Question

Two particles execute 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. ${90}^{\circ }$
• B. ${120}^{\circ }$
• C. ${180}^{\circ }$
• D. ${135}^{\circ }$

Answer 1

The correct option is A ${90}^{\circ }$

$\begin{array}{c}ForparticleinS.H.M\\ y=A\sin wtphase=\varphi =\omega t\\ y=\frac{A}{\sqrt{2}}for{P}_1\\ \frac{A}{\sqrt{2}}=A\sin \omega t\\ \left(\omega ,t\right)={\varphi }_1=\frac{\pi }{4}----\left(1\right)\\ For{p}_2-inoppositedirection\\ phase{\varphi }_2=\left(\pi -\frac{\pi }{4}\right)=\left(\frac{3\pi }{4}\right)---\left(2\right)\\ \Delta \varphi =\left({\varphi }_2-{\varphi }_1\right)=\frac{3\pi }{4}-\frac{\pi }{4}=\frac{\pi }{2}={90}^0\end{array}$

Hence, the option $A$ the correct answer.