where $x$ and $y$ are in metres and $t$ is in seconds. The ratio of maximum particle velocity to the wave velocity is
$y=3 \sin \frac{\pi}{2}(50 t-x)$
$y=3 \sin \left(25 \pi t-\frac{\pi}{2} x\right)$ $...(i)$
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, $\left(v_{p}\right)_{\max }=75 \pi \mathrm{ms}^{-1}$
$\therefore \frac{\left(v_{p}\right)_{\max }}{v}=\frac{75 \pi}{50}=\frac{3}{2} \pi$
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