A sinusoidal transverse wave traveling in the negative direction of anaxis has amplitude of 1.00 cm, a frequency of 550 Hz, and a speed of 330 m/s. If the wave equation is of the form$\mathit{y}\left(x,t\right)\mathbf{=}{\mathit{y}}_{\mathbf{m}}\mathbf{}\mathit{s}\mathit{i}\mathit{n}\left(kx\pm \omega t\right)$, What are (a)${\mathit{y}}_{\mathbf{m}}$, (b)$\mathit{\omega }$, (c)$\mathit{k}$, and (d) the correct choice of sign in front of$\mathit{\omega }$?

Amplitude of the wave,$y_m=1.00\mathrm{cm}$

Frequency of the wave,$f=550\mathrm{Hz}$

Speed of the wave,$v=330\mathrm{m}/\mathrm{s}$

We have to compare the given values with the general equation of the wave.

Formula:

The general equation of wave, $y\left(x,t\right)=y_m\sin \left(kx-\omega t\right)........\left(1\right)$

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

The wavenumber of the wave, $k=\frac{\omega }{v}........\left(3\right)$

Comparing with the wave equation (1), it can be seen that here$y_m$is amplitude of the wave.

So value of $y_m$is$1.00\mathrm{cm}$

The frequency is given, so the angular frequency using equation (2) is given as follows:

$$\begin{gathered} \omega =2\times \mathrm{\pi }\times 550 \\ =3.46\times {10}^3\mathrm{rad}/\sec \end{gathered}$$

Hence, the value of$\omega$ is$3.46\times {10}^3\mathrm{rad}/\sec$

The velocity of a wave is given and hence, using equation (3), we get the wavenumber as follows:

$$\begin{gathered} k=\frac{3.46\times {10}^3}{330} \\ =10.5\mathrm{rad}/\mathrm{m} \end{gathered}$$

Hence, the value of$k$ is$10.5\mathrm{rad}/\mathrm{m}$

Comparing with general equation (1), it can be seen that the sign to angular frequency must be plus since the wave is travelling in the negative direction.