A point simple harmonic oscillation of the period and the equation of motion is given by `x a sin (omega t + pi //6)` after the step of friction of the time period the velocity of the part 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

Velocity is the time derivative displacement wriitting the given equatiob of a point performing `SHM`
`x = a sin (omega t + (pi)/(6))`…(i)
Differentating Eq (i) w.r.t. time we obtain
`v = (dx)/(dt) = a omega cos "((omega t + )pi)/(6)`
it is given `v = (a omega )/(2)` so then
`(a omega)/(2) = a omega cos "(omega t + (pi))/(6)`
or `(1)/(2) = cos (omega t + (pi)/(6))`
or `cos "(pi)/(5) = cos (omegat + (pi)/(6))`
or `omega t = (pi)/(6) = (pi)/(3) rArr omega t + (pi)/(6)`
`rArr t = (pi)/(6 omega) = (pi xx T)/(6 xx 2 pi) = (T)/(12)`
Thus at `1//2` velocity of the pioint will be equal to half of its maximum velocity.