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(B) Let the displacements of particles $P$ and $Q$ be $x_1 = a \sin(\omega t + \phi_1)$ and $x_2 = a \sin(\omega t + \phi_2)$.The distance between the

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$\frac{\pi}{2}$

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(B) Let the displacements of particles $P$ and $Q$ be $x_1 = a \sin(\omega t + \phi_1)$ and $x_2 = a \sin(\omega t + \phi_2)$.The distance between them is $d = |x_1 - x_2| = |a \sin(\omega t + \phi_1) - a \sin(\omega t + \phi_2)|$.Using the formula $\sin C - \sin D = 2 \sin(\frac{C-D}{2}) \cos(\frac{C+D}{2})$,we get $d = |2a \sin(\frac{\phi_1 - \phi_2}{2}) \cos(\omega t + \frac{\phi_1 + \phi_2}{2})|$.The maximum distance occurs when $|\cos(\omega t + \frac{\phi_1 + \phi_2}{2})| = 1$,so $d_{max} = |2a \sin(\frac{\Delta \phi}{2})|$,where $\Delta \phi = \phi_1 - \phi_2$.Given $d_{max} = \sqrt{2} a$,we have $\sqrt{2} a = 2a \sin(\frac{\Delta \phi}{2})$.$\sin(\frac{\Delta \phi}{2}) = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}}$.$\frac{\Delta \phi}{2} = \frac{\pi}{4}$,which gives $\Delta \phi = \frac{\pi}{2}$.

Two particles $P$ and $Q$ perform $S.H.M.$ of same amplitude $a$ and frequency along the same straight line. At a particular instant,the maximum distance between the two particles is $\sqrt{2} a$. The initial phase difference between them is

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