Question

For waves on the surf ace of water, the behaviour of long wavelengths is dominated by gravitational effects-a liquid "seeking its own level." Short wavelengths are dominated by surface tension effects. Taking both into account, the dispersion relation is$\mathit{\omega }\mathbf{=}\sqrt{\mathbf{g}\mathbf{k}\mathbf{+}\mathbf{(}\mathbf{\gamma }\mathbf{/}\mathbf{\rho }\mathbf{)}{\mathbf{k}}^{\mathbf{3}}}$. where$\mathit{\gamma }$is the surface tension,p is the density of water, and gis, of course, the gravitational acceleration?

1. Make a qualitative sketch of group velocity versus wave number. How does it behave for very large k? For very small k?
2. (b) Find the minimum possible group velocity and the wavelength at which it occurs.Use$\mathit{\gamma }\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{072}\mathbf{ }\mathbf{\text{N/m}}$,$\mathit{\rho }\mathbf{=}{\mathbf{10}}^{\mathbf{3}}\mathbf{ }{\mathbf{\text{kg/m}}}^{\mathbf{\text{3}}}$and$\mathit{g}\mathbf{=}\mathbf{9}\mathbf{.}\mathbf{8}\mathbf{ }{\mathbf{\text{m/s}}}^{\mathbf{\text{2}}}$.

Short answer

1. A qualitative sketch of group velocity versus wave number is

b. The group velocity is $0.177 \text{m/s}$and the wavelength is$0.0433 \text{m}$

Step-by-step solution

Concept involved

Group velocity is the velocity of envelop in which the wave is contained, it can be found by finding the differential of angular frequency.

Wavelength can be calculated by

$\lambda =\frac{2\pi }{\sqrt{\frac{\rho g}{\gamma }\left(\frac{2\sqrt{3}}{3}-1\right)}}$

Where, gacceleration due to gravity, yis the surface tension andp is the density of water

Given equation

$\omega =\sqrt{gk+(\gamma /\rho )k^3}$

(a) Plotting Graph of Group Velocity Vs Wave Number

The group velocity ( $v_{group}$) is found by taking the derivative of the angular frequency with respect to wave number

$\begin{array}{rcl}{v}_{group}& =& \frac{d\omega }{dk}\\ & =& \frac{1}{2}{\left(gk+(\gamma /\rho )k^3\right)}^{-\frac{1}{2}}\left(g+\left(3\gamma k^2/\rho \right)\right)\\ & =& \frac{g+\frac{3\gamma k^2}{\rho }}{2\sqrt{gk+\left(\gamma k^2/\rho \right)}}\\ & & \end{array}$

This graph has Group Velocity on its y-axis and Wave Number on its x-axis.

As clear from the formula that the Group Velocity approaches infinity when k approaches zero or infinity, it can also be seen in the above-mentioned graph.

(b) Determining minimum the group velocity

The extreme values of the group velocity will occur when the derivative of the group velocity with respect to is equal to 0

$\begin{array}{rcl}\frac{d{v}_{group}}{{\displaystyle dk}}& =& 0\\ \frac{\frac{6\gamma }{\rho }k\left(2\sqrt{gk+\gamma k^3/\rho }\right)-2\left(g+\frac{3\gamma }{\rho }{k}^2\right)\frac{1}{2}{\left(gk+\gamma k^3/\rho \right)}^{-1/2}\left(g+3\gamma k^3/\rho \right)}{4\left(gk+\gamma k^3/\rho \right)}& =& 0\\ \frac{3\gamma k}{\rho \sqrt{gk+\gamma k^3/\rho }}-\frac{{\left(g+\frac{3\gamma }{\rho }{k}^2\right)}^2}{4{\left(gk+\gamma k^3/\rho \right)}^{3/2}}& =& 0\\ \frac{12\gamma k\left(gk+\gamma k^3/\rho \right)}{\rho }& =& {\left(g+\frac{3\gamma k^2}{\rho }\right)}^2\\ \frac{12\gamma k\left(gk+\gamma k^3/\rho \right)}{\rho }& =& g^2+6\frac{g\gamma k^2}{\rho }+9\frac{{\gamma }^2{k}^2}{{\rho }^4}\\ \frac{12\gamma k^2}{\rho }+\frac{12{\gamma }^2{k}^4}{{\rho }^2}& =& g^2+6\frac{g\gamma k^2}{\rho }+9\frac{{\gamma }^2{k}^2}{{\rho }^2}\\ 3\frac{{\gamma }^2{k}^4}{{\rho }^2}+6\frac{g\gamma k^2}{\rho }-g^2& =& 0\end{array}$

