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

If maximum velocity of a particle executes simple harmonic motion v_m

1.	If maximum velocity of a particle executes simple harmonic motion v_m then its average velocity will be……….
Answer» `(pi)/(2)v_(m)` `(2)/(pi)v_(m)` `(pi)/(4)v_(m)` `(v_(m))/(sqrt(2))` Solution :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)`.