Question

Two particles are in SHM in a straight line about same equilibrium position. Amplitude $A$ and time period $T$ of both the particles are equal. At time $t=0$, one particle is at displacement $y_1=+A$ and the other at $y_2=-\frac{A}{2}$, and they are approaching towards each other. After what time they cross each other?

• A. $\frac{T}{3}$
• B. $\frac{T}{4}$
• C. $\frac{5T}{6}$
• D. $\frac{T}{6}$

Answer 1

The correct option is D $\frac{T}{6}$


Angle covered to meet $\theta ={60}^{\circ }$
$=\frac{\pi }{3}\text{rad}$.
if they cross each other at time t then,
$t=\frac{\theta }{2\pi }T=\frac{\pi }{3\times 2\pi }T=\frac{T}{6}$