Two particles move parallel to `x-`axis about the origin with same amplitude 'a' and frequency `omega`. At a certain instant they are found at a distance `a//3` from the origin on opposite sides but their velocities are in the same direction. What is the phase difference between the two?

Explanation

Let equation of two S.H.M. be
`x_(1) = A sin omega t` (i)
`x_(2) = A sin (omega t + phi)` …(ii)
Given that `(A)/(3) = A sin omega t and - (A)/(3) = A sin (omega t + phi)`
Which gives, `sin omega t = (1)/(3)` ….(iii)
`sin (omega t + phi) = - (1)/(3)` ....(iv)
`sin omega t cos phi + cos omega t sin phi = - (1)/(3)`
`rArr (1)/(3) cos phi + sqrt(1- (1)/(9)) sin phi = - (1)/(3)`
or `9 cos^(2) phi + 2 cos phi -7 = 0`
or `cos phi = -1, 7//9`
`rArr phi = pi or cos^(-1) (7//9)`
Differentation (i) and (ii) we obtain
`v_(1) = A omega cos omega t and v_(2) = A omega cos (omega t + phi)`
If we put `phi = pi`, we find `v_(1) and v_(2)` are of opposite signs. Hence `phi = pi` is not acceptable `phi = cos^(-1) ((7)/(9))`