Oscillations of fluid in a U tube

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Homework Help Overview

The discussion revolves around the oscillations of a nonviscous, incompressible fluid in a U tube, specifically focusing on determining the frequency of small oscillations about the equilibrium position. The problem involves concepts from fluid dynamics and harmonic motion.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the relationship between pressure, force, and displacement in the context of simple harmonic motion (SHM). There are attempts to apply Newton's second law and derive equations of motion, with some participants questioning the role of gravitational force and the correct expression for mass in the system.

Discussion Status

Several participants have offered insights and corrections regarding the relationships between pressure, force, and displacement. There is an ongoing exploration of the correct formulation for the frequency of oscillation, with some noting discrepancies in earlier calculations. The discussion reflects a collaborative effort to clarify the underlying physics and mathematics.

Contextual Notes

Participants note potential confusion regarding the definitions and assumptions in the problem, particularly concerning the displacement used in the frequency formula and the mass of the liquid in the U tube. There is acknowledgment that the problem statement may lack clarity.

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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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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
 
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
 
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
 
OK is freq=√(ρgA/m)/2∏ for this problem?
 
That is what I got !
 
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:
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
 

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