Q7.16 Water oscillations in a U-tube (H)
Consider a column of liquid (density ρ) of length ℓ 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.- Construct expressions for the potential and kinetic energies of the liquid.
- Hence show that the
column will oscillate with a period of
Hint
Reveal
Solution
Reveal

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. |
We will write down the kinetic and potential energy separately.
First let us think about the KE.
The volume of liquid is ℓA, so its mass is
If the liquid surfaces remains flat during the oscillation every part of the liquid moves with the same speed (though not the same velocity)
![\[ v= \mid \frac{dy}{dt} \mid \]](mastermathpng-1.png)
![\[ K= \frac{1}{2} M v^2 =\frac{1}{2} \rho A \ell \left( \frac{dy}{dt} \right)^2 \]](mastermathpng-2.png)
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
The total energy follows as
![\[ E =\frac{1}{2} \rho A \ell \left( \frac{dy}{dt} \right)^2 + \rho A g y^2 \]](mastermathpng-3.png)
This is of the same form as the result for the mass-spring system:
![\[ E= \frac{1}{2}m \left( \frac{dx}{dt} \right)^2 + \frac{1}{2}kx^2 \]](mastermathpng-4.png)
![\[ m \rightarrow \rho A \ell \hspace{0.5cm}\mbox{\rm and}\hspace{0.5cm} k \rightarrow 2\rho A g \]](mastermathpng-5.png)
![\[ \omega= \sqrt{\frac{k}{m}} \]](mastermathpng-6.png)
![\[ \omega= \sqrt{\frac{2\rho A g} {\rho A \ell}} =\sqrt{\frac{2g}{\ell}} \hspace{0.5cm}\mbox{\rm and}\hspace{0.5cm} T = 2\pi / \omega = \pi\sqrt{\frac{2\ell}{g}} \]](mastermathpng-7.png)
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).