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Determining the speed of sound from a half-open open pipe

  1. Dec 8, 2012 #1
    1. The problem statement, all variables and given/known data

    An observer listens to a vibrating string using a cardboard tube of length 1.4m placed close to his ear (one end closed). The string is excited so as to vibrate at its fundamental frequency, and the tension is increased slowly. The intensity of the sound heard by the observer increases at 130N and 360N (I assume these are two resonant frequencies for the tube then). The linear density of the string is 1.6g/m and its length is 40cm. Deduce the speed of sound in this air.


    2. Relevant equations

    v=sqrt(tension/linear density)
    f(string)=nv/2L
    f(tube)=nv/4L

    That's all I've been using.

    3. The attempt at a solution

    My reasoning was: Find the two frequencies of the string, find the difference, and this is the fundamental of the tube; put it through the f(tube) equation to get v.

    v(string)= sqrt(130N/1.6E-3kg/m)= 285.04 m/s

    Thus f(string)= nv(2L) where n=1 because it's at its fundamental, f=285.04m/s/0.8m= 356.3Hz

    v(string)2= (same method) = 474.34m/s
    Thus f(string)2= (same method) = 592.9 Hz

    The difference in the two is 236.6 Hz; I took this to be the fundamental frequency and things start to feel queasy:

    v=λf= (4L)*f = (5.6)(236.6).... This comes out at 1324.96 m/s. This is, what, 4 times too big.

    I tried this too:

    356.3= (n*v)/5.6 592.9=(n+1)*v/5.6
    v=1995/n 3320=(n+1)*v = (n+1)(1995/n)

    3320n=(n+1)(1995)=1995n+1995
    1325n=1995
    n=.... 1.5. Nonsense.

    So I'm a bit stuck!
     
  2. jcsd
  3. Dec 8, 2012 #2

    mfb

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    592.9 Hz/356.3Hz=1.664 or ~5/3
    To get this ratio in the tube with one open end, 356.3 Hz have to correspond to 3/4 wavelengths in the tube and 592.9 Hz are 5/4 wavelengths.

    In formulas:
    3/4v=L*356.3Hz
    5/4v=L*592.9Hz
    Taking the difference:
    1/2v=L*236.6 Hz
    With L=1.4m, this gives 662.48m/s.
    Hmm, still a factor of 2 away.

    Cross-check: With the known speed of sound of ~(340+-10)m/s, 593Hz correspond to a wavelength of ~57cm or 2.44+-0.07 wavelengths in the tube and 356.3 Hz correspond to 1.46 wavelengths in the tube.
    This makes sense if both ends are closed (or open) and 2.5 and 1.5 wavelengths are inside.
     
  4. Dec 8, 2012 #3

    Andrew Mason

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    Since the tube is open at both ends, the tube represents an even number of half wavelengths. The two resonances are consecutive ie difference in the number of half wavelengths between two standing waves (at 356.3 Hz and 592.9 Hz respectively) is 1. But we cannot assume that the first one was a fundamental. It was some number of half wavelengths and the second was that number of half wavelengths + 1.

    So let n = the first number of half wavelengths:

    1/2 = L

    (n+1)λ2/2 = L

    and, since vair is the same in both:

    f1λ1 = f2λ2

    I think you can work out what n must be from that. That enables you to find v of the air.

    AM
     
  5. Dec 8, 2012 #4
    I think I can. I'll have to come back to it in a couple of hours- revision process moving ever onwards- but I'll update once I've figured it out