(a) Write an equation describing a sinusoidal transverse wave traveling on a cord in the positive direction of a yaxis with an angular wave number of 60 ${\mathrm{cm}}^{-1}$, a period of 0.20 s, and an amplitude of 3.0 mm. Take the transverse direction to be thedirection. (b) What is the maximum transverse speed of a point on the cord?

Wave number,$k=60{\mathrm{cm}}^{-1}\mathrm{or}6\times {10}^3{\mathrm{m}}^{-1}$

Period of oscillation, T = 0.20 s

Amplitude of the wave,$z_m=3.0\mathrm{mm}\mathrm{or}3.0\times {10}^{-3}\mathrm{m}$

We write an equation describing a sinusoidal transverse wave traveling on a cord in the positive direction of the y axis from the given wavenumber, period, and amplitude. By differentiating this equation with time, we can find the maximum transverse speed of a point on the cord.

Formula:

The general expression of the wave, $z\left(y,t\right)=z_m\sin \left(ky-\omega t\right).......\left(1\right)$

The transverse velocity of the wave, $u=dz/dt.........\left(2\right)$

The angular frequency of the wave, $\omega =2\pi /T.........\left(3\right)$

The equation traveling on a cord in the positive direction of y axis is given using equation (1).

From the given values, we get

$\begin{array}{l}k=60{\mathrm{cm}}^{-1}\\ Z_n=3.0\mathrm{mm}\end{array}$

Hence, the angular frequency of the wave using equation (3) is given as:

$\begin{array}{l}\omega =\frac{2\pi }{0.20s}\\ =10\mathrm{\pi }{s}^{-1}\end{array}$

Thus, the equation of wave after substituting the above values in equation (1), we get

$\mathrm{z}\left(\mathrm{y},\mathrm{t}\right)=\left(30\mathrm{mm}\right)\sin \left[\left(60{\mathrm{m}}^{-1}\right)\mathrm{y}-\left(10\mathrm{\pi }{\mathrm{s}}^{-1}\right)\mathrm{t}\right]$

Therefore, an equation describing a sinusoidal transverse wave traveling on a cord in the positive direction of y axis is$\mathrm{z}\left(\mathrm{y},\mathrm{t}\right)=\left(30\mathrm{mm}\right)\sin \left[\left(60{\mathrm{m}}^{-1}\right)\mathrm{y}-\left(10\mathrm{\pi }{\mathrm{s}}^{-1}\right)\mathrm{t}\right]$

Using equation (2) and the displacement equation, we get the transverse speed as:

$$\begin{gathered} u=\frac{d}{dt}\left[z_m\sin \left(ky-\omega t\right)\right] \\ =\omega z_m \\ =\left(10\pi s^{-1}\right)\left(3.0\mathrm{mm}\right) \\ =94\mathrm{mm}/\mathrm{s} \end{gathered}$$

Therefore, the maximum transverse speed of a point on the cord is 94 mm/s