊(𝒑𝒕−𝝋) , … … … … … … … (𝟓)
√𝟒𝝀𝟐 𝒑𝟐 +(𝝎𝟐 −𝒑𝟐 )𝟐

𝜔′
The first part of R.H.S is the initial damped oscillation of frequency with its
2𝜋
amplitude decaying exponentially to zero.
𝑝
The second part of R.H.S is the forced vibration of frequency and a constant
2𝜋
amplitude 𝑨.

Resonance:

The increase in the amplitude when the driving force is close to the natural
frequency of oscillation, is called resonance.

The amplitude of the driven or forced oscillator is,

𝒇
𝑨= … … … … … …(1),
√(𝝎𝟐 −𝒑𝟐 )𝟐 +𝟒𝝀𝟐 𝒑𝟐

The value of amplitude will be maximum when the denominator has minimum
value. i.e., when,
𝑑
[(𝜔2 − 𝑝2 )2 + 4𝜆2 𝑝2 ] = 0
𝑑𝑝

Or, 2(𝜔2 − 𝑝2 )(−2𝑝) + 8𝜆2 𝑝 = 0

Or, −4𝑝[(𝜔2 − 𝑝2 ) − 2𝜆2 ] = 0

Or, (𝜔2 − 𝑝2 ) − 2𝜆2 = 0 [Since 𝑝 ≠ 0]

Or, 𝑝2 = 𝜔2 − 2𝜆2 , … … … (2)

∴ 𝑝 = √𝜔 2 − 2𝜆2 = 𝑝0 [∵ 𝜔2 > 2𝜆2 ]

Cyclic resonance frequency,

 √𝜔2 −2𝜆2
𝑓𝑟 = , for which 𝐴 = 𝐴𝑚𝑎𝑥
2𝜋

𝑓
𝐴𝑚𝑎𝑥 = [Substituting 𝑝2 from equation (2) to equation (1)]
2𝜆√𝑝2 +𝜆2

So, 𝐴𝑚𝑎𝑥 depend upon the damping factor 𝜆.

Sharpness of Resonance:

Sharpness of resonance refers to the fall in amplitude with a change in frequency

on each side of the maximum amplitude.

(i) For low damping constant 𝜆, 𝑝 = 𝜔

𝑓
𝐴𝑚𝑎𝑥 = , and resonance curve is sharper.
2𝜆𝑝

𝜆 = 0.2

𝜆 = 0.5

𝜆=1

𝑝<𝜔 𝑝=𝜔 𝑝>𝜔

(ii) For large damping constant𝜆, the resonance curve is flat.

(iii) When the frictional force is absent (i.e., 𝜆 = 0), resonance is infinite.

The features of forced harmonic oscillators apply to a variety of systems:

1. When you tune a radio, for example, you are adjusting its resonant
frequency so that it only oscillates to the desired station’s broadcast
(driving) frequency.
 2. Magnetic resonance imaging (MRI) is a widely used medical diagnostic tool
in which atomic nuclei (mostly hydrogen nuclei) are made to resonate by
incoming radio waves (on the order of 100 MHz).

3. A child on a swing is driven by a parent at the swing’s natural frequency to

achieve maximum amplitude.

The destructive effects of forced harmonic oscillations:

In 1940, the Tacoma Narrows Bridge in Washington state collapsed due to heavy
cross wind. Heavy cross winds drove the bridge into oscillations at its resonance
frequency.

Before After [Source: Internet]

The bridge vibrates with larger amplitude and collapses due to resonance.

Quality factor: The quality factor, 𝑄, also referred to as the figure of merit, of a
harmonic oscillator is defined as the ratio of the response (or amplitude) of the
oscillator when driving frequency is equal to the resonance frequency to the
response when the driving frequency is zero or negligibly small. It is a
dimensionless quantity.

𝑓 Τ2𝜆𝜔 𝜔
∴𝑄= =
𝑓 Τ𝜔 2 2𝜆

The quality factor measures the quality of a harmonic oscillator. The less the
damping, the better the quality of the harmonic oscillator.
TWO BODY OSCILLATIONS
In a two-body oscillation, a spring connects two objects, each of which is free to
move. When the objects are displaced and released, they both oscillate.

