CLASS XII CHAPTER 5: OSCILLATION (PHYSICS - MAHARASHTRA STATE BOARD)

1. CHAPTER 5
Oscillations
CLASSXII
PHYSICS
MAHARASHTRASTATEBOARD

2. Can you recall?
1. What do you mean by linear motion and angular
motion?
2. Can you give some practical examples of oscillations in
our daily life?
3. What do you know about restoring force?
4. All musical instruments make use of oscillations, can
you identify, where?
5. Why does a ball floating on water bobs up and down,
if pushed down and released?

3. MOTION
“Motion is the phenomenon in which an object changes its
position over time.”
It is described by
• displacement, distance
• velocity, speed
• acceleration
• time
Types of
motion
Rotational
motion
Oscillatory
motion
Linear
motion
Reciprocating

4. “A force acting opposite to displacement
to bring the system back to equilibrium
i.e. at rest position.”
RESTORING FORCE
Periodic motion
“Any motion which repeats itself after a
definite interval of time is called periodic
motion.”

5. OSCILLATION
“Oscillation is defined as the process of repeating vibrations of
any quantity or measure about its equilibrium value in time.”
Or
“Oscillation refers to any periodic motion at a distance about the
equilibrium position and repeat itself and over for a period of
time.”
Oscillation is periodic motion
Displacement, acceleration and velocity for oscillatory motion
can be defined by Harmonic function.
• Sine
• Cosine

6. Linear simple harmonic motion (s.h.m.)
When we pull block right side from mean
position the spring will pull object toward itself
i.e. force produced by spring is opposite.
f ∝ −𝒙
f = - k x
f = m a
∴ 𝒂 =
𝒇
𝒎
Linear S.H.M. is defined as the linear periodic motion of a body,
in which force (or acceleration) is always directed towards the
mean position and its magnitude is proportional to the
displacement from the mean position.

7. A complete oscillation is when the object goes from one extreme to
other and back to the initial position.
The conditions required for simple harmonic motion are:
1. Oscillation of the particle is about a fixed point.
2. The net force or acceleration is always directed towards the fixed
point.
3. The particle comes back to the fixed point due to restoring force.
Harmonic oscillation is that oscillation which can be expressed in terms
of a single harmonic function, such as x = a sin wt or x = a cos wt
Non-harmonic oscillation is that oscillation which cannot be expressed
in terms of single harmonic function.
It may be a combination of two or more harmonic oscillations such as x
= a sin ωt + b sin 2ωt , etc.

8. Differential Equation of S.H.M.
Consider, f → Restoring force, x → Displacement done by the block.
f = - k x ……….(i)
According to newtons second law of motion, f = m a
∴ m a = - k x ……….(ii)
Also, velocity → Rate of change of displacement
∴ v =
𝒅𝒙
𝒅𝒕
Acceleration → Rate of change of velocity
∴ a =
𝒅𝒗
𝒅𝒕
=
𝒅
𝒅𝒙
𝒅𝒕
𝒅𝒕
a =
𝒅𝟐𝒙
𝒅𝒕𝟐
m x
𝒅𝟐𝒙
𝒅𝒕𝟐 = - k x
∴ m
𝒅𝟐𝒙
𝒅𝒕𝟐 + k x = 0
i.e.
𝒅𝟐𝒙
𝒅𝒕𝟐 + 𝒘𝟐 x = 0, Where,
𝒌
𝒎
= 𝝎𝟐

9. Example
A body of mass 0.2 kg performs linear S.H.M. It experiences a
restoring force of 0.2 N when its displacement from the mean
position is 4 cm.
Determine (i) force constant (ii) period of S.H.M. and (iii) acceleration
of the body when its displacement from the mean position is 1 cm.
Solution:
(i) Force constant, k = f / x = (0.2)/ 0.04 = 5 N/m
(ii) Period T = 2𝝅
𝒎
𝒌
= 2𝝅
𝟎.𝟐
𝟓
= 0.4 𝝅 s
(iii)Acceleration
a = - 𝝎𝟐
x = −
𝒌
𝒎
𝒙 = −
𝟓
𝟎.𝟐
× 0.04 = - 1 m 𝒔−𝟐

