Question

For wavelengths less than about 1 cm, the dispersion relation for waves on the surface of water is $\mathbf{\omega }\mathbf{=}\sqrt{\left(\gamma /p\right){\mathbf{k}}^{\mathbf{3}}}$, whereandare the surface tension and density of water. Given$\mathbf{\gamma }\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{072}\mathbf{N}\mathbf{/}\mathbf{m}$and$\mathbf{p}\mathbf{=}{\mathbf{10}}^{\mathbf{3}}\mathbf{kg}\mathbf{/}{\mathbf{m}}^{\mathbf{3}}$, calculate the phase and group velocities for a wave of 5mm wavelength.

Short answer

The phase velocity for a wave is 0.3 m/s

The group velocity for a wave is 0.45 m/s

Step-by-step solution

 Concept involved

Waves of different wavelengths travel at different phase speeds. Water waves on the surface propagate with gravity and surface tension as the restoring forces.

 Formulae used        

The dispersion relation for waves on the surface of water is

$\omega =\sqrt{\left(\gamma /p\right)k^3}$

Where,$\gamma$and p are the surface tension and density of water

The phase velocity is,

$v_{phase}=\frac{\sqrt{\left(\gamma /p\right)k^3}}{k}$

The group velocity is,

$v_{group}={\overline{)\frac{d\omega }{dk}}}_{k_0}$

 Determining the wave number

The wave number, k is given by

$k=\frac{2\mathrm{\pi }}{\lambda }$

Where, $\lambda$= wavelength

$$\begin{gathered} \mathrm{k}=\frac{2\mathrm{\pi }}{5\times {10}^{-3}\mathrm{m}} \\ =1.265\times {10}^3{\mathrm{m}}^{-1} \end{gathered}$$

 Calculating the angular frequency

Calculate the angular frequency$\omega$, using the wave number k, surface tension of water$\gamma$, and the density of water p

$\omega =\sqrt{\left(\frac{\gamma }{p}\right)k^3}$

Substitute the values of $\gamma$, pand k to get $\omega$

$$\begin{gathered} \mathrm{\omega }=\sqrt{\left(\frac{0.072\mathrm{kg}/{\mathrm{s}}^2}{{10}^3\mathrm{kg}/{\mathrm{m}}^3}\right){\left(1.265\times {10}^{-3}{\mathrm{m}}^{-1}\right)}^3} \\ =377.99\mathrm{rad}/\mathrm{s} \end{gathered}$$

 Calculating the phase velocity

$$\begin{gathered} v_{phase}=\sqrt{\frac{\left(\gamma /p\right)k^3}{k}} \\ {v}_{phase}=\sqrt{\left(\frac{\gamma }{p}\right)k} \end{gathered}$$

Substitute the values of $\gamma$, pand k

$$\begin{gathered} {\mathrm{v}}_{\mathrm{phase}}=\sqrt{\left(\frac{0.072\mathrm{kg}/{\mathrm{s}}^2}{{10}^3\mathrm{kg}/{\mathrm{m}}^3}\right)\left(1.265\times {10}^{-3}{\mathrm{m}}^{-1}\right)} \\ =0.3\mathrm{m}/\mathrm{s} \end{gathered}$$

 Calculating the group velocity

Group velocity is calculated by taking a derivation of the angular frequency$\omega$with respect to the wave number k

$$\begin{gathered} {\mathrm{v}}_{\mathrm{group}}={\overline{)\frac{\mathrm{d\omega }}{\mathrm{dk}}}}_{{\mathrm{k}}_0} \\ =\frac{1}{2}{\left(\frac{{\mathrm{\gamma k}}^3}{\mathrm{p}}\right)}^{-1/2}{\overline{)\left(\frac{3{\mathrm{\gamma k}}^2}{\mathrm{p}}\right)}}_{{\mathrm{k}}_0} \\ =\frac{3}{2}\sqrt{\frac{{\mathrm{\gamma k}}_0}{\mathrm{p}}} \end{gathered}$$

Use the wave number k as the median wave number $k_0$

$$\begin{gathered} {\mathrm{v}}_{\mathrm{group}}=\frac{3}{2}\sqrt{\frac{(0.072\mathrm{kg}/{\mathrm{s}}^2)(1.265\times {10}^{-3}{\mathrm{m}}^{-1})}{{10}^3\mathrm{kg}/{\mathrm{m}}^3}} \\ =0.45\mathrm{m}/\mathrm{s} \end{gathered}$$

 Conclusion        

Therefore, phase velocity for a wave is 0.3 m/s

The group velocity for a wave is 0.45 m/s