Oscillations of fluid in a U tube

In summary, the conversation is about finding the frequency of small oscillations in a U tube filled with a nonviscous, incompressible fluid at equilibrium. The equation used is f_{app} = \frac{1}{\rho}\nabla p, where f_app is an applied force per unit mass, \rho is the density, and p is the pressure. The forces at play are the applied gravitational force and the pressure force, which should be equal at equilibrium. The pressure force is proportional to the displacement and leads to SHM. The equation for the time period is T = 2∏√(L/2g), which is independent of the density or mass of the liquid.
  • #1
VortexLattice
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0

Homework Statement



We have a U tube, like this one:

u-tube_manometer.png


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).

Homework Equations



At equilibrium:

[itex]f_{app} = \frac{1}{\rho}\nabla p [/itex]

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

The Attempt at a Solution



There are two forces at play: The applied gravitational force per unit mass, [itex]f_{app} = -g\hat{z}[/itex], 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 [itex]p = -\rho g z[/itex].

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 [itex]\ddot{z} = -k^2 z[/itex]. 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!
 
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  • #2
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
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
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
OK is freq=√(ρgA/m)/2∏ for this problem?
 
  • #6
That is what I got !
 
  • #7
RTW69 said:
OK is freq=√(ρgA/m)/2∏ for this problem?

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:
  • #8
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
 

1. What is a U tube and how does it work?

A U tube is a simple apparatus used to demonstrate the oscillations of fluids. It consists of a U-shaped tube filled with a liquid, such as water or oil. The liquid levels in the two arms of the tube are equal when the tube is at rest. When one arm is raised or lowered, the liquid in the other arm will also move, creating an oscillation.

2. What causes the oscillations in a U tube?

The oscillations in a U tube are caused by the difference in pressure between the two arms of the tube. When one arm is raised, the pressure in that arm decreases, causing the liquid to flow from the other arm to balance out the pressure. This back and forth movement creates the oscillations.

3. How does the density of the liquid affect the oscillations in a U tube?

The density of the liquid affects the speed of the oscillations in a U tube. A denser liquid will have a slower oscillation frequency, while a less dense liquid will have a faster oscillation frequency. This is because the denser liquid requires more force to move, thus slowing down the oscillations.

4. Can the oscillations in a U tube be used to measure pressure?

Yes, the oscillations in a U tube can be used to measure pressure. By measuring the height of the liquid in one arm of the tube, the pressure in that arm can be determined using the equation P = ρgh, where ρ is the density of the liquid, g is the acceleration due to gravity, and h is the height of the liquid.

5. What other factors can affect the oscillations in a U tube?

The oscillations in a U tube can also be affected by the diameter of the tube, the viscosity of the liquid, and the length of the tube. A narrower tube, thicker liquid, or longer tube will result in slower oscillations, while a wider tube, thinner liquid, or shorter tube will result in faster oscillations.

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