10. TERMS FOR S.H.M.
𝒅𝟐𝒙
𝒅𝒕𝟐 + 𝒘𝟐
x = 0
𝒅𝟐𝒙
𝒅𝒕𝟐 = - 𝒘𝟐 x
a = − 𝒘𝟐 x
For velocity,
𝒅𝟐𝒙
𝒅𝒕𝟐 + 𝒘𝟐
x = 0
𝒅𝟐𝒙
𝒅𝒕𝟐 = - 𝒘𝟐 x
𝒅𝒗
𝒅𝒕
= - 𝒘𝟐 x
∴
𝒅𝒗
𝒅𝒙
.
𝒅𝒙
𝒅𝒕
= - 𝒘𝟐
x
∴
𝒅𝒗
𝒅𝒙
. v = - 𝒘𝟐
x
Integrating both side,
𝒗 𝒅𝒗 = − 𝝎𝟐 𝒙 𝒅𝒙
∴
𝒗𝟐
𝟐
= −
𝝎𝟐𝒙𝟐
𝟐
+ 𝒄 …………..(i)

11. Now, if object is at extreme position
x = A, v = 0
∴ C =
𝝎𝟐𝑨
𝟐
From equation (i)
∴ 𝒗𝟐
= − 𝝎𝟐
𝒙𝟐
+ 𝝎𝟐
𝑨𝟐
v = ± 𝝎 𝑨𝟐 − 𝒙𝟐
For displacement,
We know that, v =
𝒅𝒙
𝒅𝒕
v = 𝝎 𝑨𝟐 − 𝒙𝟐
∴
𝒅𝒙
𝒅𝒕
= 𝝎 𝑨𝟐 − 𝒙𝟐
∴ 𝒙 = 𝑨 𝒔𝒊𝒏 (𝝎𝒕 + 𝝓)

12. Case (i) If the particle starts S.H.M. from the mean position, x = 0 at t = 0
𝒙 = 𝑨 𝒔𝒊𝒏 𝝎𝒕 + 𝝓
0= 𝑨 𝒔𝒊𝒏 (𝝎 × 𝟎 + 𝝓)
𝝓 = 𝟎 𝒐𝒓 𝝅
i.e. x = ±𝑨 𝒔𝒊𝒏 𝒘𝒕
Case (ii) If the particle starts S.H.M. from the extreme position, x = ± A at t = 0
𝒙 = 𝑨 𝒔𝒊𝒏 (𝝎𝒕 + 𝝓)
A= 𝑨 𝒔𝒊𝒏 (𝝎 × 𝟎 + 𝝓)
1 = 𝒔𝒊𝒏 𝝓
𝝓 =
𝝅
𝟐
or
𝟑𝝅
𝟐
𝒙 = ± 𝑨 𝒄𝒐𝒔 𝝎𝒕

13. Expressions of displacement (x),velocity(v) and acceleration(a) at timet
𝒙 = 𝑨 𝒔𝒊𝒏 𝝎𝒕 + 𝝓
∴ 𝒗 =
𝒅𝒙
𝒅𝒕
=
𝒅 [𝑨 𝐬𝐢𝐧 𝝎𝒕 + 𝝓 ]
𝒅𝒕
v = 𝑨 𝒄𝒐𝒔 (𝝎𝒕 + 𝝓).(w+0)
v = Aw cos (𝝎𝒕 + 𝝓)
∴ 𝒂 =
𝒅𝒗
𝒅𝒕
=
𝒅 [𝑨𝒘 𝐜𝐨𝐬 𝝎𝒕 + 𝝓 ]
𝒅𝒕
a = −𝑨𝒘𝟐
𝐬𝐢𝐧 𝝎𝒕 + 𝝓
Extreme values of displacement (x), velocity(v) and acceleration(a):
1) Displacement: 𝒙 = 𝑨 𝒔𝒊𝒏 (𝝎𝒕 + 𝝓)
At mean position, (𝝎𝒕 + 𝝓) = 0 or 𝝅
∴ 𝒙𝒎𝒊𝒏 = 𝟎
At extreme position, (𝝎𝒕 + 𝝓) =
𝝅
𝟐
𝒐𝒓
𝟑𝝅
𝟐
∴ 𝒙 = ±𝑨 𝒔𝒊𝒏
𝝅
𝟐
∴ 𝒙𝒎𝒂𝒙 = ±𝑨

