91Ó°ÊÓ

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, $y\left(x,t\right)=y_m\sin \left(kx-\mathrm{Ó\neg }t\right)........\left(1\right)$

The angular frequency of the wave, $\mathrm{Ó\neg }=2\mathrm{Ï€}f.........\left(2\right)$

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

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:

$$\begin{gathered} \mathrm{Ó\neg }=2×\mathrm{Ï€}×550 \\ =3.46×{10}^3\mathrm{rad}/\sec \end{gathered}$$

Hence, the value of$\mathrm{Ó\neg }$ is$3.46×{10}^3\mathrm{rad}/\sec$

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

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

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.