Class 12 Physics Periodic motion Notes

Class 12 Physics Periodic motion Notes
UNIT 1
CLASS 12 PHYSICS • MECHANICS

Periodic Motion

Chapter 2
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Simple harmonic motion

The motion of the body is said to be simple harmonic motion if its acceleration is directed towards the mean position and is directly proportional to the displacement from that position.

i.e. In simple harmonic motion,

acceleration ∝ displacement
acceleration = −k × displacement

where k is constant and the negative sign shows that acceleration is directed in opposition to the motion of object.

Characteristics of Simple Harmonic Motion

Relation between the acceleration and displacement of the particle executing S.H.M.

Let us consider a body having mass ‘m’ moving in a circular path of radius ‘r’ with constant speed ‘v’. Suppose body is initially at point A and after time ‘t’, it reaches point B describing angular displacement ‘θ’. Draw perpendicular BN on vertical diameter yy′. Also, let when the particle is at point B, the displacement be ‘y’ such that ON = y.

Body rotating in a circular path showing SHM projection y O A B N r θ y y y′
Fig: Body rotating in circular path

From figure,

sinθ = y/r
or, y = r sinθ   — (i)

If ‘ω’ be the angular velocity, then θ = ωt

∴ y = r sinωt   — (ii)

This is the displacement equation for a particle of S.H.M.

Velocity of S.H.M.

v = dy/dt
= d(r sinωt)/dt
∴ v = rω cosωt   — (iii)

Equation (iii) can also be written as,

v = rω√(1 − sin2ωt)
v = ω√(r2 − r2sin2ωt)
v = ω√(r2 − y2)   — (iv)

This equation shows that velocity is not uniform.

(i) When y = 0; v = ωr — i.e. at mean position, the velocity is maximum.
(ii) When y = r; v = 0 — i.e. at extreme position, the velocity is minimum.

Acceleration of S.H.M.

a = dv/dt
= d(rω cosωt)/dt
= −rω2 sinωt
a = −ω2y   — (v)

This is the equation for the acceleration in S.H.M.

(i) When y = 0; a = 0 — i.e. at mean position, the acceleration is zero.
(ii) When y = r; a = −ω2r — i.e. at extreme position, the acceleration is maximum.

Simple Pendulum

A simple pendulum is a heavy point mass suspended by an inextensible, weightless and flexible string from a rigid support and which is free to oscillate in a vertical plane.
Simple pendulum with tension and components of weight at displaced position O P Q θ l T mg y mg sinθ mg cosθ
Fig: Simple pendulum

Let m be the mass of the bob of a simple pendulum and ‘l’ be its effective length (distance between point of suspension to the C.G. of the body). When the bob is not oscillating, the position of bob is called mean position which lies at point ‘P’ below the point of suspension.

When the bob is displaced from its mean position, it oscillates along the path OPQ. Also, let at any instant it be at point Q with angular displacement θ. The forces on the bob at point Q are:

  1. Weight mg of the bob acting vertically downward.
  2. Tension in the string acting towards point of suspension.

The weight mg of the bob can be resolved into two components: one is mg cosθ opposite to the tension and the other is mg sinθ towards mean position. So mg cosθ balances tension and mg sinθ provides restoring force.

Restoring force, F = −mg sinθ
For small θ, sinθ ≈ θ
∴ F = −mgθ   — (i)

If a be the acceleration of the bob, then,

F = ma = −mgθ
a = −gθ   — (ii)

Let y be the displacement from mean position. From figure,

θ = y/l   — (iii)

Using eqn (iii) in (ii), we get

a = −g(y/l)
a = −(g/l)y   — (iv)

Since g/l is a constant for a given pendulum at a given place, a ∝ y. This shows that motion of simple pendulum is S.H.M.

Also, acceleration in S.H.M. is,

a = −ω2y   — (v)

Comparing eqn (iv) and (v),

ω2 = g/l
ω = √(g/l)
2π/T = √(g/l)
T = 2π√(l/g)   — (vi)

This is the required relation for time period of a simple pendulum and this relation shows that time period of a simple pendulum is independent of mass of the bob but depends upon the effective length and acceleration due to gravity at that place.

Simple pendulum at the top of mountain

Question: What happens to the time period of simple pendulum if it is taken to the top of mountain?

Since, time period for simple pendulum is given by,
T = 2π√(l/g)
So, on the top of mountain the value of ‘g’ decreases and time period T ∝ 1/√g.
Therefore, the time period of simple pendulum increases when it is taken to the top of the mountain.

Oscillation of a Loaded Spring

(a) Vibration of a particle in horizontal spring

Horizontal spring attached to a wall and a mass, shown in natural and extended positions m m F l
Fig: Horizontal spring

Let us consider a spring of negligible mass whose one end is attached to wall and other end is attached to an object of mass ‘m’. The spring and the object lie on horizontal table as shown in figure. When the mass is pulled, the spring extends and let ‘l’ be the elongation produced. Then by Hooke’s law restoring force set up in the spring is,

F ∝ l
F = −kl   — (i)

where k is constant called force constant of spring (force per unit extension). Negative sign shows that restoring force acts opposite to the displacement of mass.

