1 PHY-101 (Waves & Oscillations) Waves & Oscillations

Course Teacher: Sahadat Jaman

Periodic Motion:

If a body moves in such a way that it crosses a certain point from the same direction after a certain period
of time, the motion is called periodic motion.

Earths motion around the sun, motion of hands (hour, minute) of a

clock, motion of planets are examples of periodic motion.

Oscillation:

A periodic back and forth motion when the body moves half of its time period in one direction and the
other half of its time period in the opposite direction, is called oscillation.

Motion of simple pendulum, vibration of tuning fork, motion

of a spring etc. are examples of oscillation.
2 Waves & Oscillations

Simple Harmonic Motion:

If the oscillation is such that the oscillating body experiences a restoring force when it is displaced from
its equilibrium position and the restoring force is proportional to the displacement, then the motion is
called Simple Harmonic Oscillation, and the system is called Simple Harmonic Oscillator.

For example motion of a spring is a SHM, oscillations of a floating cylinder, swinging of child on
playground, simple pendulum, an electron in a wire carrying AC etc.

Differential Equation of SHM:

If 𝑭 be the restoring force acting on a particle of mass 𝒎 executing in SHM at any time 𝒕 and 𝒙
is its displacement from the equilibrium position, then according to the definition of SHM we
have,

𝐹 ∝ −𝑥

∴ 𝐹 = −𝑘𝑥 …….. (1)

Where, 𝑲 is a proportional constant known as

force constant.

According to the Newton’s 2nd law of motion,

𝐹 = 𝑚𝑎
𝑑2 𝑥
∴ 𝐹 = 𝑚 𝑑𝑡 2 ……… (2)

𝑑2 𝑥
Where, 𝑎 = = acceleration of the particle.
𝑑𝑡 2

So, combining Equations (1) and (2) we get,

𝑑2 𝑥
𝑚 𝑑𝑡 2 = −𝑘𝑥

𝑑2 𝑥
⇒ 𝑚 𝑑𝑡 2 + 𝑘𝑥 = 0

𝑑2 𝑥 𝑘
∴ 𝑑𝑡 2
+ 𝑚 𝑥 = 0 ……………. (3)

This is the differential equation of SHM. The general solution of this equation can be considered as,

𝑥 = 𝑥𝑚 cos(𝜔𝑡 + 𝜙)
3 Waves & Oscillations

Physical significance of 𝝎:

The solution of the differential equation of SHM is,

𝑥 = 𝑥𝑚 cos(𝜔𝑡 + 𝜙) …………… (4)

Differentiating with respect to 𝑡 we get,

𝑑𝑥
= −𝜔𝑥𝑚 sin(𝜔𝑡 + 𝜙)
𝑑𝑡
Differentiating again,

𝑑2 𝑥
𝑑𝑡 2
= −𝜔2 𝑥𝑚 cos(𝜔𝑡 + 𝜙) ………… (5)

Putting these value in equation (3) we get,

𝑘
−𝜔2 𝑥𝑚 cos(𝜔𝑡 + 𝜙) + 𝑥 cos(𝜔𝑡 + 𝜙) = 0
𝑚 𝑚
𝑘
(−𝜔2 + ) 𝑥 cos(𝜔𝑡 + 𝜙) = 0
𝑚 𝑚
𝑘
𝜔2 =
𝑚

𝑘
𝜔=√
𝑚

𝑘
i.e. if we chose the constant 𝜔 = √𝑚, then equation (4) is in fact a solution of the differential equation of
SHM.
2𝜋
Now, if we increase the time 𝑡 by 𝜔
in equation (4), the function becomes,

2𝜋
𝑥 = 𝑥𝑚 cos [𝜔 (𝑡 + ) + 𝜙]
𝜔

𝑥 = 𝑥𝑚 cos(𝜔𝑡 + 2𝜋 + 𝜙)

𝑥 = 𝑥𝑚 cos(𝜔𝑡 + 𝜙)
2𝜋 2𝜋
That is, the function merely repeats itself after a time 𝜔
. Therefore 𝜔
is the period of the motion.

2𝜋
𝑇=
𝜔
𝑚
𝑇 = 2𝜋√
𝑘
4 Waves & Oscillations

The frequency 𝑓 of the oscillator is,

1 𝜔
𝑓= =
𝑇 2𝜋

⇒ 𝜔 = 2𝜋𝑓

The quantity 𝜔 differs from the frequency 𝑓 by a factor of 2𝜋. Its unit is 𝑟𝑎𝑑𝑖𝑎𝑛/𝑠𝑒𝑐. This quantity is
called angular frequency in SHM.

Various Parameters of Oscillator:

Amplitude:

It is the maximum displacement from the mean position. The displacement will be maximum when
cos(𝜔𝑡 + 𝜙) = ±1. Hence,

Maximum displacement = ±𝑥𝑚

So, the amplitude of the oscillator is

+𝑥𝑚 𝑎𝑛𝑑 − 𝑥𝑚 .

