Question

Two particles are executing simple harmonic motion of the same amplitude $A$ and frequency $co$ 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. $\frac{\pi }{2}$
• B. $\frac{\pi }{3}$
• C. $\frac{\pi }{4}$
• D. $\frac{\pi }{6}$

Answer 1

The correct option is A $\frac{\pi }{2}$

Let $\varphi$ be the phase difference between the two systems. Then equations will be:
$x_1\left(t\right)=Asin\left(\omega t\right)$
$x_2\left(t\right)=Asin(\omega t+\varphi )+x_0$
Since both systems have same frequency, the path difference is constant throughout the motion.
So, for t=0
$x_2\left(0\right)-x_1\left(0\right)=x_0+A$
$Asin\varphi +x_0=x_0+A$
$sin\varphi =1$
$\varphi =\frac{\pi }{2}$