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(D) Let the positions of the two particles be $x_1$ and $x_2$ relative to their respective mean positions.$x_1 = A \sin(\omega t + \phi_1)$$x_2 = A \s

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$\frac{\pi}{3}$

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(D) Let the positions of the two particles be $x_1$ and $x_2$ relative to their respective mean positions.$x_1 = A \sin(\omega t + \phi_1)$$x_2 = A \sin(\omega t + \phi_2)$The absolute positions are $X_1 = x_1$ and $X_2 = X_0 + x_2$.The separation between them is $S = X_2 - X_1 = X_0 + x_2 - x_1$.$S = X_0 + A[\sin(\omega t + \phi_2) - \sin(\omega t + \phi_1)]$.Using the identity $\sin C - \sin D = 2 \cos(\frac{C+D}{2}) \sin(\frac{C-D}{2})$:$S = X_0 + 2A \cos(\omega t + \frac{\phi_1 + \phi_2}{2}) \sin(\frac{\phi_2 - \phi_1}{2})$.The maximum separation is $S_{max} = X_0 + |2A \sin(\frac{\phi_2 - \phi_1}{2})|$.Given $S_{max} = X_0 + A$,we have $|2A \sin(\frac{\Delta\phi}{2})| = A$,where $\Delta\phi = \phi_2 - \phi_1$.$\sin(\frac{\Delta\phi}{2}) = \frac{1}{2}$.$\frac{\Delta\phi}{2} = \frac{\pi}{6} \implies \Delta\phi = \frac{\pi}{3}$.

Two particles are executing simple harmonic motion of the same amplitude $A$ and frequency $\omega$ along the $x$-axis. Their mean positions are separated by a distance $X_0$ $(X_0 > A)$. If the maximum separation between them is $(X_0 + A)$,the phase difference between their motions is:

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