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Water column resonance

  1. Mar 27, 2014 #1

    utkarshakash

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    1. The problem statement, all variables and given/known data
    There is a cylindrical vessel of cross-sectional area 20cm^2 and length 1m is initially filled with water to a certain height as shown. There is a very small pinhole of cross-sectional area 0.01cm^2 at the bottom of the cylinder which is initially plugged with a pin cork. The air column in the cylinder is resonating in fundamental mode with a tuning fork of frequency 340Hz. Suddenly the pin cork is pulled out and water starts flowing out of the cylinder. Find the time interval after which resonance occurs again.



    3. The attempt at a solution
    The setup is equivalent to a closed organ pipe. The fundamental frequency is given by nv/4L.
    Let the initial height of water column be h1. At second resonance let the height change to h2.

    h1-h2=v/4*340

    The velocity of water coming out of hole is given by [itex]\sqrt{2gx}[/itex]

    Thus, rate of decrease of water column = 0.01v/20.

    [itex]\dfrac{dx}{dt} = \dfrac{0.01\sqrt{2gx}}{20} [/itex]

    If I integrate this from h1 to h2, I am left with the expression [itex]\sqrt{h_1}-\sqrt{h_2}[/itex].

    I already know the difference between the heights but not the difference between their square roots. How do I calculate it?
     
  2. jcsd
  3. Mar 27, 2014 #2

    TSny

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    I don't think this is correct. If you haven't already done so, draw a diagram showing the nodes and antinodes of the standing sound wave for each of the two resonances.

    This looks good except you should think about whether or not a negative sign needs to be included (x is decreasing as time increases).

    From your diagrams of nodes and antinodes, you should be able to determine the heights h1 and h2.

    By the way, what are you supposed to use for the speed of sound?
     
  4. Jul 4, 2014 #3
    Here is my attempt at the problem .

    I am using g = 9.8ms-2 and speed of sound = 330 m/s

    The length of air column l = nλ/4 = nv/4f , n=1,3,5...

    l1 = 0.24m
    l2 = 0.72m

    Consequently h1 = 0.76m and h2 = 0.28m

    ##\dfrac{dx}{dt} = - \dfrac{0.01\sqrt{2gx}}{20}##

    Integrating , t = 902.93 (√h1 - √h2)

    Putting values of h1 and h2 , I am getting t ≈ 307 sec .

    TSny ,are you getting the same answer ?

    Many Thanks
     
  5. Jul 4, 2014 #4

    TSny

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    That looks good. I get slightly different results for some of the numbers in your calculation. My final answer for t = 315 s ≈ 320 s to 2 significant figures.
     
  6. Jul 4, 2014 #5
    Thanks a lot :smile:
     
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