compare the given equation with above equation
$\omega=2 \pi \mathrm{f} \quad \Rightarrow 60=2 \pi \mathrm{f}$
$\mathrm{K}=2 \quad \mathrm{f}=\frac{30}{\pi}$
$\mathrm{V}=\frac{\omega}{\mathrm{K}} \quad \Rightarrow \mathrm{V}=\frac{60}{2}=30 \mathrm{\,m} / \mathrm{sec}$
$\lambda=\frac{V}{f}=\frac{30}{30 / \pi}=\pi$ $meter$
If the distances are expressed in cms and time in seconds, then the wave velocity will be ...... $cm/sec$

