A point performs simple harmonic oscillation of period T and the equation of motion is given by `x = a sin (omega t + (pi)/(6))`. After the elapse of what fraction of the time period, the velocity of the point will be equal to half of its maximum velocity ?

Options

A. `(T)/(8)`
B. `(T)/(6)`
C. `(T)/(3)`
D. `(T)/(12)`

Answer

D

Explanation

Writing the given equation of a point perfroming SHM
`x = a sin (omega t + (pi)/(6))`
Differentiating Eq. (i), w.r.t time, we obtain
`v = (dx)/(dt) = a omega cos (omega t + (pi)/(6))`
It is given that `v = (a omega)/(2)`, so that
`(a omega)/(2) = a omega cos (omega t + (pi)/(6))`
or `(1)/(2) = cos (omega t + (pi)/(6))`
or `"cos" (pi)/(3) = cos (omega t + (pi)/(6))`
or `(pi)/(3) = omega t + (pi)/(6) rArr omega t = (pi)/(6)`
or `t = (pi)/(6 omega) = (pi xx T)/(6 xx 2 pi) = (T)/(12)`
Thus, at `(T)/(12)` velocity of the point will be equal to
`therefore" "y = (.8)/((2pi)^(2)) = 0.25 m`
half of its maximum velocity.