b
(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 \({(v)_{wave}} = \frac{\omega }{k} = \frac{{2\pi f}}{{2\pi /\lambda }} = f\lambda \)

\(\because \,\,\,{({v_{\max }})_{Particle}} = 4{v_{Wave}}\)==> \({Y_0} \times 2\pi f = 4f\lambda \) ==> \(\lambda = \frac{{\pi {Y_0}}}{2}\).