Paper 4, Section II, D

Fluid Dynamics | Part IB, 2009

An inviscid incompressible fluid occupies a rectangular tank with vertical sides at x=0,ax=0, a and y=0,by=0, b and a horizontal bottom at z=hz=-h. The undisturbed free surface is at z=0z=0.

(i) Write down the equations and boundary conditions governing small amplitude free oscillations of the fluid, neglecting surface tension, and show that the frequencies ω\omega of such oscillations are given by

ω2g=ktanhkh, where k2=π2(m2a2+n2b2)\frac{\omega^{2}}{g}=k \tanh k h, \text { where } k^{2}=\pi^{2}\left(\frac{m^{2}}{a^{2}}+\frac{n^{2}}{b^{2}}\right)

for non-negative integers m,nm, n, which cannot both be zero.

(ii) The free surface is now acted on by a small external pressure

p=ϵρghsinΩtcosmπxacosnπyb,p=\epsilon \rho g h \sin \Omega t \cos \frac{m \pi x}{a} \cos \frac{n \pi y}{b},

where ϵ1\epsilon \ll 1. Calculate the corresponding oscillation of the free surface when Ω\Omega is not equal to the quantity ω\omega given by (1).

Why does your solution break down as Ωω?\Omega \rightarrow \omega ?

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