A particle performs SHM with a period `T` and amplitude a. The mean velocity of particle over the time interval during which it travels `a//2` from the extreme position is

Options

A. `(A)/(T)`
B. `(2A)/(T)`
C. `(3A)/(T)`
D. `(A)/(2T)`

Answer

C

Explanation

Time interval is `(T)/(6)` let the equation of simple harmonic motion be `x=Acosomegat`,`v=-Asinomegat`, where `omega=(2pi)/(T)`
Mean velocity over the required interval is
`lt v gt =(int_0^((T)/(6))vdt)/((T)/(6))`
`=(int_0^((T)/(6))-Aomegasinomegatdt)/((T)/(6))=(6A)/(T)cosomegat|_(0)^((T)/(6))=(-3A)/(T)`