Question

Two particles execute simple harmonic motions of same amplitude and frequency along the same straight line. They cross one another when going opposite directions. The phase difference between them when their displacements are one half of their amplitudes is then

• A. ${60}^0$
• B. ${30}^0$
• C. ${120}^0$
• D. ${150}^0$

Answer 1

The correct option is C ${120}^0$

The equation for SHM is, $y=Asin(\omega +\varphi )$
As the displacement is half of the amplitudes $(y=\frac{A}{2})$, or $\frac{A}{2}=Asin(\omega t+\varphi )orsin(\omega t+\varphi )=\frac{1}{2}$
$\therefore \omega t+\varphi ={30}^{0}or{150}^0$.
Since the two particles are going in opposite directions, the phase of one is ${30}^0$ and that of the other ${150}^0$. Hence the phase difference between the two particles = ${150}^0-{30}^0={120}^0$.