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  4. In a transverse progressive wave of amplitude A, the maximum particle velocity is four times its wave velocity. The wavelength of the wave is:-

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Q. In a transverse progressive wave of amplitude $A$, the maximum particle velocity is four times its wave velocity. The wavelength of the wave is:-

Solution:

$V_{\max }=A \omega=4\left(\frac{\omega}{k}\right)$
$\Rightarrow A=\frac{4}{k}=\frac{4 \lambda}{2 \pi}$
$\Rightarrow \lambda=\frac{\pi A}{2}$