import numpy as np
import matplotlib.pyplot as plt
%matplotlib notebook
Assignment 5 Key¶
Q2¶
Consider a hydrostatic wave being forced at the mouth of a rectangular fjord of depth $H$ and length $L$. The sea-surface height at the mouth of the fjord is prescribed by the sea-surface height in the ocean $\eta(0,t) = \eta_O \cos (\omega t)$, where $\omega$ is the tidal frequency.
Q2.1¶
Derive an expression for the the sea-surface height $\eta$ as a function of $x$ in the fjord assuming that there is no energy dissipation in the fjord. Describe the response of sea-surface height at the head of the fjord as a function of the length of the fjord, $L$. Also note that there are sometimes nulls in the response in the fjord. Where are they?
A2.1¶
The wave has a general solution:
$$ \eta(x, t) = \mathbb{R}[A e^{i(kx - \omega t)} + B e^{i(-kx - \omega t)}]$$We need two boundary conditions, one at the head of the fjord, and one at the mouth. If we call the mouth $x=L$ then
$$ \eta_0 = \mathbb{R}[A e^{i kL} + B e^{-ikL}]$$At $x=0$, we must have $u=0$. Recall that for hydrostatic waves
$$\partial u/\partial t = -g \partial \eta / \partial x$$So at $x=0$, we have $\partial \eta / \partial x = 0$, or $kA + kB = 0$, or $A = -B$. Back into the expresion above, and we have that
$$ A = \frac{\eta_0}{2\cos{kL}}$$and we have a full expression:
$$ \eta(x, t) = \eta_0 \frac{\cos{kx}}{\cos{kL}} \cos{\omega t}$$This is a standing wave, and the whole fjord rises or falls in phase (or 180 degress out of phase)
Alternate:¶
If you called the mouth $x=0$ and the head $x=L$ then we have at the head:
$$ ik\left[Ae^{ikL} + B e^{-ikL}\right] = 0$$and
$$B = A e^{2ikL}$$At the mouth, we have
$$ \mathbb{R} \left[ A e^{-i\omega t } (1 + e^{2ikL})\right] = \eta_0 \cos{\omega t} $$or
$$ A = \frac{\eta_0 e^{-ikL}}{e^{-ikL} + e^{+ikL}} = \frac{\eta_0 e^{-ikL}}{2\cos{kL}}$$so we have the expression:
$$ \eta(x, t) = \eta_0 \frac{\cos{k(x-L)}}{\cos{kL}} \cos{\omega t}$$which is exactly the same as above, except $x$ is displaced by $L$.
But I really hate complex numbers....¶
Let just do $x=0$ is the head of the fjord and let
$$ \eta = a\cos(kx - \omega t) + b\sin(kx - \omega t) + c\cos(kx + \omega t) + d\sin(kx + \omega t) $$Note that you have four unknowns and this gets you any combination of wave phases you want. If we note that $\partial \eta / \partial x = 0$ at $x=0$, then:
$$ 0 = -ka \sin\left(\omega t\right) - bk \cos(\omega t) + c \sin{\omega t} - d \cos{\omega t}$$In order for that to be true for all values of t, we must have: $a=c$ and $b=-d$, and now
$$\eta = b \left(\cos(kx-\omega t) + \cos(kx + \omega t) \right)$$Now use the other BC at $x=L$:
$$\eta(x=kL) = \eta_0 \cos(\omega t) = b \left(\cos(kL-\omega t) + \cos(kL + \omega t) \right)$$$$\eta_0 \cos(\omega t) = b \left(\cos(kL)\cos(\omega t) + \sin(kL)\sin(\omega t) + \cos(kL)\cos(\omega t) - \sin(kL)\sin(\omega t) \right)$$and we get the same result as before:
$$ \eta(x, t) = \eta_0 \frac{\cos{kx}}{\cos{kL}} \cos{\omega t}$$Hold it, I know this is a standing wave....¶
i.e the amplitude of the left going wave is the same as the right-going so
$$ \eta = a \cos(kx+\phi)\cos(\omega t) $$But please note the phase! You don't know the phase....
