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(B) Let the positions of the two particles be $x_1(t) = X_1 + A \sin(\omega t + \phi_1)$ and $x_2(t) = X_2 + A \sin(\omega t + \phi_2)$.Given the mean

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$\pi$

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(B) Let the positions of the two particles be $x_1(t) = X_1 + A \sin(\omega t + \phi_1)$ and $x_2(t) = X_2 + A \sin(\omega t + \phi_2)$.Given the mean positions are separated by $X_0$,we have $X_2 - X_1 = X_0$.The separation between the particles is $\Delta x = x_2 - x_1 = X_0 + A[\sin(\omega t + \phi_2) - \sin(\omega t + \phi_1)]$.Using the trigonometric identity $\sin C - \sin D = 2 \sin(\frac{C-D}{2}) \cos(\frac{C+D}{2})$,the maximum value of the term $A[\sin(\omega t + \phi_2) - \sin(\omega t + \phi_1)]$ is $2A \sin(\frac{\phi_2 - \phi_1}{2})$.We are given that the maximum separation is $X_0 + 2A$.Therefore,$2A \sin(\frac{\Delta \phi}{2}) = 2A$,which implies $\sin(\frac{\Delta \phi}{2}) = 1$.This gives $\frac{\Delta \phi}{2} = \frac{\pi}{2}$,so $\Delta \phi = \pi$.

Two particles are executing $SHM$ of the same amplitude $A$ and frequency $\omega$ along the $x$-axis. Their mean positions are separated by $X_0$ (where $X_0 > A$). If the maximum separation between them is $X_0 + 2A$,then the phase difference between their motion is:

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