Question

Two particles are executing simple harmonic motion of the same amplitude $A$ and frequency $\omega$ along the x-axis. Their mean position is separated by distance $X_0(X_0>A)$. If the maximum separation between them is $(X_0+A)$, the phase difference between their motion is

• A. $\pi /2$
• B. $\pi /3$
• C. $\pi /4$
• D. $\pi /6$

Answer 1

The correct option is B $\pi /3$

$\begin{array}{c}{x}_1=A\sin \left(\omega t+{\varphi }_1\right)\\ x_2=A\sin \left(\omega t+{\varphi }_2\right)\\ x_1-x_2=A\left[2\sin \left[\omega t+\frac{{\varphi }_1+{\varphi }_2}{2}\right]\sin \left[\frac{{\varphi }_1-{\varphi }_2}{2}\right]\right]\\ A=2A\sin \left(\frac{{\varphi }_1-{\varphi }_2}{2}\right)\\ \frac{{\varphi }_1-{\varphi }_2}{2}=\frac{\pi }{6}\\ {\varphi }_1=\frac{\pi }{3}\end{array}$

Hence,

option $\left(B\right)$ is correct answer.