Delhi Public School, Sonepat

CLASS-11
SUBJECT-PHYSICS
Chapter: Oscillations
Periodic Motion: A motion which is repeated after a fixed interval of time (called time period)
is called periodic motion. Example: motion of planets around sun.

Oscillatory Motion: The to and fro and back and forth motion of an object is called oscillatory
motion. Example: Pendulum, Loaded spring.

All oscillatory motions are periodic but all periodic motions are not oscillatory.

Types of Oscillations: There are two types of oscillations:

1. Harmonic Oscillations Oscillations which are expressed in terms of single harmonic

functions (trigonometric functions) are called harmonic oscillations.
Example: y = A sin ωt
2. Non – Harmonic Oscillations Oscillations expressed as combination of two or more
harmonic functions are called non – harmonic oscillations.
Example: y = A sin ω1t + B sin ω2t

Sin and Cos functions are harmonic functions because they are both periodic and bounded.

Simple Harmonic Oscillations: Harmonic oscillations are called simple harmonic when they
have a fixed amplitude i.e. amplitude (maximum displacement of particle) is constant and the
oscillation has single frequency.

Terms used in Simple Harmonic Motion (SHM):

a) Time Period (T) Time taken to complete one oscillation i.e. the time after which a
periodic motion is repeated. SI Unit: second
b) Frequency (υ) Number of oscillations taking place in unit time i.e. one second.
SI Unit: hertz (Hz), sec-1

Time taken to complete υ oscillations = 1 sec

T=
Time taken to complete one oscillation = =T

c) Angular Frequency (ω) Also known as angular velocity, it is the rate of change of
angular displacement. ω = 2πυ

d) Displacement (y) The physical quantity which changes when SHM is taking place is
called displacement. Example: The movement of loaded spring, Density and Pressure of a

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 medium changes when sound waves travel through it due to compressions and
rarefactions.
e) Amplitude (A) Maximum displacement which takes place from the mean position on
either side of the line of zero disturbance.
f) Phase (φ) That physical quantity which carries complete information about the
position and direction of motion of a particle undergoing SHM at a particular instant of
time.
OR
It is the argument part of sine or cosine function of an equation representing SHM.
g) Initial Phase (φ0) Initial phase denotes the position and direction of motion of a
particle undergoing SHM if it is not at mean position at t = 0.
Example: y = A sin (ωt + φ0)
h) Phase Difference The lack of harmony in vibrations of two objects in SHM is called
phase difference.
Whenever the phase difference is equal to an even multiple of π, we say that the particles
are moving in phase or oscillations are in phase, i.e. φ1 – φ2 = 0, 2π, 4π etc.
Whenever the phase difference is equal to odd multiples of π/2 or odd multiple of π,
oscillations are out of phase, i.e. φ1 – φ2 = π/2, π, 3π/2 etc.

Requirements for SHM:

1. Restoring force which brings the particle back to its mean position. F = - ky where k is
the spring constant whose unit is N/m.
2. Inertia of motion to cross the mean position and go in opposite direction.

Geometric Interpretation of SHM: SHM can be interpreted / represented as projection of

uniform circular motion undertaken by a reference particle along any diameter of a reference
circle having radius equal to the amplitude of SHM.

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If angular velocity of an object undergoing SHM is assumed to be ω and amplitude A, then draw
a reference circle with radius equal to A and angle made by the reference particle i.e. the object
undergoing SHM is θ = ωt.

Along x – axis the projection is x = A cos ωt while along y – axis the projection is y = A sin ωt.

The projection along x – axis moves from extreme left to extreme right and back as the particle
moves from initial point back to it. On the other hand projection along y – axis moves from mean
position to extreme position along positive y – axis, then back to mean and then moves towards
negative y – axis and back again.

Special Cases:

1. If the particle is behind / below the mean position θ = ωt - φ0 where φ0 is the initial phase.
Hence, y = A sin (ωt - φ0)
2. If the particle is ahead / above the mean position θ = ωt + φ0 and y = A sin (ωt + φ0)

Characteristics of SHM:

a) Displacement Displacement along x – axis is given by x = A cos ωt and along y –

axis is given by y = A sin ωt if initial phase is zero.
If initial phase φ0 ≠ 0 then, x = A cos (ωt ± φ0) and y = A sin (ωt ± φ0)
b) Amplitude If y = A sin ωt then, y will be maximum when sin ωt will be maximum i.e.
sin ωt = ± 1
ωt = , , , ---
ωt = (2n+1) where n = 1, 2, 3, ---
y=±A
c) Velocity If y represents displacement then velocity, v =

v= =A = A cos ωt = Aω cos ωt = v --- (i)

Also, v = Aω 1

Now, y = A sin ωt => sin ωt = => =

v = Aω 1
v=ω --- (ii)

Special Cases:

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 1. At mean position, y = 0 => v = ± Aω (maximum). Hence, velocity is maximum at
mean position.
2. At extreme position, y = A => v = 0 (minimum). Hence, velocity is minimum at
extreme position.
!
d) Acceleration (a) If y is displacement and v is velocity then a =
"# "#
a= = Aω = - Aω sin ωt = - Aω2 sin ωt = - ω2y

a = - ω2y

At mean position, y = 0 => a = 0 (minimum)

At extreme position, y = A => a = - ω2A (maximum)

The negative sign of acceleration represents that the acceleration always acts in a
direction opposite to the direction of displacement.

