b
Transverse wave equation
$y=a \sin (\omega t+k x)$
Next crest appears at time interval of $0.4 \mathrm{s}$
So, the time period $T=0.4$
$\therefore$ Angular Frequency $\omega=\frac{2 \pi}{T}=\frac{2 \pi}{0.4}$
$\therefore$ Speed of the particle
$\nu=\frac{d y}{d t}=a \omega \sin (\omega t+k x)$
$\nu_{\max }=a \omega$
$a=1 / 2 c m$
$=\frac{1}{2} \times \frac{2 \pi}{0.4}=\frac{2 \pi}{8} \times 10$
$=\frac{5 \pi}{2} \mathrm{cm} / \mathrm{s}$