Many examples of two-body oscillations are found in nature. In the microscopic

world, many objects, such as nuclei, atoms, molecules, etc., execute oscillations
that are approximately SHM.

Diatomic molecules: In a diatomic molecule, two atoms are bonded together with
a force. Above absolute zero temperature, the atoms vibrate continuously about
their equilibrium positions.

We can compare such a molecule with a system where the atoms can be
considered as two particles with different masses connected by a spring.

The relative motion can be represented by the oscillation of a single body having a
reduced mass.

Let the molecules be represented by two masses, m1 and m2, connected to each
other by a spring of force constant k, as shown in Fig. (a).

The motion of the system can be described in terms of the separate motions of
the two particles, which are located relative to the origin O by the two
coordinates x1 and x2 in Fig. (a).

The relative separation (x1 - x2) gives the length of the spring at any time. The
unstretched length of the spring is L.

The change in length of the spring is given by,

x = (x1 - x2) – L (1)

The magnitude of the force that the spring exerts on each particle is,

𝐹 = −𝑘𝑥, … … … … … … … … … … (2)

If the spring exerts a force -F⃗ on m1, then it exerts a force F⃗⃗⃗ on m2.
Taking the force component along the X-axis, let us apply Newton’s 2nd law of
motion separately to the two particles,
𝑑 2 𝑥1
𝑚1 = −𝑘𝑥 … … … … … … … … (3)
𝑑𝑡 2

𝑑 2 𝑥2
𝑚2 = 𝑘𝑥 … … … … … … … … …(4)
𝑑𝑡 2

Multiplying equation (3) by m2 and equation (4) by m1

𝑑 2 𝑥1
𝑚1 𝑚2 = −𝑚2 𝑘𝑥, … … … … …(5)
𝑑𝑡 2

𝑑 2 𝑥2
𝑚1 𝑚2 = 𝑚1 𝑘𝑥, … … … … … …(6)
𝑑𝑡 2

Subtracting, equation (6) from equation (5),

𝑑 2 𝑥1 𝑑 2 𝑥2
𝑚1 𝑚2 − 𝑚1 𝑚2 = −𝑚2 𝑘𝑥 − 𝑚1 𝑘𝑥
𝑑𝑡 2 𝑑𝑡 2

𝑑2
⇒ 𝑚1 𝑚2 (𝑥1 − 𝑥2) = −𝑘𝑥(𝑚1 + 𝑚2 )
𝑑𝑡 2

𝑚1𝑚2 𝑑2
⇒ (𝑥1 − 𝑥2) = −𝑘𝑥, … … … … … …(7)
(𝑚1 + 𝑚2 ) 𝑑𝑡 2
 𝑚1𝑚2
The quantity has the dimension of mass. This quantity is known as the
(𝑚1 + 𝑚2 )
reduced mass of the system and it is denoted by μ.
𝑚1𝑚2
𝜇= , … … … … … … … …(8)
(𝑚1 + 𝑚2 )

Reduced mass of a system is always smaller than either of the masses of the
system. (μ<m1 and μ<m2)

Since the un-stretched length of the spring is constant, the derivative of (x1-x2) is
the same as the derivative of x.
𝑑2 𝑑2 𝑑2𝑥
(𝑥1 − 𝑥2) = (𝑥 + 𝐿) = [from equation (1)]
𝑑𝑡2 𝑑𝑡2 𝑑𝑡2

So, from equation (7) we get,

𝑑2𝑥
𝜇 = −𝑘𝑥
𝑑𝑡 2

𝑑2𝑥 𝑘
⇒ + 𝑥 = 0, … … … … … … … … (9)
𝑑𝑡 2 𝜇

𝑘 𝜇
Here, angular frequency is, ω = √ ; So, time period, T=2𝜋√
𝜇 𝑘

Equation (9) is identical to the differential equation of SHM of a single body

oscillator.

Problem-1: Two blocks of masses 12 kg and 6 kg are connected by a spring of

force constant 100𝜋 2 N/m. The system is placed on a frictionless horizontal
surface. Initially, the spring is at rest. Now, the blocks are pulled apart by a small
distance and released. Find the time period of the resultant oscillation.

