If maximum velocity of a particle executes simple harmonic motion `v_m` then its average velocity will be……….

Options

A. `(pi)/(2)v_(m)`
B. `(2)/(pi)v_(m)`
C. `(pi)/(4)v_(m)`
D. `(v_(m))/(sqrt(2))`

Answer

B

Explanation

Suppose at time t, the velocity of a particle is `v= v_(m) sin(omega t+phi)`
The average velocity on half cycle of period
`lt v gt = (1)/((T)/(2)) int_(0)^((T)/(2)) v_(m) sin (omega t+phi)dt`
`= lt v gt = (2v_m)/(T) int_(0)^((T)/(2)) sin (omega t+phi)dt`
`= -(2v_(m))/(T omega) [cos (omega t+phi)]_(0)^((T)/(2))`
`= -(2v_(m))/(T omega) [cos (omega xx (T)/(2)+phi)-cos (0+phi)]`
`= -(2v_(m))/(T omega) [cos(pi-phi)-cos phi]`
`= -(2v_(m))(Txx(2pi)/(T))[-cos phi -cos phi]`
`= +(2v_(m))/(pi)[cos phi]`
If `phi = 0" then "lt v gt = (2v_(m))/(pi)`.