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Two particles are executing SHM in a straight line with same amplitude A and time period T. At time $t = 0$ , one particle is at displacement $x_{1} = + A$ and the other at $x_{2} = - A/2$ and they are approaching towards each other. After what time they cross each other.

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\frac{T}{3}

\frac{T}{4}

\frac{T}{6}

\frac{5T}{6}

answer is D.

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Detailed Solution

x_{1} = A \cos ω t

$x_{2} = A \sin (ω t - \frac{π}{6})$

$x_{1} = x_{2}$

$A \cos ω t = A \sin (ω t - \frac{π}{6})$

$\cos ω t = \sin ω t \cos 30 ° - \cos ω t \sin 30 °$

$\tan ω t = \sqrt{3}$

$ω t = \frac{π}{3}$

$\frac{2 π}{T} t = \frac{π}{3}$

$t = \frac{T}{6}$

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