Question

Two particles are in SHM with same angular frequency and with amplitudes of $A$ and $2A$ respectively along same straight line with same mean position. They cross each other at position A/2 distance from mean position in opposite direction. The phase difference between them is (If $si{n}^{-1}\left(1/4\right)={14.48}^o$)

Answer 1

$x=\frac{A}{2}\Rightarrow \frac{A}{2}=A\sin \omega t\sin \omega t=\frac{1}{2}\Rightarrow \omega t=\frac{\pi }{6}So,t=\frac{T}{12}(\therefore \omega =\frac{2\pi }{T})$
$\frac{A}{2}=2A\sin (\omega t+\varphi )\frac{1}{4}=\sin (\frac{2\pi }{T}\times \frac{T}{12}+\varphi )Phasedifference=\varphi =|\frac{\pi }{6}-{\sin }^{-1}(\frac{1}{4}\left)\right|=15.{52}^0$