Question

For wavelengths greater than about,$\mathbf{20}\mathbf{}\mathbf{ }\mathbf{\text{cm}}$ the dispersion relation for waves on the surface of water is$\mathit{\omega }\mathbf{=}\sqrt{\mathbf{g}\mathbf{k}}$

(a) Calculate the phase and group velocities for a wave ofwavelength.

(b) Will the wave spread as it travels? Justify your answer.

Short answer

a. The phase velocity is$2.79m/s$

The group velocity is$1.40m/s$

b. The wave will spread because the dispersion relationship is non – linear so the second derivative, the dispersion coefficient is zero

Step-by-step solution

Concepts involved

Phase velocity is the velocity in which a wave propagates in a medium.

${\nu }_{phase}=\frac{\omega }{k}$

Group velocity is the speed at which the whole envelop of the wave moves. It can be found by the derivative of angular frequency.

${\nu }_{group}=\frac{d\omega }{dk}$

If second derivative of the function is zero, the function is a constant function.

If the second derivative of a function is not zero, then the function will go either concave upwards or concave downwards. That will not be a constant function.

Given quantities and equations

Wavelength,$\lambda =5 \text{m}$

The dispersion relation for wavelength greater than$20 \text{cm}$

$\omega =\sqrt{gk}$

(a) Calculating the phase velocity

As mentioned in the question, dispersion relation for waves on the surface of water is

$\omega =\sqrt{gk}$

Where,

g is acceleration due to gravity and k is wave vector.

The phase velocity is equal to

$\begin{array}{rcl}{\nu }_{phase}& =& \frac{\omega }{k}\\ & =& \frac{\sqrt{gk}}{k}\\ & =& \sqrt{\frac{g}{k}}\\ & =& \sqrt{\frac{g\lambda }{2\pi }}\\ & & \end{array}$

$\left[Since, k=\frac{2\pi }{\lambda }\right]$

Where,$\lambda$ = Wavelength

Substitute the value of g and $\lambda$in the above equation

$\begin{array}{rcl}{v}_{phase}& =& \sqrt{\frac{\left(9.8 {\text{m/s}}^{\text{2}}\right)(5 \text{m})}{2\pi }}\\ & =& 2.79 \text{m/s}\\ & & \end{array}$

(a) calculating the group velocity

The group velocity is equal to

${\nu }_{group}=\frac{d\omega }{dk}$

Now substituting the value offrom the given equation$\omega =\sqrt{gk}$

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

$v_{phase}=\sqrt{\frac{g}{k}}$, so, the above equation can be rewritten as

$v_{group}=\frac{1}{2}{v}_{phase}$

Substitute the value of $v_{phase}$in above equation

$v_{group}=1.40 \text{m/s}$

The phase velocity is$2.79 \text{m/s}$

The group velocity is$1.40 \text{m/s}$

(b) Determining spreading of wave as it travels

The dispersion relation$\omega =\sqrt{gk}$is non – linear so the second derivative, which is also known as the dispersion coefficient is not zero.

Hence, yes, the wave will spread.