14. 2) Velocity : v = ± 𝝎 𝑨𝟐 − 𝒙𝟐
At mean position, 𝒙 = 0
∴ 𝒗𝒎𝒂𝒙 = ±𝑨𝝎
At extreme position, 𝒙 = ±𝑨
∴ 𝒙𝒎𝒊𝒏 = 𝟎
3) Acceleration: a = 𝝎𝟐
𝒙
At mean position, 𝒙 = 0
∴ 𝒂𝒎𝒊𝒏 = 𝟎
At extreme position, 𝒙 = ±𝑨
∴ 𝒂𝒎𝒂𝒙 = ∓𝝎𝟐𝑨
Amplitude
The maximum displacement of a particle performing S.H.M. from its mean position is
called the amplitude of S.H.M.
𝒙 = 𝑨 𝒔𝒊𝒏 (𝝎𝒕 + 𝝓)
For maximum displacement 𝒔𝒊𝒏 (𝝎𝒕 + 𝝓) = ±1
i.e. x = ±𝐀

15. Period of S.H.M.
The time taken by the particle performing S.H.M. to complete one oscillation is
called the period of S.H.M.
Displacement of the particle at time t,
𝒙 = 𝑨 𝒔𝒊𝒏 𝝎𝒕 + 𝝓
After some time,
𝒙 = 𝑨 𝒔𝒊𝒏 [𝝎 𝒕 +
𝟐𝝅
𝝎
+ 𝝓]
𝒙 = 𝑨 𝒔𝒊𝒏 𝝎𝒕 + 𝟐𝝅 + 𝝓
𝒙 = 𝑨 𝒔𝒊𝒏 𝝎𝒕 + 𝝓
Where
𝟐𝝅
𝝎
= 𝑻
k = m 𝝎𝟐
∴ 𝝎𝟐 =
𝒌
𝒎

16. ∴ 𝝎𝟐 =
𝑭𝒐𝒓𝒄𝒆 𝒑𝒆𝒓 𝒖𝒏𝒊𝒕 𝒅𝒊𝒔𝒑𝒍𝒂𝒄𝒆𝒎𝒆𝒏𝒕
𝒎𝒂𝒔𝒔
∴ 𝝎𝟐
=
𝒂
𝒙
Now, T =
𝟐𝝅
𝒂𝒄𝒄𝒆𝒍𝒆𝒓𝒂𝒕𝒊𝒐𝒏 𝒑𝒆𝒓 𝒖𝒏𝒊𝒕 𝒅𝒊𝒔𝒑𝒍𝒂𝒄𝒆𝒎𝒆𝒏𝒕
T = 2 𝝅 X
𝒎
𝒌
Frequencyof S.H.M.
The number of oscillations performed by a particle performing S.H.M. per unit time is called the
frequency of S.H.M.
n =
𝟏
𝑻
=
𝝎
𝟐𝝅
=
𝟏
𝟐𝝅
𝒌
𝒎
PHASEIN S.H.M.
Phase in S.H.M. is basically the state of oscillation.
Requirements to know the state of oscillation
- Position of particle (displacement)
- Direction of velocity
- Oscillation number

17. PHASEIN S.H.M.
Phase in S.H.M. is basically the state of oscillation.
Requirements to know the state of oscillation
- Position of particle (displacement)
- Direction of velocity
- Oscillation number
Commonly, 𝜽 = 𝝎𝒕 + 𝝓
Expressions of displacement (x), velocity (v) and acceleration(a) at time t
𝒙 = 𝑨 𝒔𝒊𝒏 (𝝎𝒕 + 𝝓)
v =
𝒅𝒙
𝒅𝒕
= 𝑨𝝎 𝒄𝒐𝒔 (𝝎𝒕 + 𝝓)
a =
𝒅𝒗
𝒅𝒕
= 𝑨𝝎𝟐
𝒔𝒊𝒏 (𝝎𝒕 + 𝝓)

