= Solution
For a <Newtonian fluid> with <velocity> $u(z)\widehat{\mathbf x}$, $\tau=\eta\,\partial_z u$ 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 $\mathrm{Pa\,s}$. Faster layers transfer positive $x$ <momentum> toward slower neighbours; the internal <shear stress> resists relative sliding.
\Image[/past-exam-of-the-physics-b-course-of-the-university-of-cambridge/2015/paper-1-viscous-flow.png]
{title=Shear-traction convention, Couette-Poiseuille profiles, and oscillatory viscous penetration}
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 $-A p_xdz$. 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>:
$$
\boxed{\rho\frac{\partial u}{\partial t}=-\frac{\partial p}{\partial x}
+\eta\frac{\partial^2u}{\partial z^2}.}
$$
For steady flow, integrate $\eta u''=p'$ twice and impose the <no-slip boundary condition> at each plate, $u(0)=0$ and $u(a)=u_0$. The <Couette-Poiseuille flow> profile is
$$
\boxed{u(z)=u_0\frac za+\frac{p'}{2\eta}z(z-a).}
$$
The linear term is <Couette flow> and the parabolic term is <plane Poiseuille flow>. A negative $p'$ drives flow in $+x$. Integrating the profile across the gap gives the volumetric flow per unit width,
$$
\boxed{q=\int_0^au(z)\,dz=\frac{u_0a}{2}-\frac{p'a^3}{12\eta}.}
$$
It has units $\mathrm{m^2\,s^{-1}}$; multiplying by the plate width gives the actual volume per time.
With zero <pressure gradient>, substituting $u=\operatorname{Re}[A e^{i(kz-\omega t)}]$ gives $-i\rho\omega=-\eta k^2$. Select the root that decays into fluid occupying $z>0$:
$$
\boxed{k=(1+i)\sqrt{\frac{\rho\omega}{2\eta}}=\frac{1+i}{\delta},\qquad
\delta=\sqrt{\frac{2\eta}{\rho\omega}}.}
$$
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> is
$$
u(z,t)=U_0e^{-z/\delta}\cos(z/\delta-\omega t).
$$
The 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,
$$
\frac{0.002}{0.10}=e^{-z/\delta},\qquad
\boxed{z=\sqrt{\frac{2\eta}{\rho(2\pi f)}}\ln50.}
$$
Using the standard approximate water <density> $\rho\simeq10^3\ \mathrm{kg\,m^{-3}}$, explicitly an approximation rather than another value printed on the paper, gives $\delta=0.564\ \mathrm{mm}$ and \b[$z\simeq2.21\ \mathrm{mm}$ 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.
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