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Oscillations of fluid in a U tube

  1. Nov 30, 2011 #1
    1. The problem statement, all variables and given/known data

    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).


    2. Relevant 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.

    3. 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!
     
  2. jcsd
  3. Dec 1, 2011 #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
     
  4. Dec 2, 2011 #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
     
  5. Dec 2, 2011 #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
     
  6. Dec 2, 2011 #5
    OK is freq=√(ρgA/m)/2∏ for this problem?
     
  7. Dec 2, 2011 #6
    That is what I got !!!!!
     
  8. Dec 2, 2011 #7
    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
  9. Dec 2, 2011 #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
     
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