Two particles execute SHM of same amplitude and frequency on parallel lines. They pass one another when moving in opposite directions each time when their displacement is half of the amplitude. The phase difference between them is

Options

A. `30^(@)`
B. `60^(@)`
C. `90^(@)`
D. `120^(@)`

Answer

D

Explanation

`x = sin omegat = A/2 rArr omegat = pi/6` or `(5pi)/(6)`
Phse difference `= (5pi)/(6) - (pi)/(6) = (2pi)/(3)` or `120^(@)`
or

from phaser
`cos theta = (A//2)/(A) = 1/2 rArr theta = 60^(@)`
phase difference `2theta = 120^(@)`