If we assume $x=0$ is the ocean, then $a \cos(\phi) = \eta_0$.
If we assume $x=L$ is the wall, then $d\eta/d x = 0$ there, and
$$ 0 = k a \sin(kL + \phi) $$So, $\phi = -kL$, and our solution is as before:
$$\eta = \frac{\eta_0}{\cos(kL)}\cos(k(x-L)) \cos(\omega t)$$fig, ax = plt.subplots()
x = np.arange(0., 1., 0.001)
ax.axhline(0., linestyle='--', color='0.7')
ax.axhline(1., linestyle='--', color='0.7')
for kL in [np.pi/2*0.7, np.pi/2*0.9, np.pi/2 * 1.2, np.pi/2 * 1.7, np.pi*0.97, 3*np.pi/2*0.97, 3*np.pi/2*1.07 ]:
omt=0.
ax.plot(x*kL, np.cos(x*kL)/np.cos(kL)*np.cos(omt), label='kL=%1.3f'%kL)
ax.legend()
ax.set_xlabel('xk')
ax.set_ylabel('$\eta / \eta_0$')
ax.set_xticks(np.arange(0, 2*np.pi+0.1, np.pi/2))
ax.set_xticklabels(['0', '$\pi/2$', '$\pi$', '$3\pi/2$', '$2\pi$' ])
At the head of the fjord,
$$\eta(x, 0) = \eta_0 \frac{\cos{\omega t}}{\cos{kL}}$$Note that as $kL = n\pi/2$ the denominator goes to zero and we have a reonant response. Also note that the sign of the response at the head switches to being in phase or 180 degrees out of phase with the mouth:
kL = np.arange(0, np.pi*2+0.03, 0.01)
fig,ax = plt.subplots()
ax.plot(kL, 1./np.cos(kL),'.')
ax.set_ylim([-4, 4])
ax.set_xlabel('kL')
ax.set_ylabel('$\eta_{head}/\eta_{mouth}$')
ax.axhline(1., linestyle='--', color='0.6', zorder=-2)
ax.axhline(-1., linestyle='--', color='0.6', zorder=-2)
ax.set_xticks(np.arange(0, 2*np.pi+0.1, np.pi/2))
ax.set_xticklabels(['0', '$\pi/2$', '$\pi$', '$3\pi/2$', '$2\pi$' ])
If $kL>\pi/2$ then there are nulls at $kx=n\pi/2$
Q2.2¶
What is the relationship between $u(x,t)$ and $\eta(x,t)$ in the fjord? What happens to the velocity at the mouth as the fjord length approaches the resonant length?
A2.2¶
$$ \frac{\partial u}{\partial t} = -g \frac{\partial \eta}{\partial x} $$or
$$ u(x, t) = -\eta_0 \frac{gk}{\omega}\frac{\sin{kx}}{\cos{kL}} \sin{\omega t}$$So u and eta are in phase.
$$ u(L, t) = -\eta_0\frac{gk}{\omega} \frac{\sin{kL}}{\cos{kL}} \sin{\omega t}$$kL = np.arange(0, np.pi*2+0.03, 0.01)
fig,ax = plt.subplots()
ax.plot(kL, -np.tan(kL),'.')
ax.set_ylim([-4, 4])
ax.set_xlabel('kL')
ax.set_ylabel('$U_{0}/ [\eta_{0}gk/\omega]$')
ax.axhline(1., linestyle='--', color='0.6', zorder=-2)
ax.axhline(-1., linestyle='--', color='0.6', zorder=-2)
ax.set_xticks(np.arange(0, 2*np.pi+0.1, np.pi/2))
ax.set_xticklabels(['0', '$\pi/2$', '$\pi$', '$3\pi/2$', '$2\pi$' ])
So, again we see that we need infinite velocity at the mouth to achieve the resonance, and that the mouth can be 180 degrees out of phase with the head.