Solution of the Dispersion Relationship

$$\omega ^{2} = gk \tanh kh$$

Property of $\tanh kh$:

$$\tanh kh = \frac{\sinh kh}{\cosh kh} = \frac{1 - e^{ - 2kh}}{... ...all { short waves or deep water}}) \\ \end{array}} \right.$$

1.

Deep water or short waves, $kh >> 1 (h > \sim \lambda/2)$

$$\begin{align}& \omega ^{2} = gk,\ \lambda = \frac{g}{2\pi }T^2 (\lambda \mbox{(... ...frac{2\pi }{g}V_p^2 \leftarrow \mbox{\ frequency dispersion} \notag \end{align}$$

2.

Shallow water or long waves, $kh << 1,\ ( h /\lambda < \sim 1 / 20 \mbox{in practice})$

$$\begin{align}& \omega ^2 \cong gk \cdot kh \to \omega = \sqrt {gh} \ k ;\ \lambd... ...{T} = \sqrt {gh} \leftarrow \mbox{no frequency dispersion } \notag \end{align}$$

3.

Intermediate depth or wavelength. Need to solve

$$\omega ^{2} = gk \tanh kh$$

given $\omega , h$ for $k$ (given $k, h$ for $\omega$ - easy!)

(a)

Use tables or graphs (e.g. JNN fig. 6.3)

$$\omega ^{2 \quad } = gk \tanh kh = gk_{\infty }, \mbox{\ then... ... }{\lambda _\infty } = \frac{V_p }{V_{p \infty} } = \tanh kh$$

(b)

Use numerical approximation (hand calculator)

i.

calculate $C = \omega ^{2}h/g$.

ii.

$$\left. \begin{array}{lcr} \mbox{If\ } C > 2, & kh \approx C(1... ...x{ (shallower)} \end{array}\right\} \mbox{about 4 decimals}$$

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