The incident <internal-wave ray> has equation $z=-rx$. Intersecting it with the bottom $z=-H_0+sx$ gives
$$
x_1=\frac{H_0}{r+s}.
$$
Its distance to the slope is consequently
$$
\boxed{\ell=\frac{x_1}{\cos\theta}
=\frac{H_0}{\sin\theta+s\cos\theta}.}
$$
The subcritically reflected ray rises through the same vertical distance at angle $\theta$, so its next free-surface reflection is at
$$
\boxed{x_2=2x_1=\frac{2H_0}{r+s}.}
$$

On the first leg, <viscous attenuation of an internal-wave beam> multiplies the <energy density> by $e^{-2k\ell/\operatorname{Re}}$. The bottom reflection multiplies it by $\gamma^2$ and changes the <wavenumber> to $\gamma k$. Since the corresponding <Reynolds number> is $\operatorname{Re}/\gamma^2$, attenuation on the second leg contributes $e^{-2\gamma^3k\ell/\operatorname{Re}}$. Ignoring boundary-layer enhancement as requested,
$$
\boxed{\overline E_2=\gamma^2\overline E_0
\exp\left[-\frac{2k\ell}{\operatorname{Re}}(1+\gamma^3)\right].}
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 345 / 1 / e / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/iii/paper-345/1/e/solution