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(D) The displacement of a particle in $SHM$ is given by $x(t) = A \cos(\omega t + \phi)$.For the first particle at $t=0$,$x_1 = A$. Thus,$A = A \cos(\

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$\frac{T}{6}$

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(D) The displacement of a particle in $SHM$ is given by $x(t) = A \cos(\omega t + \phi)$.For the first particle at $t=0$,$x_1 = A$. Thus,$A = A \cos(\phi_1) \implies \phi_1 = 0$.So,$x_1(t) = A \cos(\omega t)$.For the second particle at $t=0$,$x_2 = -A/2$. Thus,$-A/2 = A \cos(\phi_2) \implies \phi_2 = 2\pi/3$ or $4\pi/3$. Since the particle is approaching the first one (moving towards positive direction),its velocity $v_2 = -A\omega \sin(\phi_2)$ must be positive. Thus,$\sin(\phi_2)$ must be negative,so $\phi_2 = 4\pi/3$.So,$x_2(t) = A \cos(\omega t + 4\pi/3)$.They cross when $x_1(t) = x_2(t)$:$A \cos(\omega t) = A \cos(\omega t + 4\pi/3)$.Using $\cos(	heta) = \cos(2\pi - 	heta)$,we have $\omega t = 2\pi - (\omega t + 4\pi/3) = 2\pi - \omega t - 4\pi/3$.$2\omega t = 2\pi/3 \implies \omega t = \pi/3$.Since $\omega = 2\pi/T$,we get $(2\pi/T)t = \pi/3 \implies t = T/6$.

Two particles executing $SHM$ along a straight line have the same amplitude $A$ and time period $T$. At $t=0$,one particle is at a displacement $+A$ and another is at a displacement $-\frac{A}{2}$ and they are approaching towards each other. They cross each other after a time.

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