$A$ travelling harmonic wave on a string is described by $y(x, t) = 7.5 \sin (0.0050 x + 12 t + \pi / 4)$.
$(a)$ What are the displacement and velocity of oscillation of a point at $x = 1 \; cm$ and $t = 1 \; s$? Is this velocity equal to the velocity of wave propagation?
$(b)$ Locate the points of the string which have the same transverse displacements and velocity as the $x = 1 \; cm$ point at $t = 2 \; s, 5 \; s$ and $11 \; s$.

(A) The given harmonic wave is $y(x, t) = 7.5 \sin (0.0050 x + 12 t + \pi / 4)$.
$(a)$ For $x = 1 \; cm$ and $t = 1 \; s$:
$y(1, 1) = 7.5 \sin (0.0050 + 12 + \pi / 4) = 7.5 \sin (12.0050 + 0.785) = 7.5 \sin (12.79 \; rad)$.
Converting to degrees: $12.79 \times (180 / \pi) \approx 732.81^{\circ}$.
$y = 7.5 \sin (732.81^{\circ}) = 7.5 \sin (8 \times 90^{\circ} + 12.81^{\circ}) = 7.5 \sin (12.81^{\circ}) \approx 7.5 \times 0.2217 \approx 1.663 \; cm$.
The velocity of oscillation is $v = \frac{\partial y}{\partial t} = 7.5 \times 12 \cos (0.0050 x + 12 t + \pi / 4) = 90 \cos (0.0050 x + 12 t + \pi / 4)$.
At $x = 1 \; cm, t = 1 \; s$: $v = 90 \cos (12.79 \; rad) = 90 \cos (12.81^{\circ}) \approx 90 \times 0.975 = 87.75 \; cm/s$.
The wave propagation velocity $v_p = \omega / k = 12 / 0.0050 = 2400 \; cm/s$.
Since $87.75 \; cm/s \neq 2400 \; cm/s$,the oscillation velocity is not equal to the wave propagation velocity.
$(b)$ The wavelength $\lambda = 2 \pi / k = 2 \pi / 0.0050 = 1256 \; cm = 12.56 \; m$.
Points with the same displacement and velocity are separated by integer multiples of the wavelength $\lambda$. Thus,the points are $x = (1 \; cm \pm n \lambda)$,where $n = 1, 2, 3, \dots$.