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(C) Let the positions of the two particles be $x_1(t) = A \sin(\omega t + \phi_1)$ and $x_2(t) = X_0 + A \sin(\omega t + \phi_2)$.The separation betwe

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$\frac{\pi}{3}$

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(C) Let the positions of the two particles be $x_1(t) = A \sin(\omega t + \phi_1)$ and $x_2(t) = X_0 + A \sin(\omega t + \phi_2)$.The separation between them is $\Delta x = x_2 - x_1 = X_0 + A \sin(\omega t + \phi_2) - A \sin(\omega t + \phi_1)$.Using the trigonometric identity $\sin C - \sin D = 2 \sin(\frac{C-D}{2}) \cos(\frac{C+D}{2})$,we get $\Delta x = X_0 + 2A \sin(\frac{\phi_2 - \phi_1}{2}) \cos(\omega t + \frac{\phi_1 + \phi_2}{2})$.The maximum separation is given as $X_0 + A$. Therefore,the amplitude of the oscillating part must be $A$.Thus,$2A \sin(\frac{\Delta \phi}{2}) = A$,where $\Delta \phi = \phi_2 - \phi_1$.$\sin(\frac{\Delta \phi}{2}) = \frac{1}{2}$.$\frac{\Delta \phi}{2} = \frac{\pi}{6}$,which 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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