Question

Two particle executes S.H.M. of same amplitude and frequency along the same straight line. They pass one another when going in opposite directions, each time their displacement is half of their amplitude. The phase difference between them is :-

• A. ${30}^o$
• B. ${60}^o$
• C. ${90}^o$
• D. ${120}^o$

Answer 1

Answer: (D)

${120}^o$

$y=a\sim (\omega t+{\varphi }_0)$. According to the question

$y=\frac{a}{2}\Rightarrow \frac{a}{2}=a\sin (\omega t+{\varphi }_0)\Rightarrow (\omega t+{\varphi }_0)=\varphi =\frac{\pi }{6}$ or $\frac{5\pi }{6}$

Physical meaning of $\varphi =\frac{\pi }{6}$ : Particle is at point $P$ and it is going towards $B$

So phase difference $△\varphi =\frac{5\pi }{6}-\frac{\pi }{6}=\frac{2\pi }{3}={120}^o$