Question

the angular frequency of the surface waves in a liquid is given in terms of wave number k by omega=ROOt of gk+tk^3÷rho ehere g is acceleraion due to gravity rho is the density of liquid and T is the surface tension (which gives an upward force on element of rhe surgave liquid).Find phase and group velocities for limitimg cases when the surface wa es have very large wave length and very small wavelengrhs.​

Answer 1

Answer:

Select the correct alternative(s) :

a) the amplitude of component waves is 2 mm

b) the amplitude of component waves is 4mm

c) the smallest possible length of string is 0.5 m

d) the smallest possible length of string is 1.0 m

1) a, care correct 2) b, c are correct

3) a, d are correct

Answer 2

Answer:

The phase and group velocities for the given dispersion relation in

1. Very large wavelength limit:

                         $v_{ph}=\sqrt[]{\frac{g}{k} }$       $v_{g}=\frac{1}{2} \sqrt{\frac{g}{k} }$

    2. Very small wavelength limit:  

                        $v_{ph}=\sqrt[]{\frac{Tk}{\rho} }$      $v_{g}=\frac{3}{2} \sqrt{\frac{Tk}{\rho} }$

Explanation:

Dispersion Relation:

• These are equations that relate wavelengths or wave numbers to their frequency.
• In general, they are denoted by $\omega (k)$
• The phase velocity($v_{ph}$) is given by

                              $v_{ph}=\frac{\omega}{k}$

• The group velocity ($v_{g}$) is given by

                              $v_{g}=\frac{d\omega}{dk}$

The steps to find phase and group velocities for the respective limiting cases are given below

Step 1:

The relation between wavelength and wavenumber is as follows

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

From this equation, we can see the wavenumber is inversely proportional to its wavelength. Therefore for very large wavelengths, we consider very small 'k and vice versa.

Step 2:

Consider the given dispersion relation:

                            $\omega=\sqrt{gk+\frac{tk^{3} }{\rho} }$

From the RHS we take  $k^{2}$ common and bring it to the LHS

                            $\frac{\omega}{k} =\sqrt{\frac{g}{k} +\frac{tk }{\rho} }$  ----------------(1)

Step 3:

As we know the formula for phase velocity we get,

                           $v_{ph}= \frac{\omega}{k} =\sqrt{\frac{g}{k} +\frac{tk }{\rho} }$

This is the general equation for the phase velocity of surface waves in a liquid.

Step 4:

In the very large wavelength limit or very small wavenumber limit, the term $\frac{Tk}{\rho}$ will be less than the term $\frac{g}{k}$. Hence, we can neglect the smaller term and so we have,

                          $v_{ph}=\sqrt{\frac{g}{k}}$

Step 5:

In the very small wavelength limit or very large wavenumber limit, the term $\frac{Tk}{\rho}$ will be greater than the term $\frac{g}{k}$. Hence, we can neglect the smaller term and so we have,

                          $v_{ph}=\sqrt{\frac{tk }{\rho} }$

Step 5:

In order to find the group velocity we first square the dispersion relation on both sides.

                          $\omega^{2}=(gk+\frac{Tk}{\rho})$ -----------------------(2)

Step 6:

Differentiating equation (2) wrt k,

                          $2\omega \frac{d\omega}{dk} = g+\frac{3Tk^{2} }{\rho} \\\ v_{g} =\frac{d\omega}{dk}=\frac{1}{2\omega} (g+\frac{3Tk^{2} }{\rho} )$ -------------------(3)

Step 7:

By substituting $\omega$ in equation (3) for the corresponding limiting cases, we will get the group velocity for the two limiting cases.

For very large wavelength limit or very small wavenumber limit, we substitute $\frac{\omega}{k} =\sqrt{\frac{g}{k}}$ in equation (3) and neglect the term $\frac{3Tk^{2} }{\rho}$ as it is small compared to the term $g$, so we have,

                           $v_{g} =\frac{d\omega}{dk}=\frac{g}{2\omega} \\v_{g}=\frac{1}{2}\sqrt{\frac{g}{k} }$

Step 8:

In the very small wavelength limit or very large wavenumber limit, we substitute $v_{ph}=\sqrt{\frac{tk }{\rho} }$  in equation (3) and neglect the term $g$ as it is small compared to the term $\frac{3Tk^{2} }{\rho}$ , so we have,

                         $v_{g} =\frac{d\omega}{dk}=\frac{1}{2\omega} (\sqrt{\frac{3Tk^{2} }{\rho})$

                          $v_{g}=\frac{3}{2}\sqrt{\frac{Tk}{\rho} }$

Therefore,

1. Very large wavelength limit:

                         $v_{ph}=\sqrt[]{\frac{g}{k} }$       $v_{g}=\frac{1}{2} \sqrt{\frac{g}{k} }$

    2. Very small wavelength limit:  

                        $v_{ph}=\sqrt[]{\frac{Tk}{\rho} }$      $v_{g}=\frac{3}{2} \sqrt{\frac{Tk}{\rho} }$