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(D) Let the displacement of the particles be $x_1 = A \cos(\omega t)$ and $x_2 = A \cos(\omega t + \phi)$.At $t = 0$,$x_1 = A \cos(0) = A$.For the sec

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$T/6$

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(D) Let the displacement of the particles be $x_1 = A \cos(\omega t)$ and $x_2 = A \cos(\omega t + \phi)$.At $t = 0$,$x_1 = A \cos(0) = A$.For the second particle,$x_2 = A \cos(\phi) = -A/2$,which gives $\phi = 2\pi/3$ or $4\pi/3$. Since they are approaching each other,the second particle must be moving towards the equilibrium position,so we choose $\phi = 2\pi/3$.The particles cross when $x_1 = x_2$,so $A \cos(\omega t) = A \cos(\omega t + 2\pi/3)$.This implies $\omega t = -(\omega t + 2\pi/3) + 2n\pi$. For the first crossing,$2\omega t = 2\pi/3$,so $\omega t = \pi/3$.Since $\omega = 2\pi/T$,we have $(2\pi/T)t = \pi/3$,which gives $t = T/6$.

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