Past exam of the mathematics course of the University of Cambridge / 2004 / ib / Paper 3 / 10C (Fluid Dynamics)

For velocity $\mathbf{u}=\nabla \varphi$ and a conservative body force $-\nabla \chi$, the unsteady Bernoulli equation isIt follows by integrating the inviscid Euler equations in space; a purely time-dependent potential gauge can absorb $C(t)$.

Use a coordinate along the thin U-tube from the left surface to the right, and let the right surface rise by $\zeta$ while the left falls by $\zeta$. Incompressibility and uniform area give velocity $\dot{\zeta }$ along the whole fluid column. If its total length is $L$, the velocity-potential difference between the surfaces is $L\dot{\zeta }$; the column length stays constant as their displacements cancel. Both surfaces have atmospheric pressure and the same speed. Subtracting their Bernoulli balances, with $\chi =gy$, therefore yieldsIn the long-leg approximation underlying the requested formula, the base bend contributes negligible length and $L=2h$. ConsequentlyIf the horizontal connector or bend has nonnegligible length, it belongs in $L$ and the general frequency is $\sqrt{2g/L}$; the stated height alone would then not determine the inertia.