18. 𝑺𝑷𝑬𝑪𝑰𝑨𝑳 𝑪𝑨𝑺𝑬𝑺
(i) Phase 𝜽 = 0
𝜽 = 𝟑𝟔𝟎𝟎 𝒐𝒓 𝟐𝝅𝒄
(i) Phase 𝜽 = 𝟏𝟖𝟎𝟎 𝒐𝒓 𝝅𝒄
𝜽 = (𝟑𝟔𝟎 + 𝟏𝟖𝟎)𝟎 𝒐𝒓 (𝟐𝝅 + 𝝅)𝒄
(i) Phase 𝜽 = 𝟗𝟎𝟎
𝒐𝒓
𝝅
𝟐
𝒄
𝜽 = (𝟑𝟔𝟎 + 𝟗𝟎)𝟎 𝒐𝒓 (𝟐𝝅 +
𝝅
𝟐
)𝒄
(i) Phase 𝜽 = 𝟐𝟕𝟎𝟎 𝒐𝒓
𝟑𝝅
𝟐
𝒄
𝜽 = (𝟑𝟔𝟎 + 𝟐𝟕𝟎)𝟎 𝒐𝒓 (𝟐𝝅 +
𝟑𝝅
𝟐
)𝒄

19. 𝑪𝑶𝑴𝑷𝑶𝑺𝑰𝑻𝑰𝑶𝑵 𝑶𝑭 𝑻𝑾𝑶 𝑺. 𝑯. 𝑴
Consider, two S.H.M having same period and along same path.
𝒙𝟏 𝒂𝒏𝒅 𝒙𝟐 are displacements of both S.H.M.
Composition of two S.H.M.
𝒙 = 𝒙𝟏 + 𝒙𝟐
x = 𝑨𝟏 𝒔𝒊𝒏 𝝎𝒕 + 𝝓𝟏 + 𝑨𝟐 𝒔𝒊𝒏 𝝎𝒕 + 𝝓𝟐
x = 𝑨𝟏 𝒔𝒊𝒏 𝝎𝒕. 𝒄𝒐𝒔𝝓𝟏 + 𝒄𝒐𝒔 𝝎𝒕. 𝒔𝒊𝒏𝝓𝟏 + 𝑨𝟐 𝒔𝒊𝒏 𝝎𝒕. 𝒄𝒐𝒔𝝓𝟐 + 𝒄𝒐𝒔 𝝎𝒕. 𝒔𝒊𝒏𝝓𝟐
x = 𝑨𝟏𝒔𝒊𝒏 𝝎𝒕. 𝒄𝒐𝒔𝝓𝟏 + 𝑨𝟏𝒄𝒐𝒔 𝝎𝒕. 𝒔𝒊𝒏𝝓𝟏 + 𝑨𝟐 𝒔𝒊𝒏 𝝎𝒕. 𝒄𝒐𝒔𝝓𝟐 + 𝑨𝟐 𝒄𝒐𝒔 𝝎𝒕. 𝒔𝒊𝒏𝝓𝟐
x = (𝑨𝟏𝒄𝒐𝒔𝝓𝟏 + 𝑨𝟐 𝒄𝒐𝒔𝝓𝟐) 𝒔𝒊𝒏 𝝎𝒕 + (𝑨𝟏𝒔𝒊𝒏 𝝓𝟏 + 𝑨𝟐 𝒔𝒊𝒏 𝝓𝟐) 𝐜𝐨𝐬 𝝎𝒕 ……………….(i)
R cos 𝜹 = 𝑨𝟏𝒄𝒐𝒔𝝓𝟏 + 𝑨𝟐 𝒄𝒐𝒔𝝓𝟐 ………….(ii)
R sin 𝜹 = 𝑨𝟏𝒔𝒊𝒏 𝝓𝟏 + 𝑨𝟐 𝒔𝒊𝒏 𝝓𝟐 ………….(iii)
x = R cos 𝜹 . sin 𝝎𝒕 + 𝑹 𝒔𝒊𝒏 𝜹 . 𝒄𝒐𝒔 𝝎𝒕
x = R [cos 𝜹 . sin 𝝎𝒕 + 𝒔𝒊𝒏 𝜹 . 𝒄𝒐𝒔 𝝎𝒕]
x = R sin (𝝎𝒕 + 𝜹)

