Correct Answer - D
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`.