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 amplitude is

• A. 60°
• B. 30°
• C. 120°
• D. 150°

Answer 1

Answer: (C)

120°

Answer 2

The equation for SHM is

$y = A\sin(\omega t + \varphi)$

As the displacement is half of the amplitudes

$(y = \frac{A}{2})$

Or $\frac{A}{2} = A\sin(\omega t + \varphi)$

Or $\sin(\omega t + \varphi) = \frac{1}{2}$

$\therefore\omega t + \varphi = 30º$ or $150º$

Since the two particles are going in opposite direction the phase of one is $30º$ and that of the other $150º$ hence the phase difference between the two particles

$= 150º - 30º = 120º$