91Ó°ÊÓ

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

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

$x=x_m\cos (\mathrm{Ó\neg }t+f)$(i)

The general expression for the velocity of motion

$v=−x_m\mathrm{Ó\neg }\sin (\mathrm{Ó\neg }t+f)$(ii)

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

$\begin{array}{c}{x}_1=x_m\cos (\mathrm{Ó\neg }t+f_1)\\ x_2=x_m\cos (\mathrm{Ó\neg }t+f_2)\end{array}$

Asthe two waves pass each other at the time t

$\begin{array}{c}{x}_1\left(or{x}_2\right)=\frac{1}{2}{x}_m(âˆ\mu x_1=x_2)\\ x_m\cos (\mathrm{Ó\neg }t+f_1)\left(or{x}_m\cos (\mathrm{Ó\neg }t+f_2)\right)=\frac{1}{2}{x}_m\\ \cos (\mathrm{Ó\neg }t+f_1)\left(or\cos (\mathrm{Ó\neg }t+f_2)\right)=\frac{1}{2}\end{array}$

$Phasedifference=\frac{\mathrm{Ï€}}{3}$

or either one is$\frac{\mathrm{π}}{3}$and other is$−\frac{\mathrm{π}}{3}$.

At the same instant, we also have

${\text{v}}_1(=−{\text{v}}_2)≠0$

Using equation (ii), we get

$\begin{array}{c}−v_m\mathrm{Ó\neg }\sin (\mathrm{Ó\neg }t+f_1)=−v_m\mathrm{Ó\neg }\sin (\mathrm{Ó\neg }t+f_2)\\ −\text{sin}(\mathrm{Ó\neg }t+f_1)=\text{sin}(\mathrm{Ó\neg }\text{t}+f_2)\end{array}$

This shows that the phases have an opposite sign, which means one phase is $\frac{\mathrm{π}}{3}$ and the other phase is$−\frac{\mathrm{π}}{3}$ .

Therefore,

$\frac{\mathrm{π}}{3}−(−\frac{\mathrm{π}}{3})=\frac{2\mathrm{π}}{3}$

Therefore, thephase difference of the two simple harmonic motions is $\frac{2\mathrm{Ï€}}{3}$.