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(D) Let the displacement of the particles be given by $y = a \sin(\omega t + \phi_0)$.According to the problem,the displacement is half of the amplitu

Vedclass · Accepted Answer

$\frac{2 \pi}{3}$

Vedclass · Answer

(D) Let the displacement of the particles be given by $y = a \sin(\omega t + \phi_0)$.According to the problem,the displacement is half of the amplitude,so $y = \frac{a}{2}$.Substituting this into the equation: $\frac{a}{2} = a \sin(\omega t + \phi_0)$.This gives $\sin(\omega t + \phi_0) = \frac{1}{2}$.Let $\phi = \omega t + \phi_0$. Then $\phi = \frac{\pi}{6}$ or $\phi = \pi - \frac{\pi}{6} = \frac{5\pi}{6}$.Physical meaning of $\phi = \frac{\pi}{6}$: The particle is at position $P$ (displacement $a/2$) and is moving away from the mean position $O$ (towards $B$).Physical meaning of $\phi = \frac{5\pi}{6}$: The particle is at position $P$ (displacement $a/2$) and is moving towards the mean position $O$.Since the particles are moving in opposite directions while passing each other at the same displacement,one must have phase $\phi_1 = \frac{\pi}{6}$ and the other $\phi_2 = \frac{5\pi}{6}$.The phase difference is $\Delta \phi = \phi_2 - \phi_1 = \frac{5\pi}{6} - \frac{\pi}{6} = \frac{4\pi}{6} = \frac{2\pi}{3}$.

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