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 `x_(1) =a sin omegat` and `x_(2) = a sin (omegat +delta)` be two `S.H.M.`
`(a)/(3) = a sin omegat` and `(-a)/(3) = a sin (omegat + delta)`
`sin omegat = (1)/(3)` and `sin (omegat +delta) = (-1)/(3)`
Eliminating `t, (1)/(3) cos delta + sqrt(1-(1)/(9))sin delta=(-1)/(2)`
`9 cos^(2) delta +2 cos delta -7 = 0`
`cos delta =- 1 or 7//9 i.e., delta = 180^(@) or cos^(-1) (7//9)`
Now `v_(1) = a omega cos omegat` and `v_(2) = a omega cos (omegat +delta)`
If we put `delta = 180^(@)`. We find that `v_(1)` and `v_(2)` are of opposite signs. Hence `delta = 180^(@)` is not applicable.
`:. delta = cos^(-1) (7//9)`