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A particle executes SHM with time period T and ampitude A.Find the maximum possible average velocity in time T/4

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    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

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