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(C) For $S.H.M.$,the displacement is given by $x = A \sin(\omega t + \phi)$.For particle $P_1$: It is at position $x = +0.5A$ and moving towards the m

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$\pi/3$

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(C) For $S.H.M.$,the displacement is given by $x = A \sin(\omega t + \phi)$.For particle $P_1$: It is at position $x = +0.5A$ and moving towards the mean position (negative direction). The angle $	heta$ such that $\sin 	heta = 0.5$ is $\pi/6$. Since it is in the first quadrant moving towards the mean position,its phase is $\phi_1 = \pi - \pi/6 = 5\pi/6$.For particle $P_2$: It is at position $x = -0.5A$ and moving towards the mean position (positive direction). The angle $	heta$ such that $\sin 	heta = -0.5$ is $-\pi/6$ or $11\pi/6$. Since it is in the third quadrant moving towards the mean position,its phase is $\phi_2 = \pi + \pi/6 = 7\pi/6$.The phase difference is $\Delta \phi = \phi_2 - \phi_1 = 7\pi/6 - 5\pi/6 = 2\pi/6 = \pi/3$.Wait,re-evaluating the figure: $P_1$ is at $+0.5A$ moving left,so $\phi_1 = \pi - \pi/6 = 5\pi/6$. $P_2$ is at $-0.5A$ moving right,so $\phi_2 = \pi + \pi/6 = 7\pi/6$. The difference is $\pi/3$.

Two bodies performing $S.H.M.$ have the same amplitude and frequency. Their positions and directions of motion at a certain instant are as shown in the figure. The phase difference between them is

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