question_answer

Two particles executes S.H.M. of same amplitude and frequency along the same straight line. They pass one another when going in opposite directions, and each time their displacement is half of their amplitude. The phase difference between them is                                                      [MP PMT 1999]

A)            30°

B)            60°

C)            90°

D)            120°

Correct Answer: D

Solution :

                   \[y=a\sin (\omega t+{{\varphi }_{0}})\]. 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 Physical meaning of \[\varphi =\frac{5\pi }{6}\] : Particle is at point P and it is going towards O                    So phase difference \[\Delta \varphi =\frac{5\pi }{6}-\frac{\pi }{6}=\frac{2\pi }{3}=120{}^\circ \]