Standard equation of a progressive wave is travelling along positive $x$-direction.
$$
y=a \sin (\omega t-k x+\phi)
$$
As given equation is
$$
y=5 \sin (100 \pi t-0.4 \pi x)
$$
Comparing equation (i) with the standard equation
(a) Amplitude $(a)=5 \mathrm{~m}$
(b) wavelength $(\lambda)$
$$
k=\frac{2 \pi}{\lambda}=0.4 \pi
$$
$\therefore$ Wavelength, $\lambda=\frac{2 \pi}{k}=\frac{2 \pi}{0.4 \pi}=\frac{20}{4}=5 \mathrm{~m}$
(c) $\omega=100 \pi$, from eq. (i), $v=\frac{\omega}{2 \pi}$
$$
\therefore \text { Frequency } v=\frac{100 \pi}{2 \pi}=50 \mathrm{~Hz}
$$
(d) Wave velocity
$$
v=v \lambda=50 \times 5=250 \mathrm{~m} / \mathrm{sec}(\because \lambda=5 \mathrm{~m})
$$
(e) Particle (medium) velocity in the direction of amplitude at a distance $x$ from source.
$$
y=5 \sin (100 \pi t-0.4 \pi x)
$$
So, differentiation equation w.r.t. ( $t$ )
Particle velocity
$$
\frac{d y}{d t}=5(100 \pi) \cos [100 \pi t-0.4 \pi x]
$$
For maximum velocity of particle is at its mean position.
So, particle velocity amplitude $\left(\frac{d y}{d t}\right)_{\max }$.
then $\cos [100 \pi t-0.4 \pi x]_{\max }=1$
$\therefore$ Particle velocity amplitude
$$
\begin{aligned}
&=\left(\frac{d y}{d t}\right)_{\max .}=5(100 \pi) \times 1=500 \pi \mathrm{m} / \mathrm{s} \\
&v_{\text {max. }} \text { of medium particle }=500 \pi \mathrm{m} / \mathrm{s} .
\end{aligned}
$$