91Ó°ÊÓ

The given wave equation, $y(\mathrm{x},\mathrm{t})=(4.00)h[\left(30\right)\mathrm{x}+(6.0)\mathrm{t}]$

By comparing the given wave equation with the standard form, we can find the wavenumber and angular velocity. Using these values, we can find the speed of the wave.

Formula:

The general expression of the wave, $y\left(x,t\right)=y_m\sin \left(kx-\mathrm{Ó\neg }t\right)$ (i)

The speed of the wave, $v=\mathrm{Ó\neg }/k$ (ii)

Here, $y$is displacement,$y_m$ is the amplitude of the wave, $k$is the angular wave number, $\mathrm{Ó\neg }$is the Angular frequency of the wave,$t$ is time.

By comparingthegiven equation withthesolution of the wave equation (i)

$y(\mathrm{x},\mathrm{t})=(4.00)h[\left(30\right)\mathrm{x}+(6.0)\mathrm{t}]$

We can get,

1. Angular frequency of the wave is$\mathrm{Ó\neg }=6.0\mathrm{rad}/\mathrm{s}$
2. Angular wave number$k=30.0{\mathrm{m}}^{-1}$

Using the equation (ii), we get the speed of the wave as:

$$\begin{gathered} v=\frac{6.0}{30.0} \\ =0.2\mathrm{m}/\mathrm{s} \end{gathered}$$

Hence, the value of the speed is 0.2 m/s .