Q. Two particles execute SHM of the same amplitude and frequency along the same straight line. If they pass one another when going in opposite directions, each time their displacement is half their amplitude, the phase difference between them is
A) $\frac{\pi}{3}$
B) $\frac{\pi}{4}$
C) $\frac{\pi}{6}$
D) $\frac{2\pi}{3}$
Explanation
(d) Equation of simple harmonic wave is
$y = A\sin{(\omega t + \varnothing)}$
Here, $y = \frac{A}{2}$
$\therefore\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ A\sin{(\omega t + \varnothing)} = \frac{A}{2}\ $
So, $\delta = \omega t + \varnothing = \frac{\pi}{6}\ \ or\ \ \ \frac{5\pi}{6}$
So, the phase difference of the two particles when they are crossing each other at $y = \frac{A}{2}$ in opposite directions are
$\delta = \delta_{1} - \delta_{2} = \frac{5\pi}{6} - \frac{\pi}{6} = \frac{2\pi}{3}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.