Two particles A and B oscillates in SHM having same amplitude and frequencies `f_(A)` and `f_(B)` respectively. At `t=0` the particle A is at positive extreme position while the particle B is at a distance half of the amplitude on the positive side of the mean position and is moving away from the mean position.

Options

A. The minimum time after which the two particles will be in same phase is `(5)/(6(f_(A)-f_(B)))` if `f_(A)gtf_(B)`
B. The minimum time after which the two particles will be in same phase is `(1)/(6(f_(A)-f_(B)))` if `f_(A)gtf_(B)`
C. The minimum time after which the two particles will be in same phase in `(1)/(6(f_(B)-f_(A)))` if `f_(A)ltf_(B)`
D. The minimum time after which the two particles will be in same phase is `(5)/(6(f_(B)-f_(A)))` if `f_(A)ltf_(B)`

Answer

A, C

Explanation

`omega_(B)gtomega_(A)impliesDeltat=(pi//3)/(omega_(B)-omega_(A))=(pi)/(6(f_(B)-f_(A)))`
`omega_(B)gtomega_(A)impliesDeltat=(2pi-(pi)/(3))/(omega_(A)-omega_(B))=(5pi)/(6(f_(A)-f_(B)))`