Question

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. $\frac{2\pi }{3}$

Answer 1

The correct option is D $\frac{2\pi }{3}$

Phase difference between any two particles in a wave determines lack of harmony in the vibrating state of two particles ie, how far one particle leads the other or lags behind the other.
Equation of simple harmonic wave is
$y=A\sin (\omega t+\varphi )$
Here, $y=\frac{A}{2}$
$\therefore A\sin (\omega t+\varphi )=\frac{A}{2}$
So, $\delta =\omega t+\varphi =\frac{\pi }{6}$ 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
$\delta ={\delta }_1-{\delta }_2$
$=\frac{5\pi }{6}-\frac{\pi }{6}=\frac{2\pi }{3}$