Fluids in Action.

Consider the system when liquid on left end is displaced by a small distance $x$ . Calculate the potential energy of the system ( it is convenient to assume the base level to be at zero Potential) . We get

$\displaystyle U = \dfrac{ 5 \rho g \pi R^2 x^2 }{ 8} + C$ where $C$ is a constant.

Now to evaluate the kinetic energy :

It is mentioned that the liquid doesn't " pile up" anywhere . So , for the movement of the liquid in the horizontal part of the tube, apply equation of continuity . The total kinetic energy of the liquid comes out to be

$\displaystyle K = \dfrac{ \rho \pi R^2 ( L + \frac{5h}{2} ) v^2} {4}$ , where $v$ is the velocity of liquid in left end of the rod .

( Hints to evaluating the kinetic energy :

First evaluate kinetic energy of the liquid in horizontal part by taking a small element of length $dx$ at a distance $x$ from left end of the rod and integrate . Now, evaluate the kinetic energy of the liquid in the vertical sections and add them up to get tge total kinetic energy. )

Total energy is constant ,that is ,

$\displaystyle U + K = constant$

Differentiate the above equation wrt time to get

$a = \dfrac{ -5 g x }{ 2( L + \frac{5h}{2} ) }$ , where $a$ is the acceleration of the liquid in the left end .

Comparing with the standard equation for $\color{#3D99F6}{ S.H.M }$ ,

$\omega = \sqrt{ \dfrac{ 5g }{ 2( L + \frac{5h}{2} ) } }$ .

Substituting $L = \dfrac{ 5h}{2}$ , time period of oscillations comes out to be

$\displaystyle \color{#D61F06}{ T = 2 \pi \sqrt{ \dfrac{2h}{ g} } = 12.5663 }$