Simple Harmonic Motion , Linear SHM , Angular SHM

Linear SHM

From earlier discussion, it is now clear that when a particle is disturbed from its stable equilibrium, it oscillates to and fro about that position.

Let us consider such an oscillation on a straight line. If x represents the displacement of the particle from equilibrium position, then acceleration of the simplest possible motion can be given by

a ∝ − x . . . (1)

i.e. acceleration is always

(i) proportional to the displacement.

(ii) directed towards the equilibrium point.

A motion showing such a feature is known as simple harmonic motion.

Angular SHM

Instead of straight line motion, if particle or centre of mass of body is oscillating on a small arc of circular path, then condition for angular SHM is

Angular acceleration ∝ − (angular displacement) . . . (2)

Force law of SHM

From (1), it follows that

F = ma = − k x . . . (3)

where k is a constant of proportionality

similarly, from (2), for angular SHM

τ = I α = − C θ . . . (4)

where C is a constant of proportionality.

If a body oscillates between P and Q with O as equilibrium position in accordance with equation (3) and (4) then for force (or torque), we can get plotting as shown in the figure.

If the body is displaced to the right of O, the force points towards the left. If the body is displaced towards the left of O, the force is directed towards the right.

Such force is often called as linear restoring force.

Term linear is used to highlight the fact that force is proportional to first power of x.

Equation of Motion of S.H.M

F = – kx

$\large m \frac{d^2 x}{dt^2} + k x = 0$

Solution of this differential equation is

x = A sin(ωt + φ)

Where $\omega = \sqrt{\frac{k}{m}}$ = Angular frequency

φ = Initial Phase

Exercise : For F = − k x , motion is confined between two points in space. Is it true for
(i) F = k x (ii) F = − k√ x

Simple Harmonic Motion in terms of  Potential Energy :

Using dU = − Fdx or dU = − τ(dθ)

Potential energy of particle executing SHM is given by

$ \displaystyle U =\frac{1}{2}kx^2 $ …..(5)

Or , $ \displaystyle U =\frac{1}{2}k\theta^2 $ …..(6)

For this reference zero of potential energy is taken at x = 0 (or θ = 0)

Exercise : For a particle in motion, it is known that its potential energy is directly proportional to x2 where x is the displacement from a fixed point. Can we say that it will always execute SHM ?

Illustration : The linear displacement (x) of a particle varies with time as x = A sinωt + Bcosωt . State whether the particle is executing SHM or not.

Solution : $ \displaystyle a = \frac{d^2 x}{dt^2} $

= − ω2(Asinωt + Bcosωt)

= − ω2x

=> a ∝ − x

hence , Particle is executing SHM.

Also Read :

→ Stable , Unstable & Neutral Equilibrium
→ Analytical Treatment to SHM
→ Kinetic Energy & Potential Energy & Total Energy in SHM
→ Average Value of P.E. & K.E. of Harmonic Oscillator
→ SHM as a projection of circular motion
→ Simple Pendulum in Inertial & Non Inertial Frame
→ Time period of a Long Pendulum
→ SHM of Spring Mass System
→ Physical Pendulum & Torsional Pendulum
→ Undamped & Damped simple harmonic oscillations

Next Page → 

← Back Page

Leave a Reply