Q7.16 Water oscillations in a U-tube (H)

Consider a column of liquid (density ρ) of length confined in a U-tube of uniform cross-sectional area A. Suppose that the water level on one side is pushed down a small amount and then released.
  1. Construct expressions for the potential and kinetic energies of the liquid.
  2. Hence show that the column will oscillate with a period of
    \[ T = \pi \sqrt{\frac{2\ell}{g}} \]
Hide

Hint

You want to avoid thinking about forces here. Key Point 6.8 shows the way.
 

Solution

Reveal
Hide

Solution

Let y be the displacement (with respect to its equilibrium level) of the liquid in (say) the right hand part of the tube; the displacement in the left hand side is then -y.
Graphic - No title - newubend

We will write down the kinetic and potential energy separately.

First let us think about the KE.

The volume of liquid is A, so its mass is

M=ρA

If the liquid surfaces remains flat during the oscillation every part of the liquid moves with the same speed (though not the same velocity)

\[ v= \mid \frac{dy}{dt} \mid \]
The kinetic energy is then
\[ K= \frac{1}{2} M v^2 =\frac{1}{2} \rho A \ell \left( \frac{dy}{dt} \right)^2 \]

Now the PE. An arrangement in which the level in the RH tube rises by y is identical to that generated by taking a portion of fluid of volume Ay (and mass ρAy) from the top of the LH tube and placing it on top of the column on the RHS. This requires raising the portion of fluid by y thereby increasing its potential energy by ρAgy2. Thus we identify

U=ρAgy2

The total energy follows as

\[ E =\frac{1}{2} \rho A \ell \left( \frac{dy}{dt} \right)^2 + \rho A g y^2 \]

This is of the same form as the result for the mass-spring system:

\[ E= \frac{1}{2}m \left( \frac{dx}{dt} \right)^2 + \frac{1}{2}kx^2 \]
with the identifications
\[ m \rightarrow \rho A \ell \hspace{0.5cm}\mbox{\rm and}\hspace{0.5cm} k \rightarrow 2\rho A g \]
Then the mass-spring result
\[ \omega= \sqrt{\frac{k}{m}} \]
suggests that here
\[ \omega= \sqrt{\frac{2\rho A g} {\rho A \ell}} =\sqrt{\frac{2g}{\ell}} \hspace{0.5cm}\mbox{\rm and}\hspace{0.5cm} T = 2\pi / \omega = \pi\sqrt{\frac{2\ell}{g}} \]

This is an illustration of Key Point 6.8. If you prefer something a little tighter mathematically, then you should proceed as we did in establishing that key point (namely, by differentiating the expression for the total energy with respect to time, and using the fact that E is a constant).