# Superposition of Linear Plane Progressive Waves

## Superposition of Linear Plane Progressive Waves

### Oblique Plane Waves

Consider wave propagation at an angle $\displaystyle{ \theta \, }$ to the x-axis

$\displaystyle{ \eta = A \cos ( kx\cos\theta+kz\sin\theta-\omega{t}) = A \cos (k_xx+k_zz-\omega{t}) \, }$
$\displaystyle{ \phi = \frac{gA}{\omega} \frac{\cosh k (y+h)}{\cosh k h} \sin (kx\cos\theta+kz\sin\theta-\omega t) }$
$\displaystyle{ \omega = g k \tanh k h; \ k_x=k\cos\theta, k_z = k\sin\theta, \ k=\sqrt{k_x+k_z} }$

### Standing Waves

$\displaystyle{ \eta = A \cos (kx-\omega t) + A \cos (-kx-\omega{t}) = 2 A \cos kx \cos \omega t \, }$
$\displaystyle{ \phi = - \frac{2 g A}{\omega} \frac{\cosh k (y+h)}{\cosh k h} \cos kx \sin \omega t }$
$\displaystyle{ \frac{\partial\eta}{\partial{x}} \sim \frac{\partial\phi}{\partial{x}} = \cdots \sin kx = 0 \, }$ at $\displaystyle{ x=0, \ \frac{n\pi}{k} = \frac{n\lambda}{2} \, }$

Therefore, $\displaystyle{ \left. \frac{\partial\phi}{\partial{x}} \right|_x = 0 \, }$. To obtain a standing wave, it is necessary to have perfect reflection at the wall at $\displaystyle{ x=0 \, }$.

Define the reflection coefficient as $\displaystyle{ R \equiv \frac{A_R}{A_I} (\leq 1) \, }$.

$\displaystyle{ A_I = A_R \, }$
$\displaystyle{ R = \frac{A_R}{A_I} = 1 \, }$

### Oblique Standing Waves

$\displaystyle{ \eta_I = A \cos ( k x \cos \theta + k z \sin \theta - \omega t ) \, }$
$\displaystyle{ \eta_R = A \cos ( k x \cos (\pi-\theta) + k z \sin (\pi-\theta) - \omega t ) \, }$
$\displaystyle{ \theta_R = \pi - \theta_I \, }$

Note: same $\displaystyle{ A, \ R = 1 \, }$.

$\displaystyle{ \eta_T = \eta_I + \eta_R = 2 A \cos ( k x \cos \theta ) \cos ( k z \sin \theta - \omega t ) \ , }$

and

$\displaystyle{ \lambda_x = \frac{2\pi}{k\cos\theta}; \ V_{P_x} = 0; \ \lambda_z = \frac{2\pi}{k\sin\theta}; \ V_{P_z} = \frac{\omega}{k\sin\theta} }$

Check:

$\displaystyle{ \frac{\partial\phi}{\partial{x}} \sim \frac{\partial\eta}{\partial{x}} \sim \cdots \sin (kx\cos\theta) = 0 \, }$ on $\displaystyle{ x=0 \, }$

### Partial Reflection

$\displaystyle{ \eta_I = A_I \cos ( k x - \omega t ) = A_I R_e \left\{ e^{i \ kx - \omega t} \right\} }$
$\displaystyle{ \eta_R = A_R \cos ( k x + \omega t + \delta ) = A_I R_e \left\{ e^{-i \ kx \omega t} \right\} }$

$\displaystyle{ R \, }$: Complex reflection coefficient

$\displaystyle{ R = |R| e^{-i\delta}, |R| = \frac{A_R}{A_I} \, }$
$\displaystyle{ \eta_T = \eta_I + \eta_R = A_I R_e \left\{ e^{i\ kx-\omega t} \left( 1 + R e^{-ikx} \right) \right\} }$
$\displaystyle{ |\eta_T| = A_I \left[ 1 + |R| + 2 |R| \cos ( 2 k x + \delta ) \right] \, }$

At node,

$\displaystyle{ |\eta_T| = |\eta_T| = A_I ( 1 - |R| ) \, }$ at $\displaystyle{ \cos (2 k x + \delta) = -1 \, }$ or $\displaystyle{ 2 k x + \delta = ( 2 n + 1 ) \pi \, }$

At antinode,

$\displaystyle{ |\eta_T| = |\eta_T| = A_I ( 1 + |R| ) \, }$ at $\displaystyle{ \cos (2 k x + \delta) = 1 \, }$ or $\displaystyle{ 2 k x + \delta = 2 n \pi \, }$
$\displaystyle{ 2 k L = 2 \pi \, }$ so $\displaystyle{ L = \frac{\lambda}{2} \, }$
$\displaystyle{ |R| = \frac{|\eta_T|-|\eta_T|}{|\eta_T|+|\eta_T|} = |R(k)| \, }$

### Wave Group

2 waves, same amplitude $\displaystyle{ A \, }$ and direction, but $\displaystyle{ \omega \, }$ and $\displaystyle{ k \, }$ very close to each other.

