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

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$\pi / 2$

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(B) Let the displacements of the two particles be $x_1 = a \sin(\omega t + \phi_1)$ and $x_2 = a \sin(\omega t + \phi_2)$.The distance between the particles is $|x_1 - x_2| = |a \sin(\omega t + \phi_1) - a \sin(\omega t + \phi_2)|$.Using the trigonometric identity $\sin C - \sin D = 2 \sin(\frac{C-D}{2}) \cos(\frac{C+D}{2})$,we get:$|x_1 - x_2| = |2a \sin(\frac{\phi_1 - \phi_2}{2}) \cos(\omega t + \frac{\phi_1 + \phi_2}{2})|$.The maximum value of this distance is $|2a \sin(\frac{\phi_1 - \phi_2}{2})|$.Given that the maximum distance is $a\sqrt{2}$,we have:$|2a \sin(\frac{\phi_1 - \phi_2}{2})| = a\sqrt{2}$.$|\sin(\frac{\phi_1 - \phi_2}{2})| = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}}$.Therefore,$\frac{\phi_1 - \phi_2}{2} = \frac{\pi}{4}$,which gives the phase difference $\Delta\phi = \phi_1 - \phi_2 = \frac{\pi}{2}$.

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

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