OSCILLATIONS

10/20/2023 1
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.
• Example: Earths motion around the sun, motion of
hands (hour, minute) of a clock, motion of planets are
examples of periodic motion.

10/20/2023 2
 OSCILLATION

A periodic back and Motion of simple

forth motion when pendulum, vibration
the body moves half of tuning fork,
of its time period in motion of a spring
one direction and etc. are examples of
the other half of its oscillation.
time period in the
opposite direction,
is called oscillation.

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 DIFFERENT TYPES OF
OSCILLATION

Oscillation

Simple Damped
Harmonic harmonic Free Maintained Forced
Motion motion Oscillation Oscillation Oscillation

10/20/2023 4
 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: Mechanical wave, Motion
of a spring, Oscillations of a liquid column in
a U-TUBE, Swinging of child on playground,
simple pendulum, an electron in a wire
carrying AC etc.

10/20/2023 5
CHARACTERISTICS OF SIMPLE
HARMONIC
MOTION/OSCILLATIONS

❑ Its motion is periodic

❑ At particular time interval, the motion
becomes opposite
❑ Its motion is along a straight line
❑ Its acceleration is proportional to the
displacement
❑ Acceleration is opposite to displacement
❑ Acceleration points toward the mean
position of the object.

10/20/2023 6
Various Parameters
of Oscillator
EQUILIBRIUM POSITION
An equilibrium position is a point where an oscillating
object experience zero (0) resultant forces.

COMPLETE OSCILLATION
An oscillation is said to be complete if vibrating or
oscillating object starting from a point returns to the
same point along the same direction.
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DISPLACEMENT
Let, a particle moves round a point O in a circular path of radius
A at an angular velocity ω. Let at time t, the particle is at the
position P. From P, a normal is drawn on the diameter DB. Here,
the displacement of the end point of the normal is x = ON
Now, from the figure
sin θ = ON/OP
ON = OP sin θ
x = A sin θ
Here, x is the displacement from the origin O and OP = A =
radius of the circle.
x = A sin ω t [since, θ = ω t]
Finally,
x = A sin 2𝜋 n t [ω = 2π/T and T = 1/n , T = t/N]
where, A is the amplitude of the particle which is the maximum
displacement from the mean position.

10/20/2023 8
 Graphical representation of Graphical representation of
𝑥 = 𝑥𝑚 cos 𝜔𝑡 + 𝜙 𝑦 = 𝑦𝑚 sin 𝜔𝑡 + 𝜙

10/20/2023 9
 PHASE
The state of motion of a vibrating
particle at any instant is called its
phase at that instant. State of
motion of a vibrating particle at
any instant is determined by its
displacement, velocity and
acceleration at that instant. It is
represented by 𝜔𝑡 + 𝜙 .

10/20/2023 10
Phase constant
/EPOCH
The initial phase/phase constant / epoch define the initial state ( i.e.,
the position at 𝑡=0) of the particle undergoing SHM.
(i) If we start counting time when the particle is in its mean position,
i.e., when 𝑥 = 0 at 𝑡 = 0
From equation 𝑥 = 𝑥𝑚 cos 𝜔𝑡 + 𝜙 we get 𝜙=𝜋/2
If we put 𝜙=𝜋/2 in equation 𝑥 = 𝑥𝑚 cos 𝜔𝑡 + 𝜙 , this equation reduces
to
𝜋
𝑥 = 𝑥𝑚 cos 𝜔𝑡 +
2
𝑥 = 𝑥𝑚 sin 𝜔𝑡
(ii) If we start counting time when the particle is in its extreme
position, i.e., when 𝑥 = 𝑥𝑚 at 𝑡 = 0
From equation 𝑥 = 𝑥𝑚 cos 𝜔𝑡 + 𝜙 we get 𝜙=0
If we put 𝜙=0 in equation 𝑥 = 𝑥𝑚 cos 𝜔𝑡 + 𝜙 , this equation reduces
to
𝑥 = 𝑥𝑚 cos 𝜔𝑡

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 VARIOUS PARAMETERS
OF OSCILLATOR
Amplitude
Amplitude is the maximum displacement on
both sides of an object from its equilibrium
position. The displacement will be maximum
when cos 𝜔𝑡 + 𝜙 = ±1.