If a be the acceleration produced on mass, then,

F = ma   — (ii)

From eqn (i) and (ii),

ma = −kl
a = −(k/m)l   — (iii)

This equation shows that acceleration is directly proportional to the displacement and it is directed towards mean position, so the motion of horizontal mass-spring system is S.H.M.

Also, acceleration in S.H.M. is,

a = −ω2l   — (iv)

Comparing eqn (iii) and (iv),

ω2 = k/m
ω = √(k/m)
2π/T = √(k/m)
T = 2π√(m/k)   — (v)

This is required expression for the time period of a particle in horizontal spring.

(b) Vibration of a particle in vertical spring

Vertical spring shown unloaded, loaded by extension l, and displaced further by y mg l y unloaded loaded displaced
Fig: Vertical spring

Let us consider a spring of force constant k suspended vertically from a rigid support. When a body of mass ‘m’ is attached to its lower end then suppose it is stretched by l. Then according to Hooke’s law restoring force,

F1 = −kl = mg   — (i)

Let the load be pulled down through a small distance y. Then the restoring force F2 is given by,

F2 = −k(l + y)   — (ii)

The effective restoring force which causes the oscillation is,

F = F2 − F1
= −k(l + y) − (−kl)
= −ky
As F = ma,
ma = −ky
a = −(k/m)y

Hence, the motion of a loaded vertical spring is simple harmonic.

Also, for S.H.M., a = −ω2y.

ω2 = k/m
2π/T = √(k/m)
T = 2π√(m/k)
Time period of a mass-spring system depends on:
  1. Mass of load attached
  2. Spring constant
  3. It is independent of acceleration due to gravity

Energy in S.H.M.

For a body executing S.H.M., the restoring force acting on it causes its potential energy and due to its motion it exhibits kinetic energy. The energy of particle executing S.H.M. is the sum of K.E. and P.E.

Consider a body executing S.H.M. with amplitude ‘r’ and time period ‘T’ with angular velocity ω. Let ‘m’ be the mass of particle. Now acceleration of the particle at any instant when its displacement from mean position ‘y’ is given by,

a = −ω2y

So, the force acting on the particle,

F = −mω2y

When particle is displaced by small displacement ‘dy’, then small work done,

dw = −F·dy
= mω2y dy   — (i)

Now, total amount of work done on the particle for whole displacement is,

W = ∫0y dw
= mω2∫0y y dy
= mω2y2/2
P.E. = ½mω2y2   — (ii)

Again, velocity of the particle at displacement y from mean position is,

v = ω√(r2 − y2)
K.E. = ½mv2
K.E. = ½mω2(r2 − y2)

So, total energy of the particle is given by,

E = K.E. + P.E.
= ½mω2(r2 − y2) + ½mω2y2
E = ½mω2r2

This is required relation for total energy of a particle executing S.H.M. and this relation shows that total energy remains constant for a particle executing S.H.M.

Case I: Particle at mean position, y = 0

P.E. = ½mω2(0)2 = 0
K.E. = ½mω2r2
Total energy is equal to the maximum value of K.E.

Case II: Particle at extreme position, y = r

P.E. = ½mω2r2
K.E. = 0
Total energy is equal to the maximum value of P.E.

The variation of P.E. and K.E. with displacement in S.H.M. is shown in figure below.

Variation of potential energy and kinetic energy with displacement in SHM Displacement Energy −r 0 r Total energy P.E. K.E.
Fig: Variation of K.E. and P.E. in S.H.M.

Oscillatory Motion

1) Damped oscillation

The oscillations in which amplitude gradually decreases with increase in time are called damped oscillation.
Damped oscillation with amplitude decreasing with time t
Fig: Damped oscillation

2) Sustained oscillation

The oscillations in which amplitude remains constant with time are called sustained oscillation.
Sustained oscillation with constant amplitude t
Fig: Sustained oscillation

Drawbacks of Simple Pendulum

  1. The string is considered to be weightless and inextensible but in practice there is some weight of the string and some extension in the string when bob is suspended.
  2. The approximation sinθ ≈ θ is only valid when θ < 1, but experiment is usually performed with θ > 1.
  3. The bob is considered to be a point mass but it is not so.

Pendulum Clock on the Moon

Question: A pendulum clock is taken to moon. Will it gain or lose time?

The time period of pendulum clock is given by,
T = 2π√(l/g)
T ∝ 1/√g
When the clock is taken to the moon, acceleration due to gravity ‘g’ decreases and hence time period increases.
This shows that pendulum clock loses time when it is taken to the Moon.

