$y=3 \sin \frac{\pi}{2}(50 t-x)$

$y=3 \sin \left(25 \pi t-\frac{\pi}{2} x\right)$                  $...(i)$

The standard wave equation is

$y=A \sin (\omega t-k x)$                                      $...(ii)$

Comparing $(i)$ and $(ii),$ we get

$\omega=25 \pi, \quad k=\frac{\pi}{2}$

Wave velocity,

$v=\frac{\omega}{k}=\frac{25 \pi}{(\pi / 2)}=50 \mathrm{ms}^{-1}$

Particle velocity, $v_{p}=\frac{d y}{d t}=\frac{d}{d t}\left(3 \sin \left(25 \pi t-\frac{\pi}{2}\right)\right)$

$=75 \pi \cos \left(25 \pi t-\frac{\pi}{2}\right)$

Maximum particle velocity, $\left(v_{p}\right)_{\max }=75 \pi \mathrm{ms}^{-1}$

$\therefore \frac{\left(v_{p}\right)_{\max }}{v}=\frac{75 \pi}{50}=\frac{3}{2} \pi$