# Theoretical analysis

Several theoritical expressions will be established in this part :

• the elevation of water level in a tank subject to gravity an acceleration
• the dissipation time of a wave in the tank
• the natural frequency of a wave

The main parameters of the problem are presented thereafter :

- H=1.2 m : height
​- L= [1.5 ; 3.5] m : lenght
- h=[0.2 ; 1.1] m : initial water level
- a=[1 ; 2.5] m/s2 : plane's acceleration
- g=9.81 m/s2 : gravity # Preliminary calculation

Elevation e of the fluid in a tank subject to an acceleration a

In a first time, we theoretically calculate the level of water (or retardant) in a tank subject to an acceleration and gravity. Herebelow a diagram of the problem : The blue line represent the interface between air and water and we have :

$\vec{a}=a.\vec{x}$

$\vec{g}=-g.\vec{y}$

The forces' balance gives :

$\rho \vec{a}=-\Delta P+\rho\vec{g}$

After projection, we have : $\frac{\partial P}{\partial x}=- \rho a \: and \: \frac{\partial P}{\partial y}=- \rho g$

On the surface, $P= constant$ so : $dP=0= \frac{\partial P}{\partial x} dx + \frac{\partial P}{\partial y} dy$

Therefore, $\rho a.dx + \rho g.dy = 0$

$\frac{dy}{dx} |_{surface} = - \frac{a}{g}$

$y = - \frac{a}{g}.x + C$ and we know that $\: y(x=0)=h \rightarrow C= h$ so $\: y = - \frac{a}{g}x + h$ and now we can calculate $e$, the level of water : $e + h = \left(- \frac{a}{g} \right) . \left(- \frac{L}{2} \right) + h$ and finaly we get :

$e = \frac{a.L}{2.g}$

Dispersion time T of waves

This calculation is useful to know the time which is necessary for a wave to disperse when the acceleration moves from a value $a$  to $0$ . Yet, it is important to precise that the calculation is valid only if the value of the acceleration $a$ is not too high (not over 2.5g). Herebelow is a diagram of the dimensions of the tank we need for this calculation :

Hypothesis :

$\omega^2 = (gk + \frac{\sigma}{\rho}k^2) \tanh (kh)$

Two hypothesis can be made to calculate the characteristic dispersion time :

• hypothesis of "deep water" : $\lambda << h$,  $\omega^2 = gk$,  $c=\sqrt{\frac{g \lambda}{2 \pi}}$
• hypothesis of "long waves" : $\lambda >> h$,  $\omega^2 = gk^2h$,  $c=\sqrt{gh}$

In our case, we choose to work with the hypothesis of "long waves" .

Calculation :

We have : $\frac{\partial E_c}{\partial t} = D \thicksim 2\mu S_{ij}S_{ij} \thicksim 2\mu \frac{V^2}{L^2}$

$E_c \thicksim \rho V^2 \rightarrow \mu \frac{V^2}{L^2} \thicksim \frac{\rho V^2}{T}$

$\longrightarrow T = \frac{\rho V^2 L^2}{\mu V^2} \thicksim T=\frac{\rho L^2}{\mu}$

Here, $L = e$ (calculated just before), so we have : $T \thicksim \frac{\rho}{\mu}e^2$

Thanks to the previous calculation, we know the expression of $e$ :

$e = \frac{aL}{2g}$ So we can write :

$T = \frac{\rho a^2 L^2}{4\mu g^2}$

Natural frequency f0

In the light of the framework of this problem, we can make the parallel with the problem of the oscillating pendulum and suppose, in a first time, that the natural frequency of a wave can be written as following :

$f_0 = \frac{1}{2\pi} \sqrt{\frac{g}{L}}$