To first order, vanishing tangential velocity on the <elastic plate> is simply $u(0)=0$. In the mode from part a this gives
$$
C=-kA.
$$
Consequently
$$
u=k^2Az\,e^{kz}e^{ikx},
\qquad
w=ikA(1-kz)e^{kz}e^{ikx},
$$
so $u(0)=0$, $w(0)=ikA e^{ikx}$, and $w_z(0)=0$. The dynamic pressure at the surface is
$$
p'(0)=-2i\mu k^2A e^{ikx}
=-2\mu k\,w(0).
$$

Evaluating the hydrostatic pressure at $z=\zeta$ contributes the restoring normal stress $\rho g\zeta$. Since $\nabla^4\zeta=k^4\zeta$ for this mode, the linearized normal-stress balance is
$$
\rho g\zeta-p'(0)+2\mu w_z(0)+Bk^4\zeta=0.
$$
It follows that
$$
\boxed{u(x,0,t)=0},
$$
$$
\boxed{
w(x,0,t)
=-\frac{\rho g+Bk^4}{2\mu k}\,
\epsilon(t)e^{ikx}}.
$$
The two restoring effects are buoyancy and the plate's <bending stiffness>.

Solved by gpt-5.6-sol high.

Solution

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 332 / 4 / c / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2026/iii/paper-332/4/c/solution