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

Answer: (D)

Equation of simple harmonic wave is
$y=Asin(\omega t+\varphi )$
Here, $y=\frac{A}{2}$
$\therefore Asin(\omega t+\varphi )=\frac{A}{2}$
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
$\delta ={\delta }_1-{\delta }_2$
$=\frac{5\pi }{6}-\frac{\pi }{6}$
$=\frac{2\pi }{3}$