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Two particles execute simple harmonic motion of the same amplitude and frequency along close parallel lines. They pass each other moving in opposite directions each time their displacement is half their amplitude. What is their phase difference?

Short Answer

Expert verified

The phase difference between two simple harmonic motions is 2/3.

Step by step solution

01

Stating the given data

  1. The twosimple harmonic motions have the same amplitude and frequency.
  2. Displacement is half of their amplitudes.
02

Understanding the concept of simple harmonic motion

Using the formula of displacement and velocity of the particles from the equilibrium in a simple harmonic motion, and from the given condition that the two waves pass each other moving in opposite directions each time their displacement is half their amplitude, we can find the phase difference of two simple harmonic motions.

Formulae:

The general expression for the velocity of motion

x=xmcos(t+f)(i)

The general expression for the velocity of motion

v=xmsin(t+f)(ii)

03

Calculation of phase difference

For two simple harmonic motions using equation (i), can be given as

x1=xmcos(t+f1)x2=xmcos(t+f2)

Asthe two waves pass each other at the time t

x1(orx2)=12xm(x1=x2)xmcos(t+f1)(orxmcos(t+f2))=12xmcos(t+f1)(orcos(t+f2))=12

Phasedifference=3

or either one is3and other is3.

At the same instant, we also have

v1(=v2)0

Using equation (ii), we get

vmsin(t+f1)=vmsin(t+f2)sin(t+f1)=sin(t+f2)

This shows that the phases have an opposite sign, which means one phase is 3 and the other phase is3 .

Therefore,

3(3)=23

Therefore, thephase difference of the two simple harmonic motions is 23.

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