acceleration - 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

Answer

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.