Question

Two particles are executing SHM of same amplitude and frequency along the same straight line path. They pass each other when going in opposite directions, each time their displacement is half of their amplitude. The phase difference between them is

• A.
• B.
• C.
• D.

Answer 1

The correct option is B

$y=A\sin (\omega t+\varphi )$
When $y=\frac{A}{2},\frac{A}{2}=A\sin (\omega t+\varphi )$
or $(\omega t+\varphi )={\sin }^{-1}\left(\frac{1}{2}\right)=\frac{\pi }{6}$ or $\frac{5\pi }{6}$
$\therefore$ The phases of the two particles are $\frac{\pi }{6}$ and $\frac{5\pi }{6}$ radian.
$\therefore$ Phase difference $=\frac{5\pi }{6}-\frac{\pi }{6}=\frac{2\pi }{3}$