$\displaystyle{ \eta = \Re \left( A e^{i k_1 x - \omega_1 t } \right) \, }$
$\displaystyle{ \eta = \Re \left( A e^{i k_2 x - \omega_2 t } \right) \, }$
$\displaystyle{ \omega, = \omega, ( k , ) \, }$ and $\displaystyle{ V_{P_1} \approx V_{P_2} \, }$
$\displaystyle{ \eta_T = \eta + \eta = \Re \left\{ A e^{i\ k_1x-\omega_1t} \left[ 1 + e^{i\ \delta kx - \delta\omega t} \right] \right\} \, }$ with $\displaystyle{ \delta k = k - k \, }$ and $\displaystyle{ \delta \omega = \omega - \omega \, }$
$\displaystyle{ \begin{Bmatrix} |\eta_T| = 2 |A| \ \mbox{when} \ \delta k x - \delta \omega t = 2n \pi \\ |\eta_T| = 0 \ \mbox{when} \ \delta k x - \delta \omega t = (2n+1) \pi \end{Bmatrix} x_g = V_g t, \ \delta k V_g t =0 \ \mbox{when} \ V_g = \frac{\delta\omega}{\delta k} }$

In the limit,

$\displaystyle{ \delta k, \delta\omega \to 0, \ \left. V_g = \frac{d\omega}{dk} \right|_{k_1\approx k_2\approx k} , }$

and since

$\displaystyle{ \omega = g k \tanh k h \Rightarrow \, }$
$\displaystyle{ V_g = \underbrace{\left( \frac{\omega}{k} \right)}_{V_P} \underbrace{\frac{1}{2} \left( 1+\frac{2kh}{\sinh 2kh} \right)}_n }$

$\displaystyle{ \begin{Bmatrix} & (a) \ \mbox{deep water} \ kh \gg 1 & n = \frac{V_g}{V_P} = -1 \\ & (b) \ \mbox{shallow water} \ kh \ll 1 & n=\frac{V_g}{V_P}=1 \ \mbox{no dispersion} \\ & (c) \ \mbox{intermediate depth} & -1 \lt n \lt 1 \end{Bmatrix} V_g \leq V_P }$

## Wave Energy -Energy Associated with Wave Motion.

For a single plane progressive wave:

align="center" ! Energy per unit surface area of wave
$\displaystyle{ \bullet }$ Potential energy PE $\displaystyle{ \bullet }$ Kinetic energy KE
 PE without wave $\displaystyle{ = \int_{-h} \rho g y dy = - - \rho g h \, }$ PE with wave $\displaystyle{ \int_{-h}^\eta \rho g y dy = - \rho g ( \eta - h ) \, }$ $\displaystyle{ PR_{wave} = - \rho g \eta = - \rho g A \cos ( kx - \omega t) \, }$
 $\displaystyle{ KE_{wave} = \int_{-h}^\eta dy - \rho ( u + v ) }$ Deep water $\displaystyle{ = \cdots = - \rho g A \ }$ to leading order Finite depth $\displaystyle{ = \cdots \, }$
Average energy over one period or one wavelength
$\displaystyle{ \overline{PE}_{wave} = - \rho g A \, }$ $\displaystyle{ \overline{KE}_{wave} = - \rho g A \, }$ at any $\displaystyle{ h \, }$
• Total wave energy in deep water:

$\displaystyle{ E = PE + KE = - \rho g A \left[ \cos ( k x - \omega t ) + - \right] \, }$

• Average wave energy $\displaystyle{ E \, }$ (over 1 period or 1 wavelength) for any water depth:

$\displaystyle{ \overline{E} = - \rho g A \left[ \overline{PE} + \overline{KE} \right] = - \rho g A = E_S , \, }$
$\displaystyle{ E_S \equiv \, }$ Specific Energy: total average wave energy per unit surface area.

• Linear waves: $\displaystyle{ \overline{PE} = \overline{KE} = \frac{1}{2} E_S \, }$ (equipartition).
• Nonlinear waves: $\displaystyle{ \overline{PE} \gt \overline{PE} \, }$.

## Energy Propagation - Group Velocity

Consider a fixed control volume $\displaystyle{ V \, }$ to the right of 'screen' $\displaystyle{ S \, }$. Conservation of energy:

