Question

Two particles execute SHM of same amplitude and frequency on straight lines. They pass one another when moving in opposite directions each time when their displacements is half of the amplitude. The phase difference between them is

• A. $0$
• B. $\frac{2\pi }{3}$
• C. $\pi$
• D. $\frac{\pi }{6}$

Answer 1

The correct option is B $\frac{2\pi }{3}$

Let A and B particle is moving in the circular path with uniform velocity in a clockwise direction, Motion of projection of particle on the XY line will be in SHM.

According to question,

Particle is meeting at $\frac{a}{2}$ and moving in the opposite direction, So Location of particle will be on A, B as given in the diagram.

Let Particle A making angle ${\theta }_1$ and particle B will be ${\theta }_2$

then Path diffrence, $\varphi ={\theta }_1+{\theta }_2$.......(1)

${\theta }_1=co{s}^{-1}\frac{a/2}{a}={60}^{\circ }=\frac{\pi }{3}$

${\theta }_2=co{s}^{-1}\frac{a/2}{a}={60}^{\circ }=\frac{\pi }{3}$

From equation (1)

$\varphi =\frac{2\pi }{3}$