Displacement:

𝑥 = 𝑥𝑚 cos(𝜔𝑡 + 𝜙)

Velocity:

We know the displacement of a particle executing in SHM at any time t is given by,

𝑥 = 𝑥𝑚 𝑐𝑜𝑠(𝜔𝑡 + 𝜑)
5 Waves & Oscillations

𝑑𝑥 𝑑
Hence velocity, 𝑣= = 𝑑𝑡 {𝑥𝑚 𝑐𝑜𝑠(𝜔𝑡 + 𝜑)}
𝑑𝑡

= −𝜔𝑥𝑚 𝑠𝑖𝑛(𝜔𝑡 + 𝜑)

= −𝜔𝑥𝑚 √1 − 𝑐𝑜𝑠 2 (𝜔𝑡 + 𝜑)

2 − 𝑥 2 𝑐𝑜𝑠 2 (𝜔𝑡 + 𝜑)
= −𝜔√𝑥𝑚 𝑚

∴ 𝒗 = −𝝎√𝒙𝟐𝒎 − 𝒙𝟐 .

𝒐𝒓, |𝒗| = 𝝎√𝒙𝟐𝒎 − 𝒙𝟐

This relation clearly shows that the speed is maximum at the equilibrium position (𝑥 = 0) and is zero at
the maximum displacements (𝑥 = ±𝑥𝑚 ).

Graphical representation of, 𝑣 = −𝜔𝑥𝑚 sin(𝜔𝑡 + 𝜙)

Acceleration:
𝑑𝑣
The acceleration of a particle executing in SHM at any time t is given by, 𝑎 = 𝑑𝑡

𝑑 𝑑𝑥
= 𝑑𝑡 ( 𝑑𝑡 )

𝑑
= (−𝜔𝑥𝑚 𝑠𝑖𝑛(𝜔𝑡 + 𝜑))
𝑑𝑡

= −𝜔2 𝑥𝑚 𝑐𝑜𝑠(𝜔𝑡 + 𝜑)

∴ 𝒂 = −𝝎𝟐 𝒙.

Graphical representation of 𝑎 = −𝜔2 𝑥𝑚 cos(𝜔𝑡 + 𝜙)

6 Waves & Oscillations

Time period:

Time taken for one complete oscillation of the particle executing SHM is called time period. It is denoted
by T.

We know the solution of DE of SHM is 𝑥 = 𝐴 𝑐𝑜𝑠 (𝜔𝑡 + 𝜙)

2𝜋 2𝜋
If the time is increased by 𝜔
, then 𝑥 = 𝐴 𝑐𝑜𝑠[ 𝜔(𝑡 + 𝜔
)+ 𝜙)]

𝑥 = 𝐴 𝑐𝑜𝑠 (𝜔𝑡 + 2𝜋 + 𝜙)

𝑥 = 𝐴 𝑐𝑜𝑠 {2𝜋 + (𝜔𝑡 + 𝜙)}

∴ 𝑥 = 𝐴 𝑐𝑜𝑠 (𝜔𝑡 + 𝜙)
2𝜋 2𝜋
So it is seen that after time 𝜔
, the displacement of the particle becomes same. So 𝜔
is the time period of
SHM.

2𝜋 𝑚
∴𝑇= 𝜔
= 2𝜋√ 𝐾

Frequency:

Total number of oscillations completed by an oscillating particle in one second is called its frequency.
The frequency of a particle executing in SHM with time period T is given by,
1
𝑓=𝑇

1
𝑓= 2𝜋
𝜔

𝜔
𝑓 = 2𝜋
7 Waves & Oscillations

𝟏 𝒌
∴ 𝒇 = 𝟐𝝅 √𝒎

Phase:

The term (𝜔𝑡 + 𝜙) is called the phase of oscillator. It determines the displacement as well as the direction
of motion for simple harmonic oscillator.

Phase constant:

The constant 𝜙 in the solution of SHM is called phase constant. This constant define the initial state (i.e.
the position at 𝑡 = 0) of the particle undergoing SHM.

At 𝑡 = 0,

𝑥 = 𝑥𝑚 𝑐𝑜𝑠𝜙

If 𝜙 = 0, 𝑥 = 𝑥𝑚

It means that the simple harmonic oscillation begins at the maximum displacement.
𝜋
If 𝜙 = 2
𝑥=0

i.e. the oscillations begins from the equilibrium position.

8 Waves & Oscillations

Problems:

