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

Explanation

If we assume that the particles are initially at the mean position, their equaiton for displacement.
`x = A sin omegat` But `x = (A)/(2)`
`:. (A)/(2) = Asin omegat (or) sin omegat = (1)/(2)`
Phase `=omegat = 30^(@), 150^(@)`
(`:' sin (180^(@) -theta) = sin theta, sin (180^(@) - 30^(@)) = sin 30^(@))`
One of the particles has phase of `30^(@)` and the other has phase of `150^(@)`
Phase difference between them `= 120^(@) = (2pi)/(3)` radian