The SI unit for amplitude is meter (m).

Hence,
Maximum displacement = ±𝑥𝑚
So, the amplitude of the oscillator is
+ 𝑥𝑚 𝑎𝑛𝑑 − 𝑥𝑚 .

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

𝑥 = 𝑥𝑚 𝑐𝑜𝑠 𝜔𝑡 + 𝜑
𝑑𝑥 𝑑
Hence velocity, 𝑣= = 𝑥𝑚 𝑐𝑜𝑠 𝜔𝑡 + 𝜑
𝑑𝑡 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 ). When (i) x = 𝑥𝑚 then v = 0 , (ii) when x= 0
then v = 𝑥𝑚 ω
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 ACCELERATION
The acceleration of a particle executing in SHM at any time t is given by,
𝑑𝑣
𝑎=
𝑑𝑡
𝑑 𝑑𝑥
=
𝑑𝑡 𝑑𝑡
𝑑
= −𝜔𝑥𝑚 𝑠𝑖𝑛 𝜔𝑡 + 𝜑
𝑑𝑡
= −𝜔 2 𝑥𝑚𝑐𝑜𝑠 𝜔𝑡 + 𝜑
∴ 𝒂 = −𝝎𝟐 𝒙.
When, (i) x = 0, a = 0, (ii) when, x =𝑥𝑚 then a = - ω2𝑥𝑚

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 Time taken for one complete oscillation of the particle executing SHM is called
time period. It is denoted by T. Unit: Second (sec). For one complete revolution,
the time taken is t = T, therefore
𝜔 𝑇 = 2𝜋
2𝜋
𝑇=
𝜔
We know the solution of Displacement Equation of SHM is 𝑥(𝑡) = 𝐴 𝑐𝑜𝑠 (𝜔𝑡 +
𝜙)

TIME If the time is increased by T=

𝜙)]
2𝜋
𝜔
i.e t=t+T, then 𝑥(𝑡 + 𝑇) = 𝐴 𝑐𝑜𝑠[ 𝜔(𝑡 + 𝑇) +

PERIOD
2𝜋
= 𝐴 𝑐𝑜𝑠 𝜔 𝑡 + +𝜙
𝜔

= 𝐴 𝑐𝑜𝑠 (𝜔𝑡 + 2𝜋 + 𝜙)
= 𝐴 𝑐𝑜𝑠 {2𝜋 + 𝜔𝑡 + 𝜙 }
= 𝐴 𝑐𝑜𝑠 (𝜔𝑡 + 𝜙)
= 𝑥(𝑡)
So 𝑥(𝑡 + 𝑇) =x (t)
2𝜋
It is seen that after time T= , the displacement of the particle becomes same.
𝜔
Thus, the motion repeats after one time period.

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 FREQUENCY
Total number of oscillations completed by an oscillating particle in one second is called its
frequency. Unit: Sec-1 or Hertz(Hz). 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

𝟏 𝒌
∴𝒇=
𝟐𝝅 𝒎

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 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 2 nd law of motion,

𝐹 = 𝑚𝑎
𝑑2𝑥
∴𝐹=𝑚 ……… (2)
𝑑𝑡 2

𝑑2𝑥
Where, 𝑎 = = acceleration of the particle.
𝑑 𝑡2

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

𝑑2𝑥
𝑚 = −𝑘𝑥
𝑑𝑡 2

𝑑 2𝑥
⇒𝑚 + 𝑘𝑥 = 0
𝑑𝑡 2

𝑑2𝑥 𝑘
10/20/2023 ∴ + 𝑥=0 17
𝑑 𝑡2 𝑚
10/20/2023 18
MECHANICAL A particle executing in SHM possesses both
ENERGY (E) potential and kinetic energies due to its elevation
and motion, respectively. The mechanical energy
OF A (𝐸) is the sum of potential (𝑈) and kinetic
(𝐾) energy i.e.
PARTICLE
EXECUTING (1)
𝐸 = 𝑈 + 𝐾 ……………..

