Question

Two particles perform SHM with the same amplitude and some frequency about the same mean position and along the same line. If the maximum distance between them during the motion is $A$ (A is the amplitude of either) then the phase difference between their SHM is _____.

Answer 1

$x_1=A\sin wt$
$x_2=A\sin (wt+\varphi )$ distance between them is
$(x_2-x_1)=A\sin (wt+\varphi )-A\sin \left(wt\right)$
$=A\left[2\cos \right(wt+\varphi )-A\sin (wt\left)\right]$
$=A\left[2\cos \left(wt\frac{\varphi }{2}\right)\sin \left(\frac{\varphi }{2}\right)\right]$
$(x_2-x_1)$ will be max if $\cos (wt+\frac{\pi }{2})=1$
$\therefore (x_2-x_1{)}_{max}=2A\sin \left(\frac{\varphi }{2}\right)$
$A=2A\sin \left(\frac{\varphi }{2}\right)$
$⟹\frac{\varphi }{2}={30}^o=\frac{\pi }{6}$
$\therefore \varphi ={60}^o$ or $\frac{\pi }{3}$ radian