Oscillations in U-tube U-tube, of uniform cross section $S$ and opened at both ends, is filled with a liquid column of length $L$ and density $\rho$. In equilibrium the water levels in both arms are at same height. What goes on if liquid in one arm is somehow raised, with a temporary subpressure for example, and then released?

I understand the system will try to return to and tend to oscillate about equilibrium. But I am at a loss as to what exactly drives that. Presently, I can derive the angular frequency of the oscillation as $\omega^2 = 2 g/L$ from the energy balance. But I would like to understand the Newtonian equation.
Pressure at both liquid surfaces is the same so there is no external net force on the fluid as a whole, right? There should be hydrostatic difference in pressure between points A and B equal to $\Delta p = \rho g (2 x)$ which I guess drives the liquid. What's the pressure at C? How does pressure changes from A to B to C?
(Here's link to a similar post, but which doesn't resolve my confusion.)
 A: Assuming an incompressible liquid, Bernoulli for instationary flow (neglecting friction) is
$$
\int_1^2 \frac{\partial c}{\partial t} \, \mathrm{d}s + \tfrac12 (c_2^2-c_1^2) + g(z_2-z_1) + \frac1{\rho}(p_2-p_1)=0
$$
with velocity $c$, gravitation accceleration $g$, height $z$, density $\rho$ and pressure $p$ and $1$ and $2$ denoting the two positions referring to. For incompressible liquids, conservation of mass reduces to
$$
c_1 A_1 = c_2 A_2
$$
Thus for a constant cross section area $A$, $c$ is not dependant on the position along the pipe, so $c_1=c_2$ and $\partial c/\partial t$ constant along the integral path. Since the pipes are open, $p_1=p_2$ (neglecting the influence of the air in the tube and the pressure difference in the atmosphere). As $c=\mathrm{d} x/\mathrm{d} t$, Bernoulli reduces to
$$
\frac{\mathrm{d}^2x}{\mathrm{d} t^2} + \frac{2g}{L}x = 0
$$
which is solved by $\sin(\omega t)$ and $\cos(\omega t)$ with $\omega^2=2g/L$ as you defined it.
For any two positions inside the column,
$$
L_{21} \frac{\mathrm{d}^2x}{\mathrm{d} t^2} + g(z_2-z_1) + \frac1{\rho}(p_2-p_1)=0
$$
which can be further simplified using the above differential equation ($L_{21}$ being the column length between positions $1$ and $2$). For a fixed position (not moving with the liquid),
$$
p=p_0+\rho g \left( \frac{2 x^2}{L} + \frac{2 L_P x}{L} + x - z_P\right)
$$
with $L_P$ being the length between the equilibrium position of $x$ and position $P$, and $z_P$ being the height of $P$ with respect to the equilibrium position of $x$. For point C, $L_P=\tfrac12 L$ and
$$
p_{\mathrm{C}}=p_0+\rho g \left( \frac{2 x^2}{L} + 2x-z_{\mathrm{C}}\right)
$$
The pressure for A and B can be calculated with the above equations. 
A: Suppose you do a force balance on the portion of the fluid situated between elevations z and $z +\Delta z$ in the left column.  You get:  $$p(z+\Delta z)S-p(z)S+\rho g S\Delta z=\rho S\Delta z \frac{dv}{dt}\tag{1} $$where $v$ is the downward velocity in the left column:$$v=-\frac{dx}{dt}\tag{2}$$  The latter equation is correct because the fluid is incompressible.  If we divide Eqn. 1 by $S\Delta z$ and take the limit as $\Delta z$ approaches zero, we obtain:$$\frac{\partial p}{\partial z}+\rho g=-\rho\frac{d^2x}{dt^2}\tag{3}$$
Eqn. 3 applies to the region above point A in the left column.  Similarly, for the horizontal region between points A and B, we have: $$\frac{\partial p}{\partial y}=-\rho\frac{d^2x}{dt^2}\tag{4}$$where y is the horizontal coordinate measured from A to C.  Finally, for the right hand column above point B, we have:  $$\frac{\partial p}{\partial z}+\rho g=+\rho\frac{d^2x}{dt^2}\tag{5}$$
If we integratd these equations along the continuous path (contour) between the top of the left column at z = h+x (where the pressure is atmospheric) to the top of the right column at z = h-x (where the pressure is again atmospheric), where h is the equilibrium height, we obtain:$$2\rho g x=-L\rho \frac{d^2x}{dt^2}\tag{6}$$or equivalently:$$\frac{d^2x}{dt^2}+\frac{2g}{L}x=0\tag{7}$$
If we combine Eqn. 7 with Eqn. 4, we obtain:$$p_A-p_B=\frac{2\rho g x}{L}(L-2h)\tag{8}$$
ADDEND
An even simpler way of solving this problem is to recognize that, since we are dealing here with an inviscid fluid, the sum of its Potential Energy and Kinetic Energy must remain constant, independent of time.  Taking as the datum of potential energy the base of the U, the potential energy of the fluid is given by:$$PE=\rho g S \frac{(h+x)^2}{2}+\rho g S \frac{(h-x)^2}{2}$$where h is the equilibrium height in the columns.  The kinetic energy of the fluid is given by:$$KE=\frac{\rho LS}{2}\left(\frac{dx}{dt}\right)^2$$So, $$PE+KE=\rho g S \frac{(h+x)^2}{2}+\rho g S \frac{(h-x)^2}{2}+\frac{\rho LS}{2}\left(\frac{dx}{dt}\right)^2$$If we take the time derivative of this and set the result to zero, we immediately obtain:$$\rho S\frac{dx}{dt}\left[2gx+L\frac{d^2x}{dt^2}\right]=0$$
A: There are many assumptions made in the problem. But one thing we know is that the motion is accelerated and hence the flow in unsteady. For unsteady flow you cannot use bernoullis equation either. So i guess it'll have to be experimental. I may be wrong, thete may be other methods but none within simplistic limits.