20. RESULTANT AMPLITUDE
R = (𝑹 𝒔𝒊𝒏 𝜹)𝟐+(𝑹 𝒄𝒐𝒔 𝜹)𝟐
From equation (ii) and (iii)
𝑹𝟐
= 𝑨𝟏
𝟐
+ 𝑨𝟐
𝟐
+ 𝟐𝑨𝟏𝑨𝟐𝒄𝒐𝒔(𝝓𝟏 − 𝝓𝟐)
R = 𝑨𝟏
𝟐
+ 𝑨𝟐
𝟐
+ 𝟐𝑨𝟏𝑨𝟐𝒄𝒐𝒔(𝝓𝟏 − 𝝓𝟐)
SPECIAL CASES
(i) If the two S.H.M are in phase,
(𝝓𝟏 − 𝝓𝟐) = 𝟎𝟎, ∴ 𝒄𝒐𝒔 (𝝓𝟏 − 𝝓𝟐) = 1
∴ R = 𝑨𝟏
𝟐
+ 𝑨𝟐
𝟐
+ 𝟐𝑨𝟏𝑨𝟐 = ±(𝑨𝟏 + 𝑨𝟐)
If, 𝑨𝟏 = 𝑨𝟐 = 𝑨, 𝒘𝒆 𝒈𝒆𝒕 𝑹 = 𝟐𝑨
(ii) If the two S.H.M.s are 𝟗𝟎𝟎
out of phase,
(𝝓𝟏 − 𝝓𝟐) = 𝟗𝟎𝟎
, ∴ 𝒄𝒐𝒔 (𝝓𝟏 − 𝝓𝟐) = 0
∴ R = 𝑨𝟏
𝟐
+ 𝑨𝟐
𝟐
If, 𝑨𝟏 = 𝑨𝟐 = 𝑨, 𝒘𝒆 𝒈𝒆𝒕 𝑹 = 𝟐𝑨

21. SPECIAL CASES
(iii) If the two S.H.M.s are 𝟏𝟖𝟎𝟎 out of phase,
(𝝓𝟏 − 𝝓𝟐) = 𝟏𝟖𝟎𝟎, ∴ 𝒄𝒐𝒔 (𝝓𝟏 − 𝝓𝟐) = -1
∴ R = 𝑨𝟏
𝟐
+ 𝑨𝟐
𝟐
+ 𝟐𝑨𝟏𝑨𝟐
∴ 𝑹 = I 𝑨𝟏 − 𝑨𝟐I
If, 𝑨𝟏 = 𝑨𝟐 = 𝑨, 𝒘𝒆 𝒈𝒆𝒕 𝑹 = 𝟎
Initial Phase (𝜹)
Dividing equation (ii) and (iii)
𝑹 𝒔𝒊𝒏 𝜹
𝑹 𝒄𝒐𝒔 𝜹
=
𝑨𝟏𝒔𝒊𝒏 𝝓𝟏 + 𝑨𝟐 𝒔𝒊𝒏 𝝓𝟐
𝑨𝟏𝒄𝒐𝒔𝝓𝟏 + 𝑨𝟐 𝒄𝒐𝒔𝝓𝟐
∴ 𝒕𝒂𝒏 𝜹 =
𝑨𝟏𝒔𝒊𝒏 𝝓𝟏 + 𝑨𝟐 𝒔𝒊𝒏 𝝓𝟐
𝑨𝟏𝒄𝒐𝒔𝝓𝟏 + 𝑨𝟐 𝒄𝒐𝒔𝝓𝟐
∴ 𝜹 = 𝒕𝒂𝒏−𝟏
𝑨𝟏𝒔𝒊𝒏 𝝓𝟏 + 𝑨𝟐 𝒔𝒊𝒏 𝝓𝟐
𝑨𝟏𝒄𝒐𝒔𝝓𝟏 + 𝑨𝟐 𝒄𝒐𝒔𝝓𝟐

