Two particles are executing $SHM$ in a straight line. Amplitude $'A'$ and time period $'T'$ of both the particles are equal. At time $t = 0$ one particle is at displacement $x_1 = +A$ and other at ${x_2} = \frac{{ - A}}{2}$ and they are approaching towards each other. Time after which they will cross each other is
A$\frac{T}{3}$
B$\frac{T}{4}$
C$\frac{5T}{6}$
D$\frac{T}{6}$
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D$\frac{T}{6}$
d time when they will cross each other $=\mathrm{T} / 12+\mathrm{T} / 12$
$=\mathrm{T} / 6$
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