Explanation

(a) Given that, the displacement of a particle is
Y=a sin (ωt-kx) …(i)
The particle velocity
$v_{p} = \frac{dy}{dt}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ldots(ii)$
Now, on differentiating Eq. (i) w.r.t.t,
$\frac{dy}{dt} = a\cos{(\omega t - kx).\omega}$
$\Longrightarrow \ \ \ \ \frac{dy}{dt} = a\omega\cos(\omega t - kx)\ $
From eq. (ii)
$\Longrightarrow \ \ \ \ \ \ \ \ v_{p} = a\omega\cos(\omega t - kx)$
For maximum particle velocity,
cos (ωt-kx)=1
so, $v_{p} = a\omega \times 1 \Longrightarrow v_{p} = a\omega$