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202209281002 Exercise 14.2 (Q24)

Analysis of waves in shallow water (depth much less than wavelength) yields the following wave equation:

\displaystyle{\frac{\partial^2y}{\partial x^2}=\frac{1}{gh}\frac{\partial^2y}{\partial t^2}}

where h is the water depth and g the gravitational acceleration. Give an expression for the wave speed.

Extracted from R. Wolfson. (2016). Essential University Physics.

(top-down approach)

Write a traveling sinusoidal wave in the form

y(x,t)=A\cos (kx\pm \omega t)

the wave speed v being such as

\displaystyle{v=\frac{\lambda}{T}=\frac{2\pi /k}{2\pi /\omega}=\frac{\omega}{k}}.

\begin{aligned} \textrm{LHS} & = \frac{\partial^2y}{\partial x^2} \\ & = \frac{\partial}{\partial x}\bigg(\frac{\partial}{\partial x}\big( A\cos (kx\pm\omega t)\big) \bigg) \\ & = \frac{\partial}{\partial x}\big( kA\sin (kx\pm\omega t)\big) \\ & = -k^2A\cos (kx\pm\omega t) \\ \textrm{RHS} & = \frac{1}{gh}\frac{\partial^2y}{\partial t^2} \\ & = \frac{1}{gh}\frac{\partial}{\partial t}\bigg(\frac{\partial}{\partial t}\big( A\cos (kx\pm\omega t)\big)\bigg) \\ & = \frac{1}{gh}\frac{\partial}{\partial t}\big(\omega A\sin (kx\pm\omega t)\big) \\ & = \frac{1}{gh}\big( -\omega^2A\cos (kx\pm\omega t)\big) \\ \textrm{LHS} & = \textrm{RHS} \\ \Rightarrow v = \frac{\omega}{k} & = \sqrt{gh} \\ \end{aligned}

\therefore The wave speed v is given by v=\sqrt{gh}.