\(y=3 \sin \frac{\pi}{2}(50 t-x)\)
\(y=3 \sin \left(25 \pi t-\frac{\pi}{2} x\right)\) \(...(i)\)
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, \(\left(v_{p}\right)_{\max }=75 \pi \mathrm{ms}^{-1}\)
\(\therefore \frac{\left(v_{p}\right)_{\max }}{v}=\frac{75 \pi}{50}=\frac{3}{2} \pi\)
(વાયુનો અચળાંક $R=8.3 \;JK ^{-1} mol ^{- 1}$ લો)