A transverse wave is described by $y=A\sin (kx-\omega t)$. What is the ratio of the maximum particle velocity to the wave velocity?

Solution

1. The particle velocity is $$\begin{array}{rl}{v}_p& =\frac{\partial y}{\partial t}\\ & =-A\omega \cos (kx-\omega t)\end{array}$$.
2. The maximum particle velocity is $v_{p,\text{max}}=A\omega$.
3. The wave velocity is $v_w=\frac{\omega }{k}$.
4. The ratio is $$\begin{array}{rl}\frac{v_{p,\text{max}}}{v_w}& =\frac{A\omega }{\omega /k}\\ & =Ak\end{array}$$.
5. Hence the answer is (A).