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

Answer

B

Explanation

Velocity at time t, `v=(v_("max"))/(2)`
`root(omega)(a^(2)-x^(2))= (a omega)/(2)`
`a^(2)- x^(2) = (a^2)/(4)`
`therefore a^(2)- (a^2)/(4)= x^(2)`.
`therefore x= (sqrt(3))/(2)a`
Now, `x= a sin (omega t+(pi)/(6))`
`(sqrt(3))/(2)a= a sin (omega t+(pi)/(6))`
`therefore (sqrt(3))/(2) = sin (omega t+(pi)/(6))`
`therefore omega t+(pi)/(6)= (pi)/(3)`
`therefore omega t= (pi)/(3)-(pi)/(6)`
`therefore omega t= (pi)/(6)`
`therefore t= (pi)/(6)xx (T)/(2pi)`
`therefore t= (T)/(12)`.