Simple Harmonic Motion of Class 11

Simple Pendulum

Ideally, a simple pendulum consists of a point mass suspended at the end of a massless string as shown in the fig.(13.16).

At the equilibrium position, net torque about the point ‘O’ is zero, at θ = 0.

When the pendulum is displaced through an angle θ, the restoring torque about the point O is given by

τo = −mglsinθ

For small angles sin θ ≈ θ, therefore,

τo = −mgl sin θ  


Using Newton’s Second Law,

τo = Ioα = Pendulums


Comparing the above equation with the standard differential equation (13.10), we get

ω2 = g/l

∴T = Pendulums (13.16)

Physical Pendulum

In figure (13.17), an extended body is pivoted freely about an axis that does not pass through its center of mass. Such an arrangement forms a physical pendulum that executes simple harmonic motion for small angular displacements. If d is the distance from the pivot to the center of mass, the restoring torque is –mgd sin θ.

Using Newton’s Second law, τ = Iα,

we get

- mgd sin θ = Pendulums


Where I is the moment of inertia about the given axis. If we use the small-angle approximation, sin θ ≈ θ, then


Thus,ω = Pendulums

and T = 2π Pendulums (13.17)

If the location of the center of mass and d are known, then a measurement of the period allows us to determine the moment of inertia of the body.

Torsional Pendulum

Consider a body, such as a disc or a rod, suspended at the end of a wire, as shown in the fig.(13.18).

When the end of the wire is twisted by an angle θ, the restoring torque is given by

τ = -kθ

where k is called the torsional constant.

Using Newton’s law

 I Pendulums



Thus, the angular frequency and the time period of oscillation are given by

ω = √k/I

andT = 2π √I/k (13.18)

Example: 13.11

A rod of mass M and length L is pivoted about its end O as shown in the figure(13.19). Find the period of SHM. Find the period of SHM.


Restoring torque is

τo = -Pendulums

Using Newton’s Second Law and the small angle approximation, we get

Io Pendulums

Using parallel-axes theorem


Io = Ic + Pendulums


Thus,T = Pendulums

Example: 13.12

A ball is suspended by a thread of length L at the point O on the wall PQ which is inclined to the vertical by an angle α. The thread with the ball is now displaced through a small angle β away from the vertical and also from the wall. If the ball is released, find the period of oscillation of the pendulum when

(a) β < α(b) β > α.

Assume the collision on the wall to be perfectly elastic.


The motion of simple pendulum is angular SHM; so its equation of motion will be

θ = θo sin ωt with ω = √(g/L)

(a)When β < α, i.e., when angular amplitude β is lesser than α, the pendulum will oscillate with its natural frequency, so that

T1 = 2π/ω = 2π √L/g (i)

(b)When β > α, time taken by pendulum to move from B to C and back to B,


t1 = Pendulums (ii)

Now as in case of simple harmonic motion

θ = θo sin ωt

So time taken by the pendulum to move from equilibrium position B to A,

i.e., for θ = α when θo = β, will be given by

α = β sin ωt, i.e., t = Pendulums

So time taken by pendulum to move from B to A and back to B,

t2 = 2t = Pendulums (since ω = √(g/L) )

So time period of motion

T2 = t1 + t2 = Pendulums

Example: 13.13

A simple pendulum of length L and mass m has as spring of force constant k connected to it at a distance h below its point of suspension. Find the frequency of vibrations of the system for small values of amplitude.


As shown in figure, if the pendulum is given a small angular displacement θ, the spring will also stretch by y ( = h tan θ). So there restoring torque about O will be due to both force of gravity and elastic force of the spring.

i.e. τ = -[mg(L sin θ) + k(h tan θ)h]

Now for small θ, tan θ = sin θ = θ

soτ = -(mgL + kh2)θ

i.e. restoring torque is linear, so motion is angular SHM.

Now as τ = Iα = mL2Pendulums

(as I = mL2)


with ω2 = Pendulums


This is the standard equation of angular SHM with frequency f = (ω/2π); so here

f = Pendulums

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