Two particles are performing simple harmonic motion in a straight line about the same equilibrium point. The amplitude and time period for both particle 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 is:

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$\frac{T}{3}$

$\frac{T}{6}$

$\frac{5 T}{6}$

$\frac{T}{4}$

answer is D.

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Detailed Solution

For particle at x=A, equation of SHM is given as $x_{1} = A \cos ωt$

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

When they will meet

$x_{1} = x_{2} \Rightarrow A \cos ωt = A \sin (ωt - \frac{π}{6})$

$\Rightarrow \cos ωt = \sin ωt \cos \frac{π}{6} - \cos ωt \sin \frac{π}{6}$

$\Rightarrow \cos ωt = \sin ωt \frac{\sqrt{3}}{2} - \cos ωt (\frac{1}{2})$

$\frac{3}{2} \cos ωt = \sin ωt \frac{\sqrt{3}}{2}$

$\sqrt{3} = \tan ωt;$ $ωt = \frac{π}{3}$

$\frac{2 π}{T} t = \frac{π}{3}; t = \frac{T}{6}$

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