Past exam of the physics B course of the University of Cambridge / 2015 / Paper 1 / Section C / 13 / Solution

For a Newtonian fluid with velocity $u(z)\hat{\mathbf{x}}$, $\tau =\eta {\partial }_{z}u$ is the $x$ component of traction on a plane whose normal points along $+z$. The opposite face has the opposite traction. The coefficient $\eta$ is dynamic viscosity, with units $\mathrm{Pa}\mathrm{s}$. Faster layers transfer positive $x$ momentum toward slower neighbours; the internal shear stress resists relative sliding.

Take a thin fluid element with horizontal area $A$ and thickness $dz$. The difference between its upper and lower viscous tractions is $A\eta u_{zz}dz$. The pressure gradient supplies $-A{p}_{x}dz$. Incompressibility is satisfied because $u$ has no $x$ dependence, and the convective acceleration vanishes: $(\mathbf{u}\cdot \nabla )\mathbf{u}=u{\partial }_x\mathbf{u}=0$. Equating net force to mass times acceleration therefore gives the relevant component of the Navier-Stokes equation:For steady flow, integrate $\eta u^{\prime \prime }=p^{\prime }$ twice and impose the no-slip boundary condition at each plate, $u(0)=0$ and $u(a)=u_0$. The Couette-Poiseuille flow profile isThe linear term is Couette flow and the parabolic term is plane Poiseuille flow. A negative $p^{\prime }$ drives flow in $+x$. Integrating the profile across the gap gives the volumetric flow per unit width,It has units ${\mathrm{m}}^2{\mathrm{s}}^{-1}$; multiplying by the plate width gives the actual volume per time.

With zero pressure gradient, substituting $u=\mathrm{Re}[A{e}^{i(kz-\omega t)}]$ gives $-i\rho \omega =-\eta k^2$. Select the root that decays into fluid occupying $z>0$:The other sign applies to decay in the opposite half-space. With a real wall velocity amplitude $U_0$, the established periodic solution of Stokes' second problem isThe oscillation's amplitude decays exponentially and its phase lags with distance from the plate. This is diffusion of tangential momentum, not a freely propagating undamped elastic wave. Initial transients must have decayed before this purely periodic expression applies.

The fluid-displacement amplitude and the wall-displacement amplitude have the same ratio as their velocity amplitudes, since both are divided by the common $\omega$. Thus for the specified amplitudes,Using the standard approximate water density $\rho \simeq 10^3\mathrm{kg}{\mathrm{m}}^{-3}$, explicitly an approximation rather than another value printed on the paper, gives $\delta =0.564\mathrm{mm}$ and $z\simeq 2.21\mathrm{mm}$ from either face of a sufficiently large plate. The viscous penetration depth varies as ${\rho }^{-1/2}$ if a more precise water density is used. The ideal parallel-flow solution ignores finite plate edges and tank walls.