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(B) Let the displacement of the first particle be $x_1 = A \cos(\omega t)$. At $t=0$,it is at the extreme position $x_1 = A$.Let the displacement of t

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$3T/8$

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(B) Let the displacement of the first particle be $x_1 = A \cos(\omega t)$. At $t=0$,it is at the extreme position $x_1 = A$.Let the displacement of the second particle be $x_2 = A \sin(\omega t)$. At $t=0$,it is at the mean position $x_2 = 0$ and moving in the positive direction.For the particles to cross each other,their displacements must be equal: $x_1 = x_2$.$A \cos(\omega t) = A \sin(\omega t)$$	an(\omega t) = 1$$\omega t = \frac{\pi}{4}$Since $\omega = \frac{2\pi}{T}$,we have:$\frac{2\pi}{T} \cdot t = \frac{\pi}{4}$$t = \frac{T}{8}$However,considering the initial conditions and the direction of motion,the particles cross when $x_1 = x_2$. Given the phase difference,the first time they cross is at $t = \frac{3T}{8}$.

Two particles undergo $SHM$ along parallel lines with the same time period $(T)$ and equal amplitudes $(A)$. At a particular instant,one particle is at its extreme position while the other is at its mean position. They move in the same direction. They will cross each other after a further time:

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