Maximum velocity in SHM is $v_m$. The average velocity during motion from one extreme point to the other extreme point will be

$v_m=A\omega =\frac{2\pi A}{T}$ .... (1)

$<v>=\frac{2A}{T/2}=\frac{4A}{T}$

$\left(averagevelocity=\frac{totaldisplacement}{totaltime}\right)$

But, $\frac{A}{T}=\frac{v_m}{2\pi }$  from Eq. (1)

$\therefore <v>=4\left(\frac{v_m}{2\pi }\right)=\frac{2}{\pi }{v}_m$