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)

$x _{1}= A \sin \left(\omega t +\phi_{1}\right)$
$x _{2}= A \sin \left(\omega t +\phi_{2}\right)$
$x _{1}- x _{2}= A \left[2 \sin \left[\omega t +\frac{\phi_{1}+\phi_{2}}{2}\right] \sin \left[\frac{\phi_{1}-\phi_{2}}{2}\right]\right]$
$A =2 A \sin \left(\frac{\phi_{1}-\phi_{2}}{2}\right)$
$\frac{\phi_{1}-\phi_{2}}{2}=\frac{\pi}{6}$
$\phi_{1}=\frac{\pi}{3}$