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Two particles are performing simple harmonic motion in a straight line about the same equilibrium point. The amplitude and time period for both particles are same and equal to A and T respectively. At time t= 0 one particle has displacement A while the other one has displacement $- \frac{A}{2}$ and they are moving towards each other. If they cross each other at time $t,$ then $t$ is

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$\frac{5 T}{6}$

$\frac{T}{3}$

$\frac{T}{4}$

$\frac{T}{6}$

answer is D.

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Detailed Solution

For particle at $x = A,$ equation of SHM is given as $X_{1} = A \cos w t$

For particle at $x = \frac{- A}{2}$ , equation of SHM is given as $x_{2} = A \sin (w t = \frac{π}{6})$

When they will meet

$x_{1} = x_{2}$

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

Solving further, we get

$\tan w t = \sqrt{3}$

$w t = \frac{π}{3}$

$\frac{2 π}{T} t = \frac{π}{3}$

$∴ t = T / 6$

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