1. A particle is oscillating with simple harmonic motion of amplitude 15cm and frequency 4 Hz.
Compute (i) the maximum values of acceleration and velocity, (ii) the acceleration and velocity when
the displacement is 9cm.
2. For a particle executing SHM the displacement is 8 cm at that instant the velocity is 6 cm/s and the
displacement is 6 cm at that instant the velocity is 8 cm/s. Calculate (i) amplitude(A), (ii) frequency(f)
and (iii) time period(T).
3. The amplitude and frequency of an object executing Simple harmonically are 0.01m and 12Hz
respectively. What is the velocity of the object at displacement 0.005m? What is the maximum
velocity of the object?
4. A particle executing SHM has amplitude 3cm and maximum velocity 6.24cm/s, what is the time
period of the particle?
5. The motion of a particle in SHM is given by 𝑥 = 𝑎𝑠𝑖𝑛𝜔𝑡. If it has a speed u when the displacement is
𝑥1 and speed v when the displacement is 𝑥2 , show that the amplitude of the motion is 𝑎 =
𝑣 2 𝑥12 −𝑢2 𝑥22
√( )
𝑣 2 −𝑢2
6. A particle performs SHM is given by the equation 𝑦 = 20 𝑆𝑖𝑛(𝜔𝑡 + 𝛿) , if the time period is
30 𝑠𝑒𝑐𝑜𝑛𝑑𝑠 and the particle has a displacement of 10𝑐𝑚 at 𝑡 = 0, Find (i) epoch, (ii) the phase angle
at 𝑡 = 5 𝑠𝑒𝑐𝑜𝑛𝑑𝑠 and (iii) the phase difference between two positions of the particle 15 seconds
apart.
7. A spring is hung vertically, is found to be stretched by 0.02𝑚 from its equilibrium position when a
force 4𝑁 acts on it. Then a 2kg is attached to the end of the spring and is pulled 0.04𝑚 from its
equilibrium position along the vertical line. The body is then released and it executes SHM.
a) What is the force constant of the spring?
b) What is the force executed by the spring on the 2kg body just before it is released?
c) What is the period and frequency of oscillation after released?
d) What is the amplitude of oscillation?
e) What is the maximum velocity of the oscillating body?
f) What is the mechanical energy of the oscillating system?

Mechanical energy (E) of a particle executing SHM:

A particle executing in SHM possesses both potential and kinetic energies due to its elevation and motion,
respectively. The mechanical energy (𝐸) is the sum of potential (𝑈) and kinetic (𝐾) energy i.e.

𝐸 = 𝑈 + 𝐾 …………….. (1)

Potential Energy (U):

The amount of work done for displacement 𝑥 against the restoring force will remain as potential energy in
the object.
For very small displacement 𝑑𝑥, the amount of work done against the restoring force𝑖𝑠
𝑑𝑊 = 𝐹𝑑𝑥
= 𝑘𝑥𝑑𝑥
9 Waves & Oscillations

𝑥
Total work done for displacement𝑥, ∴ 𝑊 = ∫0 𝐹⃗ . 𝑑𝑥⃗
𝑥 𝑥 𝑥
𝑥2 1
∴ 𝑈 = 𝑊 = ∫ 𝑘𝑥𝑑𝑥 = 𝑘 ∫ 𝑥𝑑𝑥 = 𝑘 [ ] = 𝑘𝑥 2
2 0 2
0 0
Let the displacement of a particle executing in SHM at any instant 𝑡 is 𝑥 which is given by,
𝑥 = 𝑥𝑚 𝑐𝑜𝑠(𝜔𝑡 + 𝜑) …………… (2)
Now, from (2) and (4) we get,
2 1
𝑈 = 2 𝑘𝑥𝑚 𝑐𝑜𝑠 2 (𝜔𝑡 + 𝜑) …… (3)

Kinetic energy (K):

If the velocity of the particle is v, then kinetic energy is given by,
1
𝐾 = 2 𝑚𝑣 2
1 2
= 𝑚(−𝜔𝑥𝑚 𝑠𝑖𝑛(𝜔𝑡 + 𝜑)) [𝑎𝑠, 𝑣 = −𝜔𝑥𝑚 𝑐𝑜𝑠(𝜔𝑡 + 𝜑)]
2
1
= 2 𝑚𝜔2 𝑥𝑚2
𝑠𝑖𝑛2 (𝜔𝑡 + 𝜑)
1 𝑘 2 𝑘
= 𝑚. 𝑥𝑚 𝑠𝑖𝑛2 (𝜔𝑡 + 𝜑) [as, 𝜔2 = ]
2 𝑚 𝑚
1 2
∴ 𝐾 = 2 𝑘𝑥𝑚 sin2 (𝜔𝑡 + 𝜙)………………. (6)

Total Energy (E):

Now, adding (5) and (6) we have the mechanical energy from (1) as follows,
1 2 1
𝐸 = 2 𝑘𝑥𝑚 𝑐𝑜𝑠 2 (𝜔𝑡 + 𝜑) + 2 𝑘𝑥𝑚
2
𝑠𝑖𝑛2 (𝜔𝑡 + 𝜑)
1 2 [𝑐𝑜𝑠 2 (𝜔𝑡
= 2 𝑘𝑥𝑚 + 𝜑) + 𝑠𝑖𝑛2 (𝜔𝑡 + 𝜑)]
𝟏
∴ 𝑬 = 𝟐 𝒌𝒙𝟐𝒎 .

Therefore, the mechanical energy of a particle executing SHM is constant or conserved, because both 𝑘
and 𝑥𝑚 constant for that particle.
10 Waves & Oscillations

Average Kinetic Energy:

