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Q.
A transverse wave is represented by $y=2 \sin (\omega t-k x)$. The wavelength for which wave-velocity is equal to the maximum particle velocity, is_____ $m$
Waves
Solution:
Wave velocity, $v=\frac{\omega}{k}$
Particle velocity, $v_{p}=\frac{d y}{d t}=A \omega \cos (\omega t-k x)$
$\therefore \left( v _{ p }\right)_{\max }= A \omega$
$A \omega=\frac{\omega}{ k } \cdots$ (given)
$\Rightarrow A =\frac{\lambda}{2 \pi}$
$\Rightarrow \lambda=2 \pi A =4 \pi=12.56\, m$