Two particles P and Q performs S.H.M. of same amplitude and frequency… | MHT CET 2021 (20 Sep, Shift 1)

Q48MHT CET 2021 · 20 SEP · SHIFT 1Medium

Periodic Motion & SHM Basics

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 $\sqrt{2}$ a. The initial phase difference between them is

$\left[\sin^{-1}\left(\frac{1}{\sqrt{2}}\right) = \cos^{-1}\left(\frac{1}{\sqrt{2}}\right) = \frac{\pi}{4}\right]$

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Answer: (b)

$x_1 - x_2 = a\sin\omega t - a\sin(\omega t + \phi) = 2a\sin\frac{\phi}{2}\cos(\dots)$
Maximum separation $= 2a\sin\frac{\phi}{2} = \sqrt{2}a$
$\sin\frac{\phi}{2} = \frac{1}{\sqrt{2}} \Rightarrow \frac{\phi}{2} = \frac{\pi}{4}$
So the answer is (b) $\frac{\pi}{2}$ .

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