Posted by Jishan on Tuesday, December 20, 2011 at 8:41am.
A particle executes SHM with time period T and ampitude A.Find the maximum possible average velocity in time T/4

physics  drwls, Tuesday, December 20, 2011 at 9:39am
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

physics  Anonymous, Wednesday, October 12, 2016 at 12:10pm
8A/T
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