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?

Options

A. `cos^(-1)((7)/(9))`
B. `cos^(-1)((5)/(9))`
C. `cos^(-1)((4)/(9))`
D. `cos^(-1)((1)/(9))`

Answer

A

Explanation

Let equation of two SHM be
`x_(1) = Asin omega t` ....(i)
`x_(2) = Asin (omega t + phi)`...(ii)
Give that `(A)/(3) = Asin omega t`
and `- (A)/(3) = Asin (omegat + phi)`
Which gives `sin omega t = (1)/(3)`...(ii)
`sin (omega t + phi) = - (1)/(3)`...(iv)
From Eq. (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)`
Solving this equation, we get
or `cos phi = -1,(7)/(9)`
`rArr phi = pi` or `cos^(-1) ((7)/(9))`
Differentiating Eqs. (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))`.