The kinetic energy of SHM at any instant is given by,
1 2
𝐾 = 𝑘𝑥𝑚 𝑠𝑖𝑛2 (𝜔𝑡 + 𝜑)
2
𝑻
∫ 𝑲 𝒅𝒕
𝑵𝒐𝒘 𝑲𝒂𝒗𝒈 = 𝟎
𝑻
𝑻
𝟏 1 2
= ∫ 𝑘𝑥𝑚 𝑠𝑖𝑛2 (𝜔𝑡 + 𝜙)𝒅𝒕
𝑻 2
𝟎
𝑇
2
𝑘𝑥𝑚
= ∫{2𝑠𝑖𝑛2 (𝜔𝑡 + 𝜙)}𝑑𝑡
4𝑇
0
𝑇
2
𝑘𝑥𝑚
= ∫{1 − Cos 2(𝜔𝑡 + 𝜙)}𝑑𝑡
4𝑇
0
2 𝑇 2 𝑇
𝑘𝑥𝑚 𝑘𝑥𝑚
= ∫ 𝑑𝑡 − ∫ cos(2𝜔𝑡 + 2𝜙) 𝑑𝑡
4𝑇 0 4𝑇 0
2 2 𝑇
𝑘𝑥𝑚 𝑇
𝑘𝑥𝑚 sin(2𝜔𝑡 + 2𝜙)
= [𝑇]0 − [ ]
4𝑇 4𝑇 2𝜔 0
2 2
𝑘𝑥𝑚 𝑘𝑥𝑚
= .𝑇 − {𝑠𝑖𝑛(2𝜔𝑇 + 2𝜙) − 𝑠𝑖𝑛2𝜙}
4𝑇 8𝜔𝑇
2 2
𝑘𝑥𝑚 𝑘𝑥𝑚 2𝜋
= − {𝑠𝑖𝑛 (2. 𝑇 + 2𝜙) − 𝑠𝑖𝑛2𝜙 }
4 8𝜔𝑇 𝑇
2 2
𝑘𝑥𝑚 𝑘𝑥𝑚
= − {𝑠𝑖𝑛(4𝜋 + 2𝜙) − 𝑠𝑖𝑛2𝜙}
4 8𝜔𝑇
2 2
𝑘𝑥𝑚 𝑘𝑥𝑚
= + {𝑠𝑖𝑛2𝜙 − 𝑠𝑖𝑛2𝜙 }
4 8𝜔𝑇
𝒌𝒙𝟐𝒎
∴< 𝑲 >=
𝟒

Average potential energy:

The potential energy 𝑈 of the particle executing SHM at a time t is given by,
2 1
𝑈 = 2 𝑘𝑥𝑚 𝑐𝑜𝑠 2 (𝜔𝑡 + 𝜑)
Hence, the average potential energy < 𝑈 > of the particle over a complete cycle or a whole time period
(T) is given by,
1 𝑇
< 𝑈 >= 𝑇 ∫0 𝑈𝑑𝑡
1 2 𝑇
= 𝑇 ∫0 {𝑘𝑥𝑚 𝑐𝑜𝑠 2 (𝜔𝑡 + 𝜑)}𝑑𝑡
𝑘𝑥𝑚2 𝑇
= 4𝑇 0
∫ {2𝑐𝑜𝑠 2 (𝜔𝑡 + 𝜑)}𝑑𝑡
𝑘𝑥𝑚2 𝑇
= 4𝑇 0
∫ {1 + cos 2(𝜔𝑡 + 𝜑)}𝑑𝑡
𝑘𝑥𝑚2 1 𝑇 1 𝑇
= 4
{𝑇 ∫0 𝑑𝑡 + 𝑇 ∫0 cos 2(𝜔𝑡 + 𝜑) 𝑑𝑡}
But the average value of both a sine and cosine function over a complete cycle or a whole time period (T)
is zero.
11 Waves & Oscillations

2 1
𝑘𝑥𝑚
∴< 𝑈 >= 4
{𝑇 [𝑇]𝑇0 − 0}

2
𝑘𝑥𝑚
= .𝑇
4𝑇
1 1 2
= . 𝑘𝑥𝑚
2 2
𝟏
∴< 𝑈 >= 𝟐 . 𝑬(𝑴𝒆𝒄𝒉𝒂𝒏𝒊𝒄𝒂𝒍 𝒆𝒏𝒆𝒓𝒈𝒚)
Therefore, the average potential energy < 𝑈 > of a particle executing SHM is half of its mechanical
energy. Similarly, the average kinetic energy < 𝐾 > of a particle executing SHM is half of its
mechanical energy.

Lissajous Figures:
“When a particle is influenced simultaneously by two SHM at right angles to each other, the resultant
motion of the particle traces a curve. These curves are called Lissajous Figures. The shape of the curve
depends on the time period, Phase difference and amplitude of the two constituent SHM”

Compisition of Two SHM of Equal time periods acting at right angles:

Let 𝑥 = 𝑎 𝑆𝑖𝑛(𝜔𝑡 + 𝜙 ….…………………. (i)

And 𝑦 = 𝑏 𝑆𝑖𝑛𝜔𝑡 …………………….. (ii)

Represents the displacements of a particle along the 𝑋 𝑎𝑛𝑑 𝑌 axes due to the influences of two SHM
acting simultaneously on a particle in perpendicular directions. Here, the two vibrations are of the same
time period but are of different amplitudes and different phase angles.