22. Energy of a Particle
Fig.: Energy in an S.H.M.
When particle performing S.H.M. then it passes both kinetic and potential energy.
Velocity of particle performing S.H.M
𝑣 = 𝜔 𝐴2 − 𝑥2 = A𝜔 cos 𝜔𝑡 + 𝜙
Kinetic Energy: 𝐸𝐾 =
1
2
𝑚𝑣2
=
1
2
𝑚[𝜔2
𝐴2
− 𝑥2
]
𝐸𝐾 =
1
2
𝑘 𝐴2
− 𝑥2
………(i)
Displacement x
𝐸𝐾 =
1
2
𝑚𝑣2
=
1
2
𝑚 [A𝜔 cos 𝜔𝑡 + 𝜙 ]2
𝐸𝐾 =
1
2
𝑘𝐴2𝑐𝑜𝑠2 𝜔𝑡 + 𝜙 ………….(ii)
External work done (dw)
dW = f (-dx)
dW = -kx (-dx)
dW = kx dx

23. Total work done on the particle,
W = 0
𝑥
𝑑𝑊 = 0
𝑥
𝑘𝑥
W =
1
2
𝑘 𝑥2
𝐸𝑃 =
1
2
𝑘 𝑥2 =
1
2
𝑚 𝜔2𝑥2
𝐸𝑃 =
1
2
𝑚 𝐴2𝜔2𝑐𝑜𝑠2(𝜔𝑡 + 𝜙) ………….(iii)
Total energy = 𝐸𝑘 + 𝐸𝑝
E=
1
2
𝑚𝜔2 𝐴2 − 𝑥2 +
1
2
𝑚 𝜔2𝑥2
E=
1
2
𝑚𝜔2 𝐴2 − 𝑥2 + 𝑥2
E=
1
2
𝑚𝜔2𝐴2
E=
1
2
𝑘𝐴2
For frequency, 𝜔 = 2 𝜋 𝑛
E=
1
2
𝑚𝜔2𝐴2 =
1
2
𝑚 (2𝜋𝑛)2𝐴2
E =
2 𝜋2 𝑚 𝐴2
𝑇2

24. Simple Pendulum
An ideal simple pendulum is a heavy particle suspended by a massless,
inextensible, flexible string from a rigid support.
A practical simple pendulum is a small heavy (dense) sphere (called bob)
suspended by a light and inextensible string from a rigid support.
In the displaced position (extreme position),
two forces are acting on the bob.
(i) Force T' due to tension in the string,
directed along the string, towards the
support and
(ii) Weight mg, in the vertically downward
direction.

25. Simple Pendulum
At the extreme positions, there should not be any net force along the
string. The component of mg can only balance the force due to tension.
Thus, weight mg is resolved into two components;
(i) The component mg cos θ along the string, which is balanced by the
tension T ' and
(ii) The component mg sin θ perpendicular to the
string is the restoring force acting on mass m
tending to return it to the equilibrium position.
∴ Restoring force, F = - mg sin θ
As θ is very small (θ < 10°), sin θ ≅ 𝜃𝑐
∴ 𝐹 ≅ −𝑚 𝑔 𝜃
Small angle, 𝜃 =
𝑥
𝐿
∴ 𝐹 = - m g
𝑥
𝐿

26. Simple Pendulum
∴ 𝐹 = - m g
𝑥
𝐿
∴ 𝐹 ∝ −x
∴ 𝑚 𝑎 = −𝑚 𝑔
𝑥
𝐿
∴
𝑎
𝑥
= −
𝑔
𝐿
For time period, T =
2 𝜋
𝜔
T =
2 𝜋
𝐴𝑐𝑐𝑒𝑙𝑒𝑟𝑎𝑡𝑖𝑜𝑛 𝑝𝑒𝑟 𝑢𝑛𝑖𝑡 𝑑𝑖𝑠𝑝𝑙𝑎𝑐𝑒𝑚𝑒𝑛𝑡
T =
2 𝜋
𝑔
𝐿
= 2 𝜋 ×
𝐿
𝑔
For frequency, n =
1
𝑇
=
1
2 𝜋
×
𝑔
𝐿

