## Q. 4.11

A wave in a tsunami has a period of 30 minutes and a height, $H_{o}$, of 0.5 m at a point where the ocean has a depth of 4 km deep. Calculate the phase speed, $c_{o}$, and wavelength, $L_{o}$, of this wave. Calculate its phase speed, $c_{i}$, wavelength, $L_{i}$, and height, $H_{i}$, in a coastal water depth of 15 m accounting for shoaling effects only.

## Verified Solution

We assume the wave is a shallow water wave so that

$c_{o}=\sqrt{gh}=\sqrt{9.81\times 4000}=198 ms^{-1}\\L_{o} =C_{o}T=198\times 30\times 60=356,400 m.$

The depth to wavelength ratio is 4000/356,400 = 0.011; thus, the wave is a shallow water wave at a depth of 4000 m. The wave steepness is extremely small and is equal to $H_{0}/L_{0} = 1.4 × 10^{−6}$. To determine the inshore wave characteristics we assume there is negligible energy dissipation and equate the wave power in deep and nearshore sites. Thus, since for shallow water waves the group velocity is equal to the phase velocity, the conservation of wave power gives

$H^{2}_{o}\sqrt{gh_{0} } =H^{2}_{i}\sqrt{gh_{i} }$

So at a depth of 15 m

$H_{i} =0.5\left\lgroup\frac{4000}{15} \right\rgroup ^{0.25} =2.02 m \\c_{i} =\sqrt{9.81\times 15} =12.13 ms^{-1} \\L_{i} =12.13\times 30\times 60= 21,834 m$

This simple example demonstrates two important points. First, the dramatic reduction in phase speed (by at least one order of magnitude). Second, the significant amplification of the height of the wave as it propagates into shallow water.