A transverse wave is described by the equation $Y = {Y_0}\sin 2\pi \left( {ft - \frac{x}{\lambda }} \right)$. The maximum particle velocity is four times the wave velocity if
IIT 1984, Diffcult
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(b) Comparing the given equation with $y = a\sin (\omega t - kx)$,
We get $a = Y_0, \, \omega = 2\ pi f, k = \frac{{2\pi }}{\lambda }$.
Hence maximum particle velocity ${({v_{\max }})_{particle}} = a\omega = {Y_0} \times 2\pi f$
and wave velocity ${(v)_{wave}} = \frac{\omega }{k} = \frac{{2\pi f}}{{2\pi /\lambda }} = f\lambda $
$\because \,\,\,{({v_{\max }})_{Particle}} = 4{v_{Wave}}$==> ${Y_0} \times 2\pi f = 4f\lambda $ ==> $\lambda = \frac{{\pi {Y_0}}}{2}$.
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