Given, $y(x, t)=7.5 \sin (0.0050 x+12 t+\pi / 4)$ $=7.5 \sin \left[0.0050\left\{\frac{12}{0.0050} t+x\right\}+\pi / 4\right] \ldots(1)$
Comparing it with standard equation of travelling wave
$$
y=a \sin \left[\frac{2 \pi}{\lambda}\{v t+x\}+\phi\right]
$$
Comparing equations (1) and (2), we get $a=7.5 \mathrm{~cm}, v=\frac{12}{0.0050} \mathrm{~cm} \mathrm{~s}^{-1}$ and $\frac{2 \pi}{\lambda}=0.0050 \mathrm{~cm}^{-1}$
(i) At $x=1 \mathrm{~cm}$ and $t=1 \mathrm{~s}$, displacement is $y=7.5 \sin (0.0050 \times 1+12 \times 1+\pi / 4)$ $=7.5 \sin 12.79$
$=7.5 \times 0.2222=1.67 \mathrm{~cm}$.
Velocity of oscillation of the particle is
$$
\begin{aligned}
&u=\frac{d y}{d t}=\frac{d}{d t}[7.5 \sin (0.0050 x+12 t+\pi / 4)] \\
&=7.5 \times 12 \cos (0.0050 x+12 t+\pi / 4) \\
&=90 \cos \left(0.0050 x+12 t+\frac{\pi}{4}\right) \\
&\text { At } x=1 \mathrm{~cm} \text { and } t=1 \mathrm{~s}, \\
&u=90 \cos (0.000050 \times 1+12 \times 1+\pi / 4) \\
&=90 \cos 12.79 \\
&=90 \times 0.9751=87.76 \mathrm{~cm} \mathrm{~s}^{-1}
\end{aligned}
$$
Velocity of wave propagation is $v=\frac{12}{0.0050}=2400 \mathrm{~cm} \mathrm{~s}^{-1}$
$\therefore$ Velocity of oscillation of a point is not equal to the velocity of wave propagation.
(ii) As $\frac{2 \pi}{\lambda}-0.0050$
$$
\therefore \lambda=\frac{2 \pi}{0.0050}=1256.64 \mathrm{~cm}
$$
All points located at distance $n \lambda$ (where $n$ is an integer) from the point $x=1 \mathrm{~cm}$ have the same transverse displacement and velocity.