From (𝑖𝑖) →
𝑦
𝑆𝑖𝑛𝜔𝑡 =
𝑏

By using (𝑖) →
𝑥
𝑆𝑖𝑛(𝜔𝑡 + 𝜙) =
𝑎
𝑥
∴ = 𝑆𝑖𝑛 𝜔𝑡 𝐶𝑜𝑠𝜙 + 𝐶𝑜𝑠𝜔𝑡 𝑆𝑖𝑛𝜙
𝑎
𝑥 𝑦
∴ = 𝐶𝑜𝑠𝜙 + 𝐶𝑜𝑠𝜔𝑡 𝑆𝑖𝑛𝜙
𝑎 𝑏
𝑥 𝑦
∴ = 𝐶𝑜𝑠𝜙 + √(1 − 𝑆𝑖𝑛2 𝜔𝑡) 𝑆𝑖𝑛𝜙
𝑎 𝑏
𝑥 𝑦
∴ − 𝐶𝑜𝑠𝜙 = √(1 − 𝑆𝑖𝑛2 𝜔𝑡) 𝑆𝑖𝑛𝜙
𝑎 𝑏

By squaring both sides →

𝑥 𝑦
∴ ( − 𝐶𝑜𝑠𝜙)2 = (1 − 𝑆𝑖𝑛2 𝜔𝑡) 𝑆𝑖𝑛2 𝜙
𝑎 𝑏
12 Waves & Oscillations

𝑥 2 𝑥 𝑦 𝑦 2 𝑦 2
∴ ( ) − 2 . 𝐶𝑜𝑠𝜙 + ( 𝐶𝑜𝑠𝜙) = {1 − ( ) } 𝑆𝑖𝑛2 𝜙
𝑎 𝑎 𝑏 𝑏 𝑏

𝑥2 𝑥 𝑦 𝑦2 𝑦2
∴ 2 − 2 . 𝐶𝑜𝑠𝜙 + 𝐶𝑜𝑠 𝜙 = 𝑆𝑖𝑛 𝜙 + 2 𝑆𝑖𝑛2 𝜙
2 2
𝑎 𝑎 𝑏 𝑏2 𝑏
2 2
𝑥 𝑥 𝑦 𝑦
∴ 2 − 2 . 𝐶𝑜𝑠𝜙 + (𝐶𝑜𝑠 2 𝜙 + 𝑆𝑖𝑛2 𝜙) = 𝑆𝑖𝑛2 𝜙
𝑎 𝑎 𝑏 𝑏2

𝑥2 𝑦2 𝑥 𝑦
∴ 2
+ 2 − 2 . 𝐶𝑜𝑠𝜙 = 𝑆𝑖𝑛2 𝜙
𝑎 𝑏 𝑎 𝑏

𝑥 2 𝑦 2 2𝑥𝑦
+ − 𝐶𝑜𝑠𝜙 = 𝑆𝑖𝑛2 𝜙 … … … … … … (𝑖𝑖𝑖)
𝑎2 𝑏 2 𝑎𝑏
This is the general equation of Lissajous figure in xy-plane.

Special Cases: Let us now consider some special cases.

(i) When 𝝓 = 𝟎 then equation (iii) becomes,

𝑥2 𝑦2 𝑥𝑦
2
+ 2−2 =0
𝑎 𝑏 𝑎𝑏
𝑥 𝑦 2
⇒( − ) =0
𝑎 𝑏
𝑥 𝑦
⇒ =
𝑎 𝑏
13 Waves & Oscillations

𝑏
∴𝑦= 𝑥
𝑎
𝑏 𝑏
This is the equation of a straight line with slope 𝑎 inclined to positive x-axis at an angle 𝜃 = tan−1 (𝑎)

(ii) When 𝝓 = 𝝅, then equation (iii) becomes,

𝑥2 𝑦2 𝑥𝑦
+ −2 (−1) = 0
𝑎2 𝑏 2 𝑎𝑏
𝑥 𝑦 2
⇒( + ) =0
𝑎 𝑏
𝑥 𝑦
⇒ + =0
𝑎 𝑏
𝑏
∴𝑦=− 𝑥
𝑎
𝑏 𝑏
This is the equation of a straight line with slope − 𝑎 inclined to negative x-axis at an angle 𝜃 = tan−1 𝑎

𝝅
(iii) When 𝜽 = 𝟐 , then equation (iii) becomes,

𝑥2 𝑦2 𝑥𝑦
+ − 2 (0) = 1
𝑎2 𝑏 2 𝑎𝑏

𝑥2 𝑦2
⇒ + =1
𝑎2 𝑏 2

This is the equation of an ellipse having major axis = 2𝑎 𝑖𝑓 (𝑎 > 𝑏) and minor axis = 2𝑏.

If 𝑎 = 𝑏, then 𝑥 2 + 𝑦 2 = 𝑎2 which is the equation of a circle of radius 𝑎 in the xy-plane.

𝝅
(iv) When 𝝓 = , then equation (iii) becomes
𝟒

𝑥2 𝑦2 𝑥𝑦 1 1
+ − 2 =
𝑎2 𝑏 2 𝑎 𝑏 √2 2

This is the equation of an oblique ellipse.

So,

“When a particle is subjected to two simple harmonic motion at right angle to each other the particle
may move along different paths. Such paths are called Lissajous figures.”
14 Waves & Oscillations

Torque: Torque is a measure of how much force acting on an object to cause rotation.

Moment of inertia:

A measure of body’s resistance to angular acceleration, equal to the product of the mass of the body and
the square of its distance from the axis of rotation.

