The displecemen-time equation of a particle execitting `SHM` is `x = A sin (omega t + phi)` At time `t = 0` position of are position is `x = A//2` and it is moving along negative `x- direction` .Then the angle `phi` can be

Options

A. `[(1)/(k_(1))+(1)/(k_(2))]^(1)`
B. `(4t - (pi)/(6))`
C. `(1)/(2pi)((k_(1) +k_(2))/(M))^(1//2)`
D. `(1)/(2pi)((k_(1) -k_(2))/(M))^(1//2)`

Answer

D

Explanation

At`t = 0, x = (A)/(2)`
`(A)/(2) = A sin phi` or `phi = (pi)/(6)` or `(5pi)/(6)`
`pi = (dx)/(dt) = A omega cos (omega t + phi)`
At `t = 0, a= A omega cos phi`
Now `cos "(pi)/(6) = sqrt(3)/(2) ans cos"(3pi)/(6) = -sqrt(3)/(2)`
As a is nagetive at `t = 0 , phi` must be `5pi//6`.