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Question on Harmonics involving Air Tube

  1. Jan 30, 2008 #1
    1. The problem statement, all variables and given/known data
    A tuning fork is set into vibration above a vertical open tube filled with water, as seen in the figure below.
    [​IMG]

    The water level is allowed to drop slowly. As it does so, the air in the tube above the water level is heard to resonate with the tuning fork when the distance from the tube opening to the water level is L1 = 0.115 m and L2 = 0.400 m. What is the frequency of the tuning fork? Assume that the speed of sound in air is 343 m/s.



    2. Relevant equations
    F=V/wavelength

    at first I assumed it was a closed tube and used:

    Fn=nv/(4L)



    3. The attempt at a solution

    I treated it like a closed tube, and tried F= V/4L
    but the computer said it was wrong.
    I fiddled around a couple of times and tried it as an open tube and still couldnt get the right answer.

    Ive tried: 746 Hz, 301 Hz, 1200 Hz, and 1490 Hz.

    Can anyone see what I am doing wrong?

    Thanks iun advance
     
  2. jcsd
  3. Jan 30, 2008 #2

    G01

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    The tube is an open/closed tube. Have you tried analyzing it by treating one end as open and one as closed? If so, can you please show your work? I can't find what your doing wrong if you don't show your work.
     
  4. Jan 30, 2008 #3
    I tried it as the top being open and the water representing the closed end of the tube.

    I used F=V/4L. L being 0.115 m and V 343 m/s

    Does L2 need to be used somehow?
     
  5. Jan 30, 2008 #4

    G01

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    Do you know that L1 is the first resonance point of the wave? If you don't, you can't assume n=1.

    If this is the case,

    HINT: Let, [tex]n=n_o[/tex] for L=L1 and n = the next odd integer after[tex]n_o[/tex](since this is an open/closed tube) for L=L2. Can you solve for F using the equations you obtain?
     
    Last edited: Jan 31, 2008
  6. Jan 31, 2008 #5

    andrevdh

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    The standing wave that forms in the air column need to have a antinode (position of maximum displacement of the oscillations of the air molecules) at the open end and a node at the bottom (top of the water). This means that a quarter of the standing wavelength will fit into the air column when the first resonance occurs and three quarters will fit into the air column when the second resonance is heard (antinode at the open end and a node at the bottom).

    In this case there is another complication. The top end of the standing is a bit above the mouth of the tube. This distance is normally referred to as the end-correction (e). The formula should therefore be modified as follows

    [tex]e + L_n = \frac{n}{4}\lambda[/tex]

    in his case we have the first resonance when n = 1 (L1) (quarter of a wavelength fits) and in the second case n = 3 (L2) (three quarters of the wavelength fits in the air column).
     
    Last edited: Jan 31, 2008
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