Two particles P and Q performs S.H.M. of same amplitude and frequency along the same straight line. At a…
Two particles P and Q performs S.H.M. of same amplitude and frequency along the same straight line. At a particular instant, maximum distance between two particles is $\sqrt{2} \mathrm{a}$. The initial phase difference between them is
$\frac{\pi}{6}$
$\frac{\pi}{2}$
zero
$\frac{\pi}{3}$
Solution
Maximum distance is $\sqrt{2} \mathrm{a}$, means one particle lies at $\mathrm{x}=\frac{\mathrm{a}}{\sqrt{2}}$ and second lies at $\mathrm{x}=-\frac{\mathrm{a}}{\sqrt{2}}$.
Hence, $x_1=a \sin \left(\omega t+\frac{\pi}{4}\right)$ and $x_2=a \sin \left(\omega t-\frac{\pi}{4}\right)$
Thus initial phase difference between the particles is $\frac{\pi}{2}$