Q. 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 :
Solution:
angle covered to meet $=60^{\circ}=\frac{\pi}{3}$ rad
$t=\frac{Q}{w}$
$=\frac{\pi}{3\times2\pi} T=\frac{T}{6}$