Torsional Pendulum:

“A torsional pendulum consists of a disk suspended by a wire attached to the center of the mass of the
disk. The other end of the wire is fixed to a rigid support”.

The disk oscillates about the line 𝑂 with an amplitude 𝜃𝑚𝑎𝑥 . When the disk is twisted through some small
angle 𝜃 , the twisted wire exerts on the body a restoring torque i.e. the wire resists such twist by
developing restoring torque. We can write hence

𝜏 ∝ −𝜃

⇒ 𝜏 = −𝜅𝜃 ………… (1)

𝜅(𝑘𝑎𝑝𝑝𝑎) is called the torsional constant

According to angular form of Newton’s 2nd law:

𝜏 = 𝐼𝛼 …………… (2)

From (1) and (2)

𝐼𝛼 = −𝜅𝜃

𝑑2 𝜃 𝑑2 𝜃
⇒𝐼 = −𝜅𝜃 [𝑎𝑠, 𝛼 =
𝑑𝑡 2 𝑑𝑡 2

𝑑2 𝜃 𝜅
∴ + 𝜃 = 0 … … … … … (3)
𝑑𝑡 2 𝐼
Equation (3) represents angular SHM. This equation is equivalent to differential equation of linear SHM
given by,

𝑑2 𝑥 𝑘
+ 𝑥 = 0 … … … … … … … . (4)
𝑑𝑡 2 𝑚
Comparing (3) and (4)

𝑥 → 𝑐𝑜𝑟𝑟𝑒𝑠𝑝𝑜𝑛𝑑𝑠 𝑡𝑜 𝜃

𝑚 → 𝑐𝑜𝑟𝑟𝑒𝑠𝑝𝑜𝑛𝑑𝑠 𝑡𝑜 𝐼

𝑘 → 𝑐𝑜𝑟𝑟𝑒𝑠𝑝𝑜𝑛𝑑𝑠 𝑡𝑜 𝜅
15 Waves & Oscillations

Therefore, the solution of equation (3) can be written as:

𝜃 = 𝜃𝑚 cos(𝜔𝑡 + 𝜙)

Where 𝜃 is the maximum angular displacement and 𝜔 is the angular frequency.

𝜅
𝜔=√
𝐼

Time period: The time period is given by,

2𝜋 𝐼
𝑇= = 2𝜋√
𝜔 𝜅

Different Types of Oscillations:

Free Oscillations: When a body vibrates with its own natural frequency then it is said to be executing
free oscillations.
Examples:
a) Vibrations of tuning fork.
b) Vibrations in stretched string.
c) Oscillations of simple pendulum.

Damped Oscillations: Most of the oscillations in air or any

medium are damped. When an oscillation occurs, some kind of
damping force may arise due to friction or air resistance offered by
the medium. So a part of the energy is dissipated in overcoming the
resistive force. Consequently, the amplitude of oscillations
decreases with time and finally becomes zero. Such oscillations are
called damped oscillations.
Examples:
a) The oscillations of a pendulum.
b) Electromagnetic damping in galvanometer.
c) Electromagnetic oscillations in tank circuit.

Maintained Oscillations: The amplitude of an oscillating

system can be made to constant by feeding some energy to the
system. If energy is fed to the system to compensate the energy
it has lost, the amplitude will be a constant. Such oscillations
are called maintained oscillations. Examples: A swing to which
energy is continuously fed to the system to maintain amplitude
of oscillation.
16 Waves & Oscillations

Forced Oscillations: When a vibrating body is maintained in the state of vibration by a periodic force of
frequency other than its natural frequency of the body, then the vibrations are called forced vibrations.
Examples:
a) Sound boards of stringed instruments execute forced vibrations.
b) Press the stem of tuning fork against table, the table suffers forced vibrations.

Resonance: In the case of forced vibration if the frequency difference is small then the amplitude will be
large. Ultimately when two frequencies are same, the amplitude becomes maximum. This condition is
known as resonance.

Advantages:
a) Using resonance, frequency of a given tuning fork can be determined.
b) In radio-TV, using tank circuit, required frequency can be obtained.
Disadvantages:
a) Resonance can cause disaster in an earthquake, if the natural frequency of the
building matches the frequency of the periodic oscillations present in the earth.
The building begins to oscillate with large amplitude thus leading to a collapse.
b) A singer maintaining a note at a resonant frequency of a glass can cause it to shatter into pieces.

Damped Harmonic Oscillations:

In many real systems, dissipative forces such as friction and air resistance retard the motion.
Consequently the mechanical energy of the system diminishes with time and the motion is said to be
damped.

The damping force can be expressed as 𝐷 = −𝑏𝑣; where 𝑏 is a constant called the damping coefficient.

