Fluid filled harmonic oscillator A vessel (preferably circular) filled with water is accelerating unidirectionally such that the level of water is higher on one end than the other. What I want to know is that if the vessel is immediately stopped, the water level will force itself back to the equilibrium point, but in doing so push the other end up. If water is an ideal fluid, what type of oscillation will occur in this situation. 
I can tell that maybe the mathematical expression for this oscillation will be complex, but please do let me know if I am thinking this correctly. 
I am considering, finding the oscillation equation for infinitesimally small u-tubes(considering a 2-d plane) and integrating it to get the final expression.
Thanks in advance 
 A: The answer of Gert is very insightful, but there is a slight error in the equation of motion. (Sorry, since I am new here, I cannot comment yet, therefore I have to write it as a separate answer.)
Instead of $y$ the variable should by $x$, which measures the height of the column from the equilibrium position (when the water levels are equal in both branches). It means, that the weight of the column, that exerts force is: 
$2\rho A g x$
The equation of motion becomes: 
$Ma+2\rho A g x=0$
The $a$ acceleration can be expressed as $a=\frac{d^2x}{dt^2}$, so 
$M\frac{d^2x}{dt^2}+2\rho A g x=0$
The important difference here is that when $x=0$, the force is indeed $0$, as it is expected. However, when in Gert's derivation $y=0$, the force is not $0$, because at that point there would be a column of height $y$ in the left branch of the tube, which would exert force on the system. $y=0$ is not an equilibrium position.
We can look for the solution in the form of $x=x_0\cos (\omega t)$, and the frequency will be: 
$\omega=\sqrt{\frac{2\rho A g}{M}}$
If the length of the entire water 'column' in the tube is $L$, this can be simplified since $M=\rho A L$: 
$\omega=\sqrt{\frac{2g}{L}}$
The problem you refer to in your question is known as 'sloshing' and it is a fairly complicated one. You can search for this term and find papers like this: 
http://link.springer.com/article/10.1007%2Fs10665-010-9397-5
Also, you can find videos, which show that the surface behaves in a very complex way in the described situation. It wouldn't behave as a flat surface acting like a seesaw. The U-tube is a crude first assumption, but it gives some insights. 
A: The answer I gave above is actually only approximately correct, if the radius of the circular bend is large compared to the radius of the U tube. The reason is that in the bent section the fluid actually rotates around its centre point, not translates as previously assumed. Look at the diagram below:

Firstly let's define a few things:
Total length of the fluid column, $L=y_1+\pi R + y_2$.
Total mass of the fluid, $M=\rho LA$.
Mass of the fluid contained in the bend, $m=\rho \pi RA$.
Force acting on the column, $F=\rho A(y_1-y_2)= \rho A(2y_1-L+\pi R)$.
This force now creates a torque about the centre of the bend that causes the fluid in the bend to experience angular acceleration $\alpha=\frac{d\omega}{dt}$, acc.:
$$\tau=FR=I \alpha,$$
where $I$ is the inertial moment of the fluid in the bend, $I=\frac{mR^2}{2}$, so:
$$FR=\frac{\rho \pi A R^3}{2}\alpha,$$
with $\alpha=\frac{d\omega}{dt}$ and $\frac{d\omega}{dt}=\frac{1}{R}\frac{dv}{dt}=\frac{1}{R}a$, then:
$$F=\frac{\rho \pi A R}{2}a$$
However, $F$ is also responsible for the translational acceleration of the fluid that is not contained in the bend, which has a mass:
$$M-m$$
$$\rho A(L-\pi R)$$
Since as all the fluid experiences the same acceleration $a$, so we can write:
$$F=\frac{\rho \pi A R}{2}a+\rho A(L-\pi R)a$$
$$F=\rho A\frac{2L-\pi R}{2}a$$
With the expression for $F$ obtained higher up, we then get the equation of motion:
$$\rho A\frac{2L-\pi R}{2}a+\rho Ag(2y_1-L+\pi R)=0,$$
$$\frac{2L-\pi R}{2}a+(2y_1-L+\pi R)g=0$$
Now substitute: $u=2y_1-L+\pi R$, so:
$du=2dy_1$ and $a=\frac{1}{2}\frac{d^2u}{dt^2}$, so:
$$\frac{(2L-\pi R)}{4}\ddot{u}+gu=0,$$
which is the DE of the Harmonic Oscillator, with solution:
$$u=u_0\cos\omega t,$$
where:
$$\large{\omega=\sqrt{\frac{4g}{2L-\pi R}}}$$
A: Consider the following diagram:

A U tube contains a fluid with higher level in the right hand side. This could be achieved as you suggested or more simply by applying a partial vacuum on the right hand side: the higher (atmospheric) pressure in the left tube then pushes the fluid up into the right side, until a level difference of $y$ is achieved.
At $t=0$ we break the vacuum. It can now be shown that the system has potential energy $U$:
$$U=mg\frac{y}{2},$$
where $m$ is the mass of the fluid column of height $y$:
$$m=\rho Ay,$$
with $\rho$ the density of the fluid and $A$ the cross-section of the tube.
So that:
$$U=\frac{\rho Agy^2}{2}.$$
Ignoring for now all viscous friction the fluid might experience during flow (we're assuming a perfectly smooth pipe) potential energy will now be converted to kinetic energy.
Kinetic energy of a column of fluid of mass $M$ (that is the total mass of the fluid in the U tube) is given by:
$$K=\frac{Mv^2}{2},$$
where $v$ is the velocity of the fluid column and $v=\frac{dy}{dt}$.
Assuming no energy losses then at all times the total energy of the system is:
$$T=U+K$$
Now let us set up a Newtonian equation of motion. The weight of the column of height $y$ exerts a force on the whole column of:
$$\rho Agy$$
So that the equation of motion becomes:
$$Ma+\rho Agy=0,$$
where $a$ is the acceleration the column of fluid experiences and $a=\frac{d^2y}{dt^2}$, so:
$$M\frac{d^2y}{dt^2}+\rho Agy=0$$
This is a the equation of motion of a simple Harmonic Oscillator and assuming at $t=0$, $v=0$ and $y=y_0$ then the solution is:
$$\large{y=y_0\cos\omega t}$$
where $\omega$ is the angular speed, $\omega=\frac{2\pi}{T}$ with $T$ the period of oscillation.
It can further be shown that:
$$\omega=\sqrt{\frac{\rho Ag}{M}}.$$
In reality, due to inevitable friction, $T=U+K$ will not be respected fully and the oscillation will be a damped harmonic oscillation.
