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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