For any motion to fall in the criteria of simple harmonic motion, two conditions are required
mathematically,

Acceleration should be directly proportional to displacement i.e. y

Acceleration should act towards the mean position i.e. it should be negative of y

e) Time Period (T) We know, ω = => T =

$
But Acceleration, a = ω2y (neglecting the negative sign which stands for opposite
&
direction), we get ω = √

()*+&,-.-/
T = 2π i.e., T = 2π
' &,,-+-0& (1/

'
Also, T = => υ =
2

Dynamics of SHM: Two important requirements for a particle to execute SHM is elasticity (so
that restoring force can act) and mass (so that inertia of motion can act).

Restoring force F α – y

F = - ky
ma = - ky
3
a=- y ---- (i)
.

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 But a = - y ---- (ii)
3
Comparing equation (i) and (ii), =
.

3
ω=
.

4 6/-0 (& 7&, 10

Time Period, T = = 2π = 2π =T
5 8*0(/9 7&, 10

5
Frequency, υ =
2 4

Energy of a particle in SHM: The particle executing SHM has energy which is partly kinetic
and partly potential. Kinetic energy (K) is possessed by the particle due to its motion and
potential energy (U) is possessed due to its position away from the mean position.

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At mean position, y = 0, Potential energy = 0 (minimum)

Also, v = Aω, Kinetic energy = ½ mv2 = ½ mω2A2 (maximum)

Total Energy = Potential energy + Kinetic energy = ½ mω2A2 at the mean position.

At extreme position, y = A, Potential energy ≠ 0

Also, v = 0, Kinetic energy = 0

We know, restoring force F = - ky

Potential energy = work done on the body against restoring force

dW = :; . =>; = - ky dy cos 180°

dW = ky dy
W = ? =@ = k ?B A =A

W=k {limit from 0 to A}

W = ½ k y2 = Potential energy at extreme position

3
Now, ω = => k = ω2m
.

Potential energy = ½ mω2A2

Total Energy = Potential energy + Kinetic energy = ½ mω2A2 at extreme position.

At any given time t, Kinetic energy = ½ mv2 = ½ mω2A2 cos2 ωt = ½ mω2 (A2 – y2) --- (i)

Potential energy = ½ ky2 = ½ k A2 sin2 ωt --- (ii)

Total energy = Potential energy + Kinetic energy

T.E. = ½ k A2 sin2 ωt + ½ mω2A2 cos2 ωt
T.E. = ½ mω2A2 sin2 ωt + ½ mω2A2 cos2 ωt
T.E. = ½ mω2A2 (sin2 ωt + cos2 ωt)
T.E. = ½ mω2A2

Also, T.E. = ½ mω2 (A2 – y2) + ½ ky2

T.E. = ½ mω2A2 - ½ mω2y2 + ½ mω2y2

T.E. = ½ mω2A2

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This implies that the total energy of a particle executing SHM is conserved as the object starts
moving away from the mean position, its Kinetic energy decreases and the Potential energy
increases in such a way that Total energy remains constant.

Types of Simple Harmonic Oscillations:

a) Damped Oscillations Oscillations in which

amplitude does not remain constant but keeps on
decreasing with time are called damped oscillations.
Example: Pendulum oscillating in a medium like air or
water.
b) Undamped Oscillations Oscillations in which
amplitude remains constant with time are called
undamped oscillations. Example: Oscillations of a
simple pendulum in vacuum.
c) Maintained Oscillations Those oscillations where
proper amount of energy or force is fed to the oscillating
system at correct instant of time to keep the amplitude
maintained constant are known as maintained oscillations. Example: Oscillations of a
pendulum in a clock with electric cell or key.
5
Natural Frequency: υ0 =
2 4
The frequency at which an object starts vibrating itself about its mean position on
giving a small amount of energy is called natural frequency.
d) Free vibrations / oscillations When a system capable of oscillating vibrates with its
own natural frequency given by the formula above, oscillations are known as free
oscillations. Example: Oscillations of a pendulum or loaded spring.
e) Forced vibrations / oscillations When a body oscillates with the help of an external
periodic force with a frequency other than its own natural frequency the oscillations are
called forced oscillations.
Amplitude does not remain constant in such oscillations and are inversely proportional to
the square of the difference of the two frequencies i.e.
Aα
C D
As the difference between the two frequencies (υ0 – υ) decreases, amplitude increases.
f) Resonant vibrations / oscillations If the frequency of the external periodic force
becomes equal to the natural frequency of a vibrating object υ0 the value of amplitude
A ∞. In this situation resonance is said to occur i.e. when a body is allowed to vibrate
or oscillate under the action of an external periodic force whose frequency matches the
natural frequency of the oscillating body. Example: A glass kept on television starts