Use the above-mentioned quadratic equation to produce two solutions for the potential values of $k^2$

$$\begin{gathered} k^2=\frac{\frac{-6gy}{\rho }\pm \sqrt{\frac{36g^2{\gamma }^2}{{\rho }^2}-4\left(\frac{3y^2}{{\rho }^2}\right)\left(-g^2\right)}}{\frac{6{\gamma }^2}{{\rho }^2}} \\ {k}^2=-\frac{\rho g}{\gamma }\pm \frac{{\rho }^2}{6{\gamma }^2}\sqrt{\frac{12g^2{\gamma }^2(3+1)}{{\rho }^2}} \\ {k}^2=-\frac{\rho g}{\gamma }\pm \frac{2\rho g}{3\gamma }\sqrt{3} \\ {k}^2=\frac{\rho g}{\gamma }\left(-1\pm \frac{2\sqrt{3}}{3}\right) \\ k=\sqrt{\frac{\rho g}{\gamma }\left(\frac{2\sqrt{3}}{3}-1\right)} \end{gathered}$$

Now use the value of k to calculate the minimum group velocity

$\begin{array}{rcl}{v}_{group}& =& \frac{g+\frac{\frac{\rho g}{r}\left(\frac{2\sqrt{3}}{3}-1\right)}{\rho }}{2\sqrt{g{\left(\frac{pg}{\gamma }\left(\frac{2\sqrt{3}}{3}-1\right)\right)}^{1/2}+\left(\gamma {\left(\frac{\rho g}{\gamma }\left(\frac{2\sqrt{3}}{3}-1\right)\right)}^{3/2}I\rho \right)}}\\ & =& \frac{g+g\left(2\sqrt{3}-3\right)}{2\sqrt{{\left(\frac{\rho g^3}{\gamma }\left(\frac{2\sqrt{3}}{3}-1\right)\right)}^{1/2}\left(1+\left(\frac{2\sqrt{3}}{3}-1\right)\right)}}\\ & =& \frac{\sqrt{3}g-g}{\sqrt{{\left(\frac{\rho g^3}{\gamma }\left(\frac{2\sqrt{3}}{3}-1\right)\right)}^{1/2}\left(\frac{2\sqrt{3}}{3}\right)}}\\ & & \end{array}$

Finally, substitute the given values of the density of water $\left(\rho \right)$, acceleration of gravity () and the surface tension of water$\left(\gamma \right)$ then

$\begin{array}{rcl}{v}_{group}& =& \frac{\sqrt{3}\left(9.8 {\text{m/s}}^{\text{2}}\right)-\left(9.8 {\text{m/s}}^{\text{2}}\right)}{\sqrt{{\left(\frac{\left({10}^6 {\text{g/m}}^{\text{3}}\right){\left(9.8 {\text{m/s}}^{\text{2}}\right)}^3}{72 {\text{g/s}}^{\text{2}}}\left(\frac{2\sqrt{3}}{3}-1\right)\right)}^{1/2}\left(\frac{2\sqrt{3}}{3}\right)}}\\ & =& 0.177 \text{m/s}\\ & & \end{array}$

(b) Determine wavelength

The wavelength corresponding to the slowest group velocity can be calculate to the wave number corresponding to the slowest group velocity

$\begin{array}{rcl}\lambda & =& \frac{2\pi }{\sqrt{\frac{\rho g}{r}\left(\frac{2\sqrt{3}}{3}-1\right)}}\\ & =& \frac{2\pi }{\sqrt{\frac{\left({10}^6 {\text{g/m}}^{\text{3}}\right)\left(9.8 {\text{m/s}}^{\text{2}}\right)}{72 {\text{g/s}}^{\text{2}}}\left(\frac{2\sqrt{3}}{3}-1\right)}}\\ & =& 0.0433 \text{m}\\ & & \end{array}$

Hence,

The group velocity is$0.177 \text{m/s}$

The wavelength is$0.0433 \text{m}$