Solution

(a) The given harmonic wave is
$$\displaystyle y(x, t)=7.5 \,sin\left(0.0050x+12t+\frac{\pi}{4}\right)$$
For $$x=1\,cm$$ and $$t=1s$$,
$$\displaystyle y(1, 1) = 7.5\, sin\left(0.0050x+12+\frac{\pi}{4}\right)$$
$$\displaystyle =7.5\,sin\left(12.0050+\frac{\pi}{4}\right)$$
$$=7.5\,sin\, \theta$$
Where,
$$\displaystyle \theta = 12.0050+\frac{\pi}{4}=12.0050+\frac{3.14}{4} = 12.79rad$$
$$=\displaystyle \frac{180}{3.14}\times 12.79=732.81^o$$
$$\therefore y = (1, 1) = 7.5\,sin(732.81^o)$$
$$=7.5\,sin(90\times 8+12.81^o)=7.5\,sin\, 12.81^o$$
$$=7.5\times 0.2217$$
$$=1.6629\approx 1.663\,cm$$
The velocity of the oscillation at a given point and times is given as:
$$\displaystyle v=\frac{d}{dt}y(x, t) = \frac{d}{dt}\left[7.5\,sin \left(0.0050x+12t+\frac{\pi}{4}\right)\right]$$

$$\displaystyle =7.5\times 12\,cos\left(0.0050x+12t+\frac{\pi}{4}\right)$$
At $$x=1\, cm$$ and $$t=1s$$
$$\displaystyle v=y(1, 1)=90\, cos\left(12.005+\frac{\pi}{4}\right)$$
$$=90\,coss(732.81^o)=90\,cos(90\times 8+12.81^o)$$
$$=90\,cos(12.81^o)$$
$$=90\times 0.975=87.75 cm/s$$
Now, the equation of a propagating wave is given by:
$$y(x, t)=a\, sin(kx+wt+\phi)$$

where,
$$k=2\pi/\lambda$$
$$\therefore \lambda = 2\pi/k$$
and, $$\omega =2\pi v$$
$$\therefore v = \omega/2\pi$$
Speed, $$v = v\lambda = \omega /k$$
Where,
$$\omega= 12\,rad/s$$
$$k=0.0050\,m^{-1}$$
$$\therefore v = 12/0.0050 = 2400\, cm/s$$
Hence, the velocity of the wave oscillation at $$x=1\,cm$$ and $$t=1s$$ is not equal to the velocity of the wave propagation.

(b) Propagation constant is related to wavelength as:
$$k=2\pi/\lambda$$
$$\therefore \lambda = 2\pi/k = 2\times 3.14/ 0.0050$$
$$=1256\,cm=12.56\,m$$
Therefore, all the points at distance $$n\lambda(n=\pm1\pm 2,...$$ and so on$$)$$, i.e. $$\pm 12.56\,m,\,\pm 25.12\,m,...$$ and so on for $$x=1\,cm$$, will have the same displacement as the $$x=1\,cm$$ points at $$t=2s, 5s$$ and $$11s$$.