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 $\mathrm{t}=0$ one particle has displacement $A$ while the other one has displacement $-\frac{\mathrm{A}}{2}$ and they are moving towards each other. If they cross each other at time $t$, then $t$ is:

• A

  $\frac{5\mathrm{T}}{6}$
• B

  $\frac{\mathrm{T}}{3}$
• C

  $\frac{\mathrm{T}}{4}$
• Correct answer
  D

  $\frac{\mathrm{T}}{6}$

Solution

Angle covered to meet $={60}^{\mathrm{o}}=\frac{\mathrm{\pi }}{3}\mathrm{r}\mathrm{a}\mathrm{d}$

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

Hence, the correct answer is option: 'D'.