Question

Two particles are executing simple harmonic motion of the same amplitude $A$ and frequency $\omega $ along the $x$ -axis. Their mean positions are separated by distance $x_{0}\left(\right.x_{0}>A\left.\right)$ . If the maximum separation between them is $\left(\right.x_{0}+A\left.\right)$ , the phase difference between their motions is

• A. $\frac{\pi }{3}$
• B. $\frac{\pi }{4}$
• C. $\frac{\pi }{6}$
• D. $\frac{\pi }{2}$

Answer 1

Answer: (A)

When the distance between the particles is maximum, their relative velocity will be zero. This means the particles are at the same distance from from their respective mean positions. Since the maximum distance is $x_{0}+A$ , it means the particles are at half the amplitude position. Also, if one of the particles is left of its mean position, then the other one would be to the right of the mean position.
$ x_{1}=A \sin \left(\omega t+\phi_{1}\right) $
At $t=0$, let $x_{1}=\frac{A}{2}$, then
$\omega t+\phi_{1}=\frac{\pi}{6}$ and the particle is moving along positive $x$-axis
$ x_{2}=A \sin \left(\omega t+\phi_{2}\right) $
At $t=0$ , $x_{2}=\frac{- A}{2}$
$\omega t+\phi_{2}=2\pi -\frac{\pi }{6}$ and this particle is also moving towards positive along positive x-axis
$\Rightarrow \phi_{2}-\phi_{1}=2\pi -\frac{\pi }{3}$
But a phase difference of $2\pi -\theta $ is same as a phase difference of $\theta $ .
So, $\phi_{2}-\phi_{1}=\frac{\pi }{3}$