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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 x₁ = +A and the other at x₂ = -A/ 2 and they are approaching towards each other. After what time they cross each other?

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$\frac{T}{3}$

$\frac{T}{6}$

$\frac{T}{4}$

$\frac{5 T}{6}$

answer is D.

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Detailed Solution

$Total time = 2 t = \frac{T}{4} + \frac{T}{12}$

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