Two particles execute SHMs of the same amplitude and frequency along the same straight line. They cross one another when going in opposite direction. What is the phase difference between them when their displacements are half of their amplitudes ?

Options

A. `60^(@)`
B. `30^(@)`
C. `120^(@)`
D. `150^(@)`

Answer

C

Explanation

The equation for SHM is, `y=Asin(omegat+phi)`
As the displacement is half of the amplitudes `(y=(A)/(2))`,
or `(A)/(2)=Asin(omegat+phi)` or `sin(omegat+phi)=(1)/(2)`
`therefore omegat+phi=30^(@)` or `150^(@)`
Since the two particles are going in opposite directions, the phase of one is `30^(@)` and that of the other `150^(@)`. hence the phase difference between the two particles `=150^(@)-30^(@)=120^(@)`