Question

Two particles are executing $S.H.M$ of same amplitude and frequency along the same straight line path. They pass each other when going in opposite directions, each time their displacement is half of their amplitude. What is the phase difference between them?

• A. $5\pi /6$
• B. $2\pi /3$
• C. $\pi /3$
• D. $\pi /6$

Answer 1

Answer: (B)

$2\pi /3$

$y=a\sin (\omega t+\varphi );\text{when}y=a/2then\frac{a}{2}=a\sin (\omega t+\varphi )$
$or\sin (\omega t+\varphi )=\frac{1}{2}=\sin \frac{\pi }{6}or\sin \frac{5\pi }{6}$
So phase of two particles is $\pi /6$ and $5\pi /6$ radians.
Hence phase difference $=\left(5\pi /6\right)-\pi /6=2\pi /3$