91Ó°ÊÓ

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

${\mathrm{ν}}_{phase}=\frac{\mathrm{Ó\neg }}{k}$

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

${\mathrm{ν}}_{group}=\frac{d\mathrm{Ó\neg }}{dk}$

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.

Wavelength,$\mathrm{λ}=5 \text{m}$

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

$\mathrm{Ó\neg }=\sqrt{gk}$

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

$\mathrm{Ó\neg }=\sqrt{gk}$

Where,

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

The phase velocity is equal to

$\begin{array}{rcl}{\mathrm{ν}}_{phase}& =& \frac{\mathrm{Ó\neg }}{k}\\ & =& \frac{\sqrt{gk}}{k}\\ & =& \sqrt{\frac{g}{k}}\\ & =& \sqrt{\frac{g\mathrm{λ}}{2\mathrm{Ï€}}}\\ & & \end{array}$

$\left[Since, k=\frac{2\mathrm{π}}{\mathrm{λ}}\right]$

Where,$\mathrm{λ}$ = Wavelength

Substitute the value of g and $\mathrm{λ}$in the above equation

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

The group velocity is equal to

${\mathrm{ν}}_{group}=\frac{d\mathrm{Ó\neg }}{dk}$

Now substituting the value offrom the given equation$\mathrm{Ó\neg }=\sqrt{gk}$

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

$v_{phase}=\sqrt{\frac{g}{k}}$, 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$1.40 \text{m/s}$

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

Hence, yes, the wave will spread.