 $\displaystyle{ \underbrace{\frac{dW}{dt}} \, }$ $\displaystyle{ = \, }$ $\displaystyle{ \underbrace{\frac{dE}{dt}} \, }$ $\displaystyle{ = \, }$ $\displaystyle{ \underbrace{\Im} \, }$ rate of work done on $\displaystyle{ S \, }$ rate of change of energy in $\displaystyle{ V \, }$ energy flux left to right

where

$\displaystyle{ \Im = \int_{-h}^\eta pu dy \ \, }$ with $\displaystyle{ \ p = - \rho \left( \frac{d\phi}{dt} + gy \right) \ }$ and $\displaystyle{ \ u = \frac{\partial\phi}{\partial x} \, }$
$\displaystyle{ \overline{\Im} = \underbrace{\left( -\rho g A \right)}_{\overline{E}} \underbrace{\underbrace{\frac{\omega}{k}}_{V_P} \underbrace{\left[-\left(1+\frac{kh}{kh}\right)\right]}_n}_{V_g} = \overline{E} (n V_P) = \overline{E} V_g }$

e.g. $\displaystyle{ A = 3m, \ T = 10\mbox{sec} \rightarrow \overline{\Im} = 400KW/m \, }$

## Equation of Energy Conservation

$\displaystyle{ \left( \overline{\Im} - \overline{\Im} \right) \Delta t = \Delta \overline{E} \Delta x \, }$
$\displaystyle{ \overline{\Im} = \overline{\Im} + \left. \frac{\partial\overline{\Im}}{\partial{x}} \right| \Delta x + \cdots \, }$
$\displaystyle{ \frac{\partial\overline{E}}{\partial{t}} + \frac{\partial\overline{\Im}}{\partial{x}} = 0 \, }$, but $\displaystyle{ \overline{\Im} = V_g \overline{E} \, }$
$\displaystyle{ \frac{\partial\bar{E}}{\partial{t}} + \frac{\partial}{\partial{x}} \left( V_g \overline{E} \right) = 0 \, }$

1. $\displaystyle{ \frac{\partial\overline{E}}{\partial{t}}=0, \ V_g \overline{E} = \ \, }$ constant in $\displaystyle{ x \, }$ for any $\displaystyle{ h(x) \, }$.

2. $\displaystyle{ V_g = \, }$ constant (i.e., constant depth, $\displaystyle{ \delta k \ll k )\, }$

$\displaystyle{ \left( \frac{\partial}{\partial t} + V_g \frac{\partial}{\partial{x}} \right) \bar{E} = 0, \ }$ so $\displaystyle{ \ \overline{E} = \overline{E} (x-V_g t) \ }$ or $\displaystyle{ \ A = A ( x - V_g t ) \, }$

i.e., wave packet moves at $\displaystyle{ V_g \, }$.

## Steady Ship Waves, Wave Resistance

• Ship wave resistance drag $\displaystyle{ D_w \, }$
Rate of work done = rate of energy increase
$\displaystyle{ D_w U + \overline{\Im} = \frac{d}{dt} (\overline{E}L) = \overline{E}U \, }$
$\displaystyle{ D_w = \frac{1}{U} ( \overline{E} U - \overline{E} U /2 ) = - \overline{E} = - \rho g A \ \Rightarrow \ D_w \propto A }$
• Amplitude of generated waves

The amplitude $\displaystyle{ A \, }$ depends on $\displaystyle{ U \, }$ and the ship geometry. Let $\displaystyle{ \ell \equiv \, }$ effective length.

To approximate the wave amplitude $\displaystyle{ A \, }$ superimpose a bow wave ($\displaystyle{ \eta_b \, }$) and a stern wave ($\displaystyle{ \eta_s \, }$).

$\displaystyle{ \eta_b = a \cos (kx) \ \, }$ and $\displaystyle{ \ \eta_S = - a \cos (k ( x+ \ell )) \, }$
$\displaystyle{ \eta_T = \eta_b + \eta_S \, }$
$\displaystyle{ A = | \eta_T | = 2 a \left|\sin (-k\ell)\right| \ \leftarrow \ }$ envelope amplitude
$\displaystyle{ D_w = - \rho g A = \rho g a \sin ( -k \ell ) \ \Rightarrow \ D_w = \rho g a \sin \left( - \frac{g\ell}{U^2} \right) \, }$
• Wavelength of generated waves To obtain the wave length, observe that the phase speed of the waves must equal $\displaystyle{ U \, }$. For deep water, we therefore have
$\displaystyle{ V_p = U \ \Rightarrow \ \frac{\omega}{k} = U \ \begin{matrix} \mbox{deep} \\ \longrightarrow \\ \mbox{water} \end{matrix} \sqrt{\frac{g}{k}} = U, \ }$ or $\displaystyle{ \lambda = 2 \pi \frac{U}{g} }$
• Summary Steady ship waves in deep water.
$\displaystyle{ U = \, }$ ship speed
$\displaystyle{ V_p = \sqrt{\frac{g}{k}} = U; \ }$ so $\displaystyle{ \ k = \frac{g}{U} \ \, }$ and $\displaystyle{ \ \lambda = 2 \pi \frac{U}{g} \, }$
$\displaystyle{ L = \, }$ ship length, $\displaystyle{ \ \ell \sim L \, }$
$\displaystyle{ D_w = \rho g a \sin \left( - \frac{g\ell}{U^2} \right) \cong \rho g a \sin \left( \frac{1}{2F_{rL}} \right) \cong \rho g \sin \left( \frac{1}{2F_{rL}} \right) }$