Question

$2$ particles are in SHM with same time period same amplitude, same mean position (parallel to each other) if maximum separation between them is $\sqrt{3}A$. Then the find phase difference between them is.

Answer 1

It is given that

$x_1=A\sin \omega t$

$x_2=A\sin (\omega t+\varphi )$

$x_2-x_1=A\left[\sin \right(\omega t+\varphi )+sin\omega t]$

$=2A\cos \left[\frac{2\omega t+\varphi }{2}\right]\sin \left[\frac{\varphi }{2}\right]..............\left(1\right)$

$\sin \frac{\varphi }{2}=cons\tan t$

So the maximum value of $x_2-x_1$ occurs when $\cos \left[\frac{2\omega t+\varphi }{2}\right]=1$

Now equation (1) becomes

$\sqrt{3}A=2A\sin \left[\frac{\varphi }{2}\right]$

$\sin 60=\sin \left[\frac{\varphi }{2}\right]$

$\varphi =120$

So, the phase difference between two waves is $\varphi =120$