Question 51 of 194

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 $-\dfrac{A}{2}$ and they are moving towards each other. If they cross each other at time $t$, then $t$ is:

A$\dfrac{5T}{6}$B$\dfrac{T}{3}$C$\dfrac{T}{4}$D$\dfrac{T}{6}$

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Answer: (D) $\dfrac{T}{6}$

Angle covered to meet $={60}^{o}=\dfrac{\pi }{3}rad$

Therefore, $t=\dfrac{\theta }{\omega }=\dfrac{\pi }{3}\times \dfrac{T}{2\pi }=\dfrac{T}{6}$