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Steel ball oscillating in a glass tube

  1. Aug 13, 2013 #1
    1. The problem statement, all variables and given/known data
    a) Find the frequency of vibration under adiabatic conditions of a column of gas confined to a cylindrical tube, closed at one end, with a well-fitting but freely moving piston of mass m.

    b) A steel ball of diameter 2 cm oscillates vertically in a precision-bore glass tube mounted on a 12-liter flask containing air at atomospheric pressure. Verify that the period of oscillation should be about 1 sec. (Assume adiabatic pressure change with [itex]\gamma[/itex] = 1.4, Density of steel = 7600 [itex]\frac{kg}{m^{3}}[/itex]



    2. Relevant equations
    from part a, we get:
    [tex]\omega = \left( \frac{\gamma p A}{m l} \right) ^{1/2}[/tex]
    where [itex]p[/itex] is the pressure, [itex]A[/itex] is the area of the tube, [itex]l[/itex] is the length from the bottom of the tube to an undisplaced piston.

    for part b, I thought it would be realistic to state the period as:
    [tex]T = 2*pi \left( \frac{m l}{A \gamma p} \right) ^{1/2}[/tex]
    before going on I suspect this may be where I'm getting hung up. I'm assuming the 12 L flask to be equivalent to a tube that is the area of the steel ball and then a given length so both volumes match. Also I'm assuming the steel ball to be about the same as a piston, only really taking into account its sphericalness when calculating the mass from the density of steel and the volume of a sphere.
    [tex]l = \frac{V}{A}[/tex]
    Are these two assumptions realistic?
     
  2. jcsd
  3. Aug 13, 2013 #2

    Simon Bridge

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    You want to know if the geometry of the ball, since it is different from the piston, has a noticable effect on the outcome?
    I don't see how to work that out without crunching the numbers.
    I'd proceed, off the context, by taking the result from (a) as useful in (b) without being modified, and see what happens.
     
  4. Aug 13, 2013 #3
    Thats what I'm trying to do with [itex]T = \frac{2 \pi}{\omega}[/itex]. But I'm not sure what to use for [itex]l[/itex], because its a flask and not a tube.
     
  5. Aug 13, 2013 #4

    Simon Bridge

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    What is the difference between the flask and the tube?
     
  6. Aug 13, 2013 #5
    It seems like the geometry of a flask and a tube are different. A tube has a uniform area all the way up and down its length. While a flask the area changes as you go up or down the length.
     
  7. Aug 13, 2013 #6

    Simon Bridge

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    Well then you need to know how the geometry changes ... does it matter if the change in volume during the oscillations is small? It's the way the volume changes that's important right? So pick l to keep the same volume.
     
  8. Aug 13, 2013 #7
    [tex]T = 2 \pi \left( \frac{m l}{A \gamma p} \right) ^{1/2}[/tex]
    [tex]m = 3.18*10^{-5} \ kg[/tex]
    [tex]l = \frac{V}{A} = \frac{0.012 \ m^{3}}{ 3.14*10^{-6} \ m^{2}} = 3820 \ m[/tex]
    [tex]A = 3.14*10^{-6} \ m^{2}[/tex]
    [tex]\gamma = 1.4[/tex]
    [tex]p = 101325 \ \frac{N}{m^{2}}[/tex]
    http://wolfr.am/16OKjiZ

    Another thing is that the ball actually needs to sink a little bit before it reaches equilibrium because it is vertical, but adding these adjustments doesn't seem to significantly change the answer.

    It should be about 1 sec. I'm not sure what part of this calculation isn't realistic but I suspect it has to do with what I'm using for my [itex]l[/itex] or what I'm using for my [itex]p[/itex].
     
  9. Aug 14, 2013 #8

    TSny

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    Check the calculation of the mass of the ball and the area ##A##. I think you have the wrong power of 10.
     
  10. Aug 14, 2013 #9
    thank you for spotting that. I had 1 cm = .001m instead of .01. With that adjustment I get the required 1 sec.
    http://wolfr.am/1cBL6ZJ
     
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