27. Second’s Pendulum
A simple pendulum whose period is two seconds is called second’s pendulum.
T = 2 𝝅
𝑳
𝒈
For a second s pendulum, 2 = 2 𝝅
𝑳𝑺
𝒈
Where, 𝑳𝑺 𝒊𝒔 𝒕𝒉𝒆 𝒍𝒆𝒏𝒈𝒕𝒉 𝒐𝒇 𝒔𝒆𝒄𝒐𝒏𝒅′𝒔 𝒑𝒆𝒏𝒅𝒖𝒍𝒖𝒎, 𝒉𝒂𝒗𝒊𝒏𝒈 𝒑𝒆𝒓𝒊𝒐𝒅 𝑻 = 𝟐 𝒔
𝑳𝑺 =
𝒈
𝝅𝟐

28. Angular S.H.M. and its Differential Equation
Thus, for the angular S.H.M. of a body, the restoring torque acting upon it, for angular
displacement θ, is
𝝉 ∝ − 𝜽 𝒐𝒓 𝝉 = −𝒄 𝜽 ……….(i)
The constant of proportionality c is the restoring torque per unit angular displacement.
𝝉 = 𝑰 𝜶
Where, 𝜶 𝒊𝒔 𝒕𝒉𝒆 𝒂𝒏𝒈𝒖𝒍𝒂𝒓 𝒂𝒄𝒄𝒆𝒍𝒆𝒓𝒂𝒕𝒊𝒐𝒏
𝑰 𝜶 = − 𝒄 𝜽
∴ 𝑰
𝒅𝟐𝜽
𝒅𝒕𝟐
+ 𝒄 𝜽 = 𝟎
𝜶 =
𝒅𝟐𝜽
𝒅𝒕𝟐 = −
𝒄 𝜽
𝑰
Since c and I are constants, the angular acceleration α is directly
proportional to θ and its direction is opposite to that of the
angular displacement.
Hence, this oscillatory motion is called angular S.H.M.

29. Angular S.H.M. is defined as the oscillatory motion of a body in which the
torque for angular acceleration is directly proportional to the angular
displacement and its direction is opposite to that of angular displacement.
The time period T of angular S.H.M. is given by,
T =
𝟐 𝝅
𝝎
T =
𝟐 𝝅
𝑨𝒏𝒈𝒖𝒍𝒂𝒓 𝒂𝒄𝒄𝒆𝒍𝒆𝒓𝒂𝒕𝒊𝒐𝒏 𝒑𝒆𝒓 𝒖𝒏𝒊𝒕 𝒂𝒏𝒈𝒖𝒍𝒂𝒓 𝒅𝒊𝒔𝒑𝒍𝒂𝒄𝒆𝒎𝒆𝒏𝒕
Magnet Vibrating in UniformMagnetic Field
If a bar magnet is freely suspended in the plane of a uniform
magnetic field.
Consider, μ be the magnetic dipole moment and
B the magnetic field.

30. The magnitude of this torque is, 𝝉 = 𝝁 𝑩 𝐬𝐢𝐧 𝜽
If θ is small, 𝐬𝐢𝐧 𝜽 ≅ 𝜽
𝝉 = 𝝁 𝑩 𝜽
Here, restoring torque is in anticlockwise
𝝉 = 𝑰 𝜶 = − 𝝁 𝑩 𝜽
Where, I – Moment of inertia
∴ 𝜶 = −
𝝁 𝑩
𝑰
𝜽
∴
𝜶
𝜽
= −
𝝁 𝑩
𝑰
T =
𝟐 𝝅
𝑨𝒏𝒈𝒖𝒍𝒂𝒓 𝒂𝒄𝒄𝒆𝒍𝒆𝒓𝒂𝒕𝒊𝒐𝒏 𝒑𝒆𝒓 𝒖𝒏𝒊𝒕 𝒅𝒊𝒔𝒑𝒍𝒂𝒄𝒆𝒎𝒆𝒏𝒕
=
𝟐 𝝅
𝜶
𝜽
T =
𝟐 𝝅
𝝁 𝑩
𝑰
𝑻 = 𝟐 𝝅
𝑰
𝝁 𝑩