posted by Jishan on .
A particle executes SHM with time period T and ampitude A.Find the maximum possible average velocity in time T/4
Maximum velocity is achieved twice in every period. The average V will depend upon which T/4 interval is selected. Since the interval is less than half the period, true maximum velocity may not be achieved during that short interval.
The maximum AVERAGE velocity in a T/4 interval will occur when it occurs in the middle of that interval.
Vav,max = Vmax*(Integralof)cos(2 pi t/T) dt/(T/4)
from t = -T/8 to T/8
Compute that integral for:
Vmax = 2 pi*(Amplitude)/T