Oscillations of fluid in a U tube

1. Nov 30, 2011

VortexLattice

1. The problem statement, all variables and given/known data

We have a U tube, like this one:

With the a nonviscous, incompressible fluid at height h at equilibrium. We're interested in finding the frequency of small oscillations about the equilibrium. The tubes have area A (though I'm guessing this falls out in the end) and are L apart from each other (this definitely falls out).

2. Relevant equations

At equilibrium:

$f_{app} = \frac{1}{\rho}\nabla p$

Where f_app is an applied force per unit mass, $\rho$ is the density (constant here), and p is the pressure.

3. The attempt at a solution

There are two forces at play: The applied gravitational force per unit mass, $f_{app} = -g\hat{z}$, and the pressure force (I think). At equilibrium, they should be equal. I'm going to say z is in the vertical direction, and z = 0 at equilibrium. Using the equation above and solving for p, we find that $p = -\rho g z$.

This is good because it's a linear term in z. Now, I know I need to apply Newton's 2nd law and get something of the form $\ddot{z} = -k^2 z$. We have p, and force on a surface is pA, where A is the area. So I almost have it, but I'm having trouble putting it all together.

I'm also confused because it seems like when I try to sum all the forces for Newton's 2nd law, I have the pressure force and the gravitational force. But the gravitational force doesn't have z in it... I guess I could do that substitution trick where I let x = z - g, but it just seems like I'm doing something simple wrong.

Can anyone help?? Thanks!

2. Dec 1, 2011

technician

The displaced liquid of height h exerts a pressure =hρg on the rest of the liquid in the U tube.
The force on the liquid is therefore =pressure/area = hρg/A.
This force is proportional to h (displacement..... actually 2 x displacement) therefore the system undergoes SHM.
If you work through the logic of SHM you should get an equation for the time period.
I hope this gets you started..... ask if you need any more help

3. Dec 2, 2011

RTW69

Isn't Force= pressure*area?

F=-ρghA where A=area

d2h/dt2=-(ρgA)h/m

but m=ρAh problem?

or are we going to set ω2=-ρgA/m?

d2h/dt2=-ω2h

4. Dec 2, 2011

technician

SORRY!!!!!RTW. My mistake f =P x A.
Acceleration = F/m where m = total mass of liquid in tube
This is the acceleration of the SHM.
Is this any help......sorry again about my careless typing error

5. Dec 2, 2011

RTW69

OK is freq=√(ρgA/m)/2∏ for this problem?

6. Dec 2, 2011

technician

That is what I got !!!!!

7. Dec 2, 2011

Stonebridge

Not quite.
In the analysis for SHM, the displacement from the equilibrium position is h/2 here
not h. Which means there is a factor of 2 missing in the formula.

Note also that m = ρLA where L is the total length of the liquid in the U-Tube

This gives T = 2∏√(L/2g) and bears a striking resemblance to another well-known formula!
It shows that the period is independent of the density or mass of the liquid.

Last edited: Dec 2, 2011
8. Dec 2, 2011

technician

I think I now agree with the displacement =h/2.
I think the question is not very clear, I left the mass of the liquid as m in the equation
Cheers