Solved Numericals

Q.1

A simple pendulum 4 m long swings with an amplitude of 0.2 m. Calculate the velocity of the pendulum at its lowest point and its acceleration at extreme ends.

Length (l) = 4 m
r = 0.2 m
T = 2π√(l/g)
= 2 × 3.14 × √(4/9.8)
T = 4.01 s

(i) Velocity at lowest point, y = 0

v = ωr
= 2πf × r
= 2 × 3.14 × (1/4.01) × 0.2
v = 0.31 m/s

(ii) At extreme point, y = r

a = ω2r
= (2πf)2 × 0.2
= 4 × 9.8596 × (1/16.08) × 0.2
a = 0.49 m/s2

Q.2

A body of mass 0.1 kg is undergoing S.H.M. of amplitude 1 m and period 0.2 s. If the oscillation is produced by a spring, what will be the maximum value of the force and the force constant of the spring?

Mass (m) = 0.1 kg
r = 1 m
T = 0.2 s
a = ω2r = (2πf)2r
= 4 × (3.14)2 × (1/0.2)2 × 1
= 985.96 m/s2
F = ma
= 0.1 × 985.96
F = 98.59 N
T = 2π√(m/k)
0.2 = 2 × 3.14 × √(0.1/k)
0.04 = 39.4384 × 0.1/k
k = 98.6 N/m

Q.3

A particle of mass 0.2 kg vibrates with a period of 2 sec. If its amplitude is 0.5 m, what is its maximum K.E.?

Mass (m) = 0.2 kg
T = 2 sec
r = 0.5 m
ω2 = (2πf)2
= 4 × 9.8596 × 1/4
= 9.8596
K.E. = ½mω2r2
= ½ × 0.2 × 9.8596 × (0.5)2
K.E. = 0.78 J

Q.4

A second pendulum is taken to the moon. If the time period on the surface of the moon is 4.90 sec, what will be the acceleration due to gravity of the moon? Take acceleration due to gravity of the moon to be 1/6th that of the earth.

Time (T) = 4.90 sec
T = 2π√(l/g)
4.90 = 2 × 3.14 × √(l/1.63)
24.01 = 39.5102 × l/1.63
l = 0.99 m

Again,

a = ω2l
= (2πf)2 × 0.99
= 4 × 9.8596 × (1/24.01) × 0.99
a = 1.62 m/s2

Q.5

The displacement y of a mass vibrating with S.H.M. is given by y = 20 sin 10πt, where y is in millimeter and time in second. What is (a) amplitude, (b) the period, (c) the velocity at t = 0?

y = 20 sin 10πt   — (i)
y = r sinωt   — (ii)

Comparing eqn (i) and (ii),

(a) r = 20 mm = 20 × 10−3 m
ω = 10π
ω = 2π/T
10π = 2π/T
(b) T = 0.2 sec

(c) Velocity at t = 0

v = dy/dt
= d(r sinωt)/dt
= rω cosωt
= 20 × 10−3 × 10π × cos(10π × 0)
= 20 × 10−3 × 10 × 3.14 × cos0°
v = 0.628 m/s

Q.6

A simple pendulum has period of 4.2 sec. When the pendulum is shortened by 1 m, the period is 3.7 sec. From these measurements calculate the acceleration of free fall and the original length of the pendulum.

T1 = 2π√(l/g)
4.2 = 2π√(l/g)
16.84 = 4π2l/g   — (i)
T2 = 2π√((l − 1)/g)
3.7 = 2π√((l − 1)/g)
The supplied scanned PDF stops this solution at this point. No final value of g or the original length is shown in the source.

Q.7

A body of mass 2 kg is suspended from a spring of negligible mass and is found to stretch the spring 0.1 m. What is the force constant and the time period?

Mass (m) = 2 kg
l = 0.1 m
Force constant (k) = ?
Time period (T) = ?
F = kl
mg = kl
k = mg/l
= (2 × 9.8)/0.1
k = 196 N/m
T = 2π√(m/k)
= 2 × 3.14 × √(2/196)
T = 0.631 second

Q.8

A glider with mass m = 2 kg sits on a frictionless horizontal air track, connected to a spring with force constant k = 5 N/m. You pull the glider, stretching the spring 0.1 m and then release it with no initial velocity. The glider begins to move back toward its equilibrium position (x = 0). What is its velocity when x = 0.080 m?

Glider connected to a horizontal spring on a frictionless track m x = 0.08 m
Glider–spring system used in the numerical
Mass (m) = 2 kg
k = 5 N/m
x = 0.08 m
r = 0.1 m
T = 2π√(m/k)
= 2 × 3.14 × √(2/5)
T = 3.97 sec
ω = 2π/T
= (2 × 3.14)/3.97
= 1.58
v = ω√(r2 − x2)
= 1.58√(0.01 − 0.0064)
= 1.58 × 0.06
v = 0.0948 m/s
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