Solution: Given, 𝑚1 = 12𝑘𝑔, 𝑚2 = 6𝑘𝑔, 𝑘 = 100𝜋 2 𝑁/𝑚

𝑚1 𝑚2 12×6
We know, Reduced mass, 𝜇 = = = 4𝑘𝑔
𝑚1 +𝑚2 12+6

𝜇
𝑜𝑟, 𝑇 = 2𝜋√
𝑘
 4
= 2𝜋√
100𝜋2

= 0.4sec (Ans.)

Problem: How much energy must the shock absorbers of a 1200-kg car dissipate
in order to damp a bounce that initially has a velocity of 0.8 𝑚/𝑠 at the
equilibrium position? Assume the car returns to its original vertical position.

Solution: From the periodic motion we know that, at the equilibrium position the
velocity of the car is maximum.
21
The kinetic energy, 𝐾 = 𝑚𝑣𝑚𝑎𝑥
2

Where, 𝐾 is the kinetic energy of the car

𝑣𝑚𝑎𝑥 is the maximum velocity of the car at the equilibrium

𝑚 is the mass of the car

Given, 𝑚 = 1200 kg, 𝑣𝑚𝑎𝑥 = 0.8 m/s

1
𝑘 = × 1200 × 0.82 = 384 J
2

So, the energy required to stop the car equals to its kinetic energy.

Simple harmonic oscillation in LC circuit

Simple harmonic oscillation in an electrical system

• The capacitor (C) gets charged upon pressing key S.

• C discharges through the inductance coil on releasing S.

𝑑𝑄
• Magnetic flux (ɸ) increases due to the current, I=
𝑑𝑡

𝑑𝐼
• ɸ induces emf (-L ), which opposes the growth of current.
𝑑𝑡

• L opposes both growth and decay of current in the circuit.

 𝑄
• Voltage drop across the capacitor =
𝐶

Since there is no external emf in the circuit (the battery being cut-off), the net emf
in the circuit is,
𝑄 𝑑𝐼
+L =0 (From Kirchhoff’s law)
𝐶 𝑑𝑡

𝑄 𝑑𝐼
⇒ + =0
𝐿𝐶 𝑑𝑡

𝑄 𝑑2𝑄
⇒ + =0
𝐿𝐶 𝑑𝑡2

𝑑2𝑄 1 1
⇒ + 𝜔2 𝑄 = 0 (Where, 𝜔2 = or, 𝜔 = )
𝑑𝑡2 𝐿𝐶 √𝐿𝐶

This equation is similar to the differential equation of SHM with y replaced by Q,

1
m replaced by L and k replaced by .
𝐶

ɛ = emf of the source

R =resistance

S= switch

L = inductance

C =capacitance

Q= charge on each plate of C

Hence, the solution can be written as,

Q= Qo sin(𝜔t+φ)
Qo =𝑎𝑚𝑝𝑙𝑖𝑡𝑢𝑑𝑒 𝑜𝑓 𝑐ℎ𝑎𝑟𝑔𝑒

The frequency of variation of charge between +Qo to –Qo is,

1
n=
2𝜋√𝐿𝐶

Time period, T= 2𝜋√𝐿𝐶

𝑑𝑄
I= = Qo 𝜔cos(𝜔t+φ);
𝑑𝑡

Since maximum value of cos(𝜔t+𝜑)=1

Maximum current, 𝐼𝑜=𝑄𝑜𝜔

𝐼=𝐼𝑜 cos(𝜔t+𝜑)

Assignment:

Problem: A 1.6kg block on a horizontal surface is attached to a spring with a force

constant of 1.0 × 103 N/m. The spring is compressed a distance of 2cm, and the
block is released from rest. (i) Calculate the speed of the block as it passes
through the equilibrium position, x=0, if the surface is frictionless. (ii) ) Calculate
the speed of the block as it passes through the equilibrium position if a constant
frictional force of 4.0 N retards its motion.(iii) How far does the block travel before
coming to rest in part (b)?

Hints:

Fig. (c) Initially, when the block is compressed against the spring, energy is stored
1
in the form of elastic potential energy. ∴ 𝑃𝐸𝑆 = 𝑘𝑥 2
2

Fig. (b) When the spring is in equilibrium position, the spring is relaxed, therefore,
1
there is no potential energy but only kinetic energy. 𝐾𝐸𝑠 = 𝑚𝑣 2
2