Again, the restoring force of the system is – 𝑘𝑥. So, we can write,

∑ 𝐹𝑥 = −𝑘𝑥 − 𝑏𝑣

𝑑2 𝑥 𝑑𝑥
∴𝑚 2
= −𝑘𝑥 − 𝑏
𝑑𝑡 𝑑𝑡

𝑑2 𝑥 𝑏 𝑑𝑥 𝑘
∴ 2+ + 𝑥=0
𝑑𝑡 𝑚 𝑑𝑡 𝑚

𝑑2 𝑥 𝑏 𝑑𝑥
∴ + + 𝜔02 𝑥 = 0 − − − − − − − (𝑖)
𝑑𝑡 2 𝑚 𝑑𝑡
This represents the Differential equation for damped oscillation. One example of
damped oscillator is a mass attached to a spring and submersed in a viscous liquid. The above equation is
in the form of a homogeneous 2nd order D.E. Let the trial solution is,

𝑥 = 𝑒 𝜆𝑡
17 Waves & Oscillations

𝑑𝑥
∴ = 𝜆 𝑒 𝜆𝑡
𝑑𝑡

𝑑2 𝑥
∴ = 𝜆2 𝑒 𝜆𝑡
𝑑𝑡 2
𝑏
Substituting these equations in (i), we get 𝜆2 𝑒 𝜆𝑡 + 𝑚
𝜆 𝑒 𝜆𝑡 + 𝜔02 𝑒 𝜆𝑡 = 0

𝑏𝜆
∴ 𝑒 𝜆𝑡 (𝜆2 + + 𝜔02 ) = 0
𝑚
𝑏𝜆 𝑘
∴ 𝑒 𝜆𝑡 (𝜆2 + + )=0
𝑚 𝑚

∴ 𝑚𝑒 𝜆𝑡 (𝑚𝜆2 + 𝑏𝜆 + 𝑘) = 0

We get following auxiliary equation, 𝑚𝜆2 + 𝑏𝜆 + 𝑘 = 0

−𝑏 ± √𝑏 2 − 4𝑚𝑘
⇒𝜆=
2𝑚

𝑏 𝑏 2 𝑘
∴𝜆=− ± √( ) − … … … (𝑖𝑖)
2𝑚 2𝑚 𝑚

𝑏 𝑏 2 𝑘 𝑏 𝑏 2 𝑘
Let 𝜆1 = − + √( ) − and 𝜆2 = − − √( ) −
2𝑚 2𝑚 𝑚 2𝑚 2𝑚 𝑚

Hence, the complete solution of equation (i) will be,

𝑥 = 𝐴𝑒 𝜆1 𝑡 + 𝐵𝑒 𝜆2 𝑡

It is clear that for oscillating motion, the damping force should be less than restoring force,

i.e. 𝑏 2 < 4𝑚𝑘

For this condition the general solution for underdamped

motion can be written as,
𝑏
𝑥 = 𝐴𝑒 −2𝑚𝑡 cos(𝜔𝑡 + 𝜙)

𝑘 𝑏 2
Where the angular frequency is 𝜔 = √𝑚 − (2𝑚)

𝑜𝑟, 𝜔 = √𝜔02 − 𝛽 2

𝑏
Where 𝜔𝑜 represents the angular frequency in the absence of retarding force. And 𝛽 = .
2𝑚
18 Waves & Oscillations

Three types of damping →

(a) Under-damped when 𝑏 2 − 4𝑚𝑘 < 0

(b) Critical damping when 𝑏 2 − 4𝑚𝑘 = 0
(c) Over damped when 𝑏 2 − 4𝑚𝑘 > 0

Problem:

1. For the damped oscillator 𝑚 = 250 𝑔𝑟𝑎𝑚, 𝑘 = 85 𝑁/𝑚 𝑎𝑛𝑑 𝑏 = 70 𝑔/𝑠.

(a) What is the time period of the motion? Answer: 0.34s
(b) How long does it take for the amplitude of the damped oscillations to drop to half its initial value?
Answer: 5.0s
(c) How long does it take for the ME to drop to one-half its initial value? Answer: 2.5s
2. A massless spring, suspended from a rigid support carries a mass of 500𝑔𝑚 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 the force constant and the relaxation time.

Forced Oscillations and Resonance:

A damped harmonic oscillator under the action of a periodic force 𝐹𝑝 = 𝐹𝑚 𝑠𝑖𝑛𝜔𝑡 is called forced
harmonic oscillator.
Let’s consider a damped harmonic oscillator under the action of periodic force 𝐹𝑚 𝑠𝑖𝑛𝜔𝑡, where 𝜔 is the
frequency of the applied force and 𝐹𝑚 is the maximum value of external force. The total force on the
oscillator is given as,
𝐹 = −𝑘𝑥 − 𝑏𝑣 + 𝐹𝑚 𝑠𝑖𝑛𝜔𝑡
𝑑2 𝑥
⇒𝑚 + 𝑏𝑣 + 𝑘𝑥 = 𝐹𝑚 𝑠𝑖𝑛𝜔𝑡
𝑑𝑡 2
𝑑2 𝑥 𝑏 𝑑𝑥 𝑘 𝐹𝑚
2
+ + 𝑥= 𝑠𝑖𝑛𝜔𝑡 … … … (1)
𝑑𝑡 𝑚 𝑑𝑡 𝑚 𝑚
This is the differential equation of forced harmonic oscillations. The solution of this can be written as,
𝐹𝑚
𝑥= sin(𝜔𝑡 + 𝜙) … … … (2)
𝐺
19 Waves & Oscillations

𝑏𝜔
−
Where, 𝐺 = √𝑚2 (𝜔02 − 𝜔 2 )2 + 𝑏 2 𝜔 2 and 𝜙 = tan−1 (𝜔2 −𝜔
𝑚
2)
0

Here, 𝜔0 is the natural frequency and 𝜔 is the driving frequency.

