A travelling harmonic wave on a string is described by $y(x, t)=7.5 \sin (0.0050 x+12 t+m / 4)$
(a) What are the displacement and velocity of oscillation of a point at $x=1 \mathrm{~cm}$, and $t=1 \mathrm{~s}$ ? 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 $x=1 \mathrm{~cm}$ point at $t=2 \mathrm{~s}, \mathbf{5} \mathrm{s}$ and $11 \mathrm{~s}$

(a) The travelling harmonic wave is given by,

$y(x, t)=7.5 \sin (0.0050 x+12 t+\pi / 4)$

At $x=1 \mathrm{~cm}$ and $\mathrm{t}=1 \mathrm{~s}$

$y(1,1)=7.5 \sin (0.0050(1)+12(1)+\pi / 4)$

$=7.5 \sin (12.0050+\pi / 4)$

$=7.5 \sin \theta$

$\theta=(12.0050+\pi / 4) \times(180 / 3.14)$

$=12.0050+(3.14 / 4)=12.79 \times(180 / 3.14)$

$\theta=733.18^{\circ}$

$y(1,1)=7.5 \sin \left(733.18^{\circ}\right)$

$=7.5 \sin \left(90 \times 8+13.18^{\circ}\right)$

$=7.5 \sin \left(13.18^{\circ}\right)$

$=7.5 \times 0.228$

$=1.71$

The velocity of oscillation,

$v=\frac{d}{d t} y(x, t)=\frac{d}{d t}\left[7.5 \sin \left(0.0050 x+12 t+\frac{\pi}{4}\right)\right]$

$=7.5 \times 12 \cos \left(0.0050 x+12 t+\frac{\pi}{4}\right)$

At $x=1 \mathrm{~cm}$ and $\mathrm{t}=1 \mathrm{~s}$

$v=90 \cos \left(12.005+\frac{\pi}{4}\right)$

$\theta=(12.005+\pi / 4)$

$=(12.005+\pi / 4) \times 180 / \pi$

$=12.79 \times 180 / 3.14$

$\theta=12.79 \times 57.32=733.18$

$v=90 \cos (733.18)$

$=90 \cos (720+13.18)$

$=90 \cos 13.18$

$=90 \times 0.97 \mathrm{~cm} / \mathrm{s}$

$=87.63 \mathrm{~cm} / \mathrm{s}$

The standard equation is given as

$y(x, t)=t \sin \left[\frac{\pi}{4}(v t+x)+\phi_{0}\right]$

We get $y(x, t)=a \sin (k x+\omega t+\phi)$

here $\lambda$ is the wavelength

The transverse displacement and velocity of all locations at a distance of $\pm \lambda, \pm 2 \lambda, \ldots$ from $x=1 \mathrm{~cm}$ will be the same. Because $\pm 12.56 \mathrm{~m}, \pm 25.12 \mathrm{~m}, \ldots$ $x=1cm$ will have the same displacement as the $x=1 cm$ points at $t=2,5,11$.