Maximum velocity of a particle executing SHM is V. Then average speed of the particle over half cycle is 

If A be the amplitude and T be the time period of oscillations, then $V_{\max }=A\omega$

$\therefore A.\frac{2\pi }{T}=V\Rightarrow \frac{2A}{T}=\raisebox{1ex}{$V$}\!\left/ \!\raisebox{-1ex}{$\pi $}\right.$

$\therefore$ Average speed over half cycle

$=\frac{2A}{T/2}=2\left(\frac{2A}{T}\right)=\frac{2V}{\pi }$