Resonance Phenomena:
Resonance is the increase in the amplitude of a vibrating body under the action of a periodic force whose
frequency is equal to the natural frequency of the body. So, for resonance 𝜔 = 𝜔0
The amplitude in equation (2) takes the form,
𝐹𝑚
𝐴𝑚𝑝𝑙𝑖𝑡𝑢𝑑𝑒 𝐴 =
𝐺
𝐹𝑚
𝐴=
√𝑚2 (𝜔02 − 𝜔 2 )2 + 𝑏 2 𝜔 2
𝐹𝑚
𝐴= [𝑎𝑠, 𝜔 = 𝜔0 ]
𝑏𝜔0
This shows that the amplitude of the oscillator
becomes infinity (∞) if there is no damping ((𝑏 = 0).
But there is always some damping. That’s why the
amplitude never becomes infinity.
In the figure shown, we have three curves for different
damping coefficients. These curves shows the
variation of amplitude with ratio 𝜔/𝜔0 . It is clear that,
𝜔
(i) When = 1 and damping is smallest,
𝜔0
then curve is sharp and amplitude is
maximum.
𝜔
(ii) When 𝜔 > 1 𝑜𝑟 < 1 and damping
0
increase, the amplitude decreases and
curve becomes wider.

Two-body Oscillations:

In a two body oscillations a spring connects two objects, each of which is free to move. When
the objects are displaced and released, they both oscillate.

Let, 𝑥1 𝑎𝑛𝑑 𝑥2 be the positions of masses 𝑚1 𝑎𝑛𝑑 𝑚2 respectively with respect to the origin, 𝑂.
The relative separation of the masses is 𝑥1 − 𝑥2 . If 𝐿 is the unstretched length of the spring then,

𝑥 = (𝑥1 − 𝑥2 ) − 𝐿 … … … … (𝑎)
20 Waves & Oscillations

is the change in length of the spring. As shown in the figure, if the spring exerts a force −𝐹⃗ on
𝑚1 , then it exerts a force +𝐹⃗ on 𝑚2 .

Let’s apply Newton’s second law separately to the two particles.

𝑑 2 𝑥1
𝑚1 = −𝑘𝑥 … … … … … (1)
𝑑𝑡 2
𝑑 2 𝑥2
𝑚2 = +𝑘𝑥 … … … … … (2)
𝑑𝑡 2
Multiplying (1) by 𝑚2 and (2) by 𝑚1 and then subtracting (2) from (1) we get,

𝑑2 𝑥1 𝑑 2 𝑥2
𝑚1 𝑚2 − 𝑚 𝑚
1 2 = −𝑚2 𝑘𝑥 − 𝑚1 𝑘𝑥
𝑑𝑡 2 𝑑𝑡 2
𝑑2 𝑥1 𝑑 2 𝑥2
𝑚1 𝑚2 ( 2 − ) = −(𝑚1 + 𝑚2 )𝑘𝑥
𝑑𝑡 𝑑𝑡 2

𝑑2
𝑚1 𝑚2 (𝑥 − 𝑥2 ) = −(𝑚1 + 𝑚2 )𝑘𝑥
𝑑𝑡 2 1
𝑚1 𝑚2 𝑑 2
(𝑥 − 𝑥2 ) = −𝑘𝑥 … … … … (3)
𝑚1 + 𝑚2 𝑑𝑡 2 1
𝑚 𝑚
The quantity 𝑚 1+𝑚2 has the dimension of mass and is known as the reduced mass, m.
1 2

𝑚1 𝑚2
𝑚=
𝑚1 + 𝑚2

From equation (a),

𝑥1 − 𝑥2 = 𝑥 + 𝐿
21 Waves & Oscillations

𝑑 𝑑
∴ (𝑥1 − 𝑥2 ) = (𝑥 + 𝐿)
𝑑𝑡 𝑑𝑡
𝑑 𝑑𝑥
⇒ (𝑥1 − 𝑥2 ) =
𝑑𝑡 𝑑𝑡
𝑑2 𝑑 𝑑𝑥
⇒ 2
(𝑥1 − 𝑥2 ) = ( )
𝑑𝑡 𝑑𝑡 𝑑𝑡

𝑑2 𝑑2𝑥
∴ 2 (𝑥1 − 𝑥2 ) = 2
𝑑𝑡 𝑑𝑡
So, the equation (3) takes the form,

𝑑2𝑥
𝑚 = −𝑘𝑥
𝑑𝑡 2
𝑑2𝑥 𝑘
∴ 2+ 𝑥=0
𝑑𝑡 𝑚
This equation is identical for single oscillating mass. Thus we can say that the system of two
masses can be replaced by a single particle as shown in figure below with a mass equal to the
reduced mass of the system.

And the time period will be,

𝑚
𝑇 = 2𝜋√ , 𝑤ℎ𝑒𝑟𝑒 𝑚 𝑖𝑠 𝑡ℎ𝑒 𝑟𝑒𝑑𝑢𝑐𝑒𝑑 𝑚𝑎𝑠𝑠.
𝑘