SHM

10/20/2023 19
 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
Total work done for displacement𝑥,
𝑥
Ԧ 𝑑 𝑥Ԧ
𝑊 = ‫׬‬0 𝐹.
𝑥 𝑥
2 𝑥
𝑥 1
∴ 𝑈 = 𝑊 = න 𝑘𝑥𝑑𝑥 = 𝑘 න 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,
1 2 𝑐𝑜𝑠 2 (𝜔𝑡 + 𝜑) …… (3)
𝑈 = 𝑘𝑥𝑚
2

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 KINETIC ENERGY (K)

If the velocity of the particle is v, then kinetic energy is given by,

1
𝐾 = 𝑚𝑣 2
2
1 2
= 𝑚 −𝜔𝑥𝑚 𝑠𝑖𝑛 𝜔𝑡 + 𝜑 𝑎𝑠, 𝑣 = −𝜔𝑥𝑚 𝑐𝑜𝑠 𝜔𝑡 + 𝜑
2
1
= 𝑚𝜔 2𝑥𝑚
2
𝑠𝑖𝑛2 (𝜔𝑡 + 𝜑)
2
1 𝑘 2 𝑘
= 𝑚. 𝑥𝑚 𝑠𝑖𝑛2 (𝜔𝑡 + 𝜑) [as, 𝜔 2= ]
2 𝑚 𝑚
1
2 sin2 (𝜔𝑡 + 𝜙)………………. (6)
∴ 𝐾 = 𝑘𝑥𝑚
2

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 TOTAL ENERGY (E)
Now, adding (5) and (6) we have the mechanical
energy from (1) as follows,
1 2 1
𝐸= 𝑘𝑥𝑚 𝑐𝑜𝑠 2 2
𝜔𝑡 + 𝜑 + 𝑘𝑥𝑚 𝑠𝑖𝑛2 (𝜔𝑡 + 𝜑)
2 2
2 1
= 𝑘𝑥𝑚 𝑐𝑜𝑠 2 𝜔𝑡 + 𝜑 + 𝑠𝑖𝑛2 (𝜔𝑡 + 𝜑)
2
𝟏
∴ 𝑬 = 𝒌𝒙𝟐𝒎 .
𝟐
Therefore, the mechanical energy of a particle
executing SHM is constant or conserved, because
both 𝑘 and 𝑥𝑚 constant for that particle.