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 vibrating when the frequency of the sound coming from the TV matches the natural
frequency of the glass.

Question: Give reason why soldiers are always advised to break their steps while crossing the
bridges?

Answer: Soldiers are always advised to break their steps while crossing the bridges because if
the frequency of their steps matches with the natural frequency of the bridge, the bridge will start
swinging violently and might collapse.

Cases of Simple Harmonic Motion:

I. Simple Pendulum Simple Pendulum can be defined as a heavy point mass

suspended from a flexible, inextensible, mass less string held from rigid support.

Forces acting on the bob at extreme position are:

a) Tension along the length of the bob towards point of

suspension,
b) Weight = mg in vertically downward direction.

Resolving mg into two perpendicular components we get (i)

mg cos θ along a direction opposite to tension balancing it
and (ii) mg sin θ towards mean position.

Hence, mg cos θ = T

Restoring force = - mg sin θ

For small oscillations, sin θ ≈ θ

Fres = - mg θ
Fres = - mg --- [since angle θ = arc y/radius l]
+
m a = - mg
+
a=-g --- (i)
+

As acceleration ‘a’ is directly proportional to displacement ‘y’ and is negative. Hence, motion is
simple harmonic.

Comparing the equation (i) obtained with the general equation of SHM i.e. a = - ω2 y, we get
9
ω2 =
+

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 9
ω=
+

E
T = 2π
F

F
υ=
2 E
Time period does not depend upon mass of the pendulum but only on its length.

II. Vertical Spring Let us consider a spring of negligible mass in comparison to the
mass ‘m’ hanging from it and length ‘l’. It has a spring constant ‘k’.
Restoring force acting on spring when mass ‘m’ is suspended from it = F1 = - k l
When the spring is pulled by a distance ‘y’, the restoring force F2 = - k (l + y)
Net Restoring Force = F2 – F1 = - k (l + y) – (-kl) = - ky
Also, Fnet = ma
ma = - ky
3
a=- y --- (i)
.

As acceleration is negative and directly proportional to ‘y’, hence the motion is simple
harmonic.

Also in SHM, a = - ω2 y --- (ii)

Comparing equation (i) and (ii) we get,

3
- ω2 y = - y
.

3
=> ω2 =
.

3
=> ω= --- (iii)
.

4 4
As Time period = T = => T = 2π and frequency = ν =
5 2 5

Special Cases: COMBINATION OF SPRINGS

(i) Springs in Series: Assuming that mass of two springs with spring constant k1 and k2
connected in series is negligible w.r.t. the mass ‘m’ hanging from them, the
deforming force acting on them remains same in series and becomes F = mg.

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 As the spring constant k1 and k2 are not equal hence the displacements are also
different.
G G
Displacement in the 1st spring y1 = - and Displacement in the 2nd spring y2 = -
3H 3
As restoring force F = - ky where y = y1 + y2
G G
y=- -
3H 3

y=-F( + ) --- (i)

3H 3
For a series combination of springs, we can also write Fnet = - ksy
Where, ks = net spring constant in series and y = net displacement
G
- =y --- (ii)
3I
Comparing (i) and (ii) we get
= +
5J 5 5

5 5
ks =
5 K 5

If k1 = k2 = k then, ks = k/2. Hence, on connecting the springs in series the net spring
constant decreases.

(ii) Springs in Parallel: Assuming that mass of two springs with spring constant k1 and
k2 connected in parallel is negligible w.r.t. the mass ‘m’ hanging from
them, the displacement ‘y’ remains same while the restoring force is
different due to different spring constants.
Restoring Force for 1st spring = F1 = - k1y
Restoring Force for 2nd spring = F2 = - k2y
Net restoring force F = F1 + F2 = - k1y- k2y = - y (k1+ k2) --- (i)
For Parallel Combination, F = - kPy --- (ii)
Comparing (i) and (ii),
kP = k1 + k2

If k1 = k2 = k then, kP = 2k. Hence, on connecting the springs in parallel the net spring
constant increases.

DO THE OTHER CASES OF SIMPLE HARMONIC MOTION FROM

BOOK.

***********************

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