10/20/2023 22
 AVERAGE KINETIC ENERGY
2 1
The kinetic energy of SHM at any instant is given by, 𝐾 = 2 𝑘𝑥 𝑚 𝑠𝑖𝑛2 𝜔𝑡 + 𝜑
𝑻
‫𝒕𝒅 𝑲 ׬‬
𝑵𝒐𝒘 𝑲𝒂𝒗𝒈 = 𝟎
𝑻
𝑻
𝟏 1 2
= න 𝑘𝑥𝑚 𝑠𝑖𝑛2 (𝜔𝑡 + 𝜙)𝒅𝒕
𝑻 2
𝟎
𝑇
2
𝑘𝑥𝑚
= න 2𝑠𝑖𝑛2 (𝜔𝑡 + 𝜙) 𝑑𝑡
4𝑇
0
𝑇
2
𝑘𝑥𝑚
= න 1 − Cos 2(𝜔𝑡 + 𝜙) 𝑑𝑡
4𝑇
0
2 𝑇 2 𝑇
𝑘𝑥𝑚 𝑘𝑥𝑚
= න 𝑑𝑡 − න 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
𝑘𝑥 𝑚 𝑘𝑥 𝑚
= .𝑇 − {𝑠𝑖𝑛 2𝜔𝑇 + 2𝜙 − 𝑠𝑖𝑛2𝜙}
4𝑇 8𝜔𝑇
2
𝑘𝑥 𝑚 𝑘𝑥𝑚2 2𝜋
= − {𝑠𝑖𝑛 2. 𝑇 + 2𝜙 − 𝑠𝑖𝑛2𝜙 }
4 8𝜔𝑇 𝑇
𝑘𝑥𝑚2 𝑘𝑥𝑚2
= − {𝑠𝑖𝑛 4𝜋 + 2𝜙 − 𝑠𝑖𝑛2𝜙}
4 8𝜔𝑇
2
𝑘𝑥 𝑚 2
𝑘𝑥𝑚
= + {𝑠𝑖𝑛2𝜙 − 𝑠𝑖𝑛2𝜙 }
4 8𝜔𝑇
𝒌𝒙𝟐𝒎
∴< 𝑲 >=
𝟒
10/20/2023 23
 AVERAGE POTENTIAL ENERGY
The potential energy 𝑈 of the particle executing SHM at a time t is given by,
1 2 𝑐𝑜𝑠 2 (𝜔𝑡 + 𝜑)
𝑈 = 𝑘𝑥𝑚
2
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 >= ‫׬‬0 𝑈𝑑𝑡
𝑇
1 𝑇
2 𝑐𝑜𝑠 2 (𝜔𝑡 + 𝜑) 𝑑𝑡
= 𝑇 ‫׬‬0 𝑘𝑥𝑚
𝑘𝑥2𝑚 𝑇
=
4𝑇 0
‫ ׬‬2𝑐𝑜𝑠 2 (𝜔𝑡 + 𝜑) 𝑑𝑡
𝑘𝑥2𝑚 𝑇
= 4𝑇 0
‫ ׬‬1 + cos 2(𝜔𝑡 + 𝜑) 𝑑𝑡
2
𝑘𝑥𝑚 1 𝑇 1 𝑇
= ‫׬‬ 𝑑𝑡 + ‫ ׬‬cos 2(𝜔𝑡 + 𝜑) 𝑑𝑡
4 𝑇 0 𝑇 0
But the average value of both a sine and cosine function over a complete cycle or a whole time period (T) is zero. ∴< 𝑈 >=
𝑘𝑥2𝑚 1 𝑇
𝑇 0 −0
4 𝑇

𝑘𝑥𝑚2
= .𝑇
4𝑇
1 1 2
= 2 . 2 𝑘𝑥𝑚
𝟏
∴< 𝑈 >= . 𝑬(𝑴𝒆𝒄𝒉𝒂𝒏𝒊𝒄𝒂𝒍 𝒆𝒏𝒆𝒓𝒈𝒚)
𝟐
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.

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 Time period 0,T, T/2
𝟑
T T/4,
𝟒

Parameter x, t Phase 𝟑𝝅 π/2

0,π,2π
𝟐
Displacement 𝑥 = 𝑥𝑚 cos 𝜔𝑡 + 𝜙 0 -𝑥𝑚 +𝑥𝑚

Velocity v = −𝜔𝑥𝑚 𝑠𝑖𝑛 𝜔𝑡 + 𝜑

2 − 𝑥2
𝑣 = −𝜔 𝑥𝑚 v = 𝜔𝑥𝑚 0 0

Acceleration 𝑎 = −𝜔2𝑥𝑚 𝑐𝑜𝑠 𝜔𝑡 + 𝜑

𝑎 = −𝜔2𝑥 0 +𝜔2𝑥𝑚 −𝜔2𝑥𝑚

Potential energy 1
𝑈 = 𝑚𝜔2𝑥𝑚
2 𝑐𝑜𝑠 2 (𝜔𝑡 + 𝜑)
2
1 1 1
𝑈 = 𝑚𝜔2𝑥 2 0 𝑈 = 𝑚𝜔2 𝑥𝑚
2
𝑈 = 𝑚𝜔2 𝑥𝑚
2
2 2 2

Kinetic energy 1 2
𝐾 = 𝑘𝑥𝑚 sin2 (𝜔𝑡 + 𝜙)
2 1
1 𝐾 = 𝑚𝜔2𝑥𝑚
2 0 0
10/20/2023 𝑘 = 𝑚𝜔2 (𝑥𝑚
2
− 𝒙𝟐 ) 2 25
2
10/20/2023 26
 MATHEMATICAL 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. 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?
10/20/2023 27
10/20/2023 28
 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 … … … … . . (3)
𝑑 2𝜃 𝑑2𝜃
⇒ 𝐼 2 = −𝜅𝜃 [𝑎𝑠, 𝛼 = ]
𝑑𝑡 𝑑𝑡 2

𝑑2𝜃 𝜅
∴ 2 + 𝜃 = 0 … … … … … (4)
𝑑𝑡 𝐼
10/20/2023 29
 𝑑 2𝜃
we can write,𝛼 = −𝜔 2 𝜃 … … … … (5)[ Here, acceleration, 𝛼 = and angular velocity,ω
𝑑𝑡 2

𝜅
Now from equation (2) and (4), − 𝜃 = −𝜔 2 𝜃
𝐼
𝜅
=𝜔2
𝐼
𝜅
𝜔=
𝐼
𝑑 2𝜃
Equation (4) can be written as∴ 2 + 𝜔2𝜃 = 0
𝑑𝑡

𝑑 2𝑥
This equation is similar to the differential equation of a simple harmonic motion 2 + 𝜔 2 𝑥 =0
𝑑𝑡

Hence the motion of a torsional pendulum is simple harmonic motion

Time period: The time period is given by,

2𝜋 𝐼
𝑇= = 2𝜋
𝜔 𝜅

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 FREE OSCILLATIONS

• When a body vibrates with its own natural

frequency then it is said to be executing free
oscillations. Examples:
• Vibrations of tuning fork.
• Vibrations in stretched string.
• Oscillations of simple pendulum.

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 DAMPED OSCILLATIONS
A damped oscillation means an oscillation
that fades away with time. Most of the
oscillations in air or any medium are
damped. This oscillation occurs due to
some kind of damping force such as
friction or air resistance offered by the
medium.

Examples: The oscillations of a

pendulum.

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MAINTAINED
OSCILLATIONS
• The oscillation in which the loss of oscillator is
compensated by the supplying energy from an external
source are known as maintained oscillation.
• Examples: A swing to which energy is continuously fed
to the system to maintain amplitude of oscillation.

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FORCED OSCILLATIONS
• Oscillations produced by an external periodic driving
force (with a frequency that's not the natural frequency)
are called forced oscillations
• Examples:
• Sound boards of stringed instruments execute forced
vibrations.
• Press the stem of tuning fork against table, the table
suffers forced vibrations.

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 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:
• Using resonance, frequency of a given tuning fork can be determined.
• In radio-TV, using tank circuit, required frequency can be obtained.

Disadvantages:
• 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.
• A singer maintaining a note at a resonant frequency of a glass can cause
it to shatter into pieces.

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 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,
σ 𝐹𝑥 = −𝑘𝑥 − 𝑏𝑣

𝑑 2𝑥 𝑑𝑥
∴ 𝑚 2 = −𝑘𝑥 − 𝑏
𝑑𝑡 𝑑𝑡
𝑑 2𝑥 𝑏 𝑑𝑥 𝑘
∴ 2+ + 𝑥 =0
𝑑𝑡 𝑚 𝑑𝑡 𝑚
𝑑 2𝑥 𝑏 𝑑𝑥
∴ 2+ + 𝜔02𝑥 = 0 −−−−−− −(𝑖)
𝑑𝑡 𝑚 𝑑𝑡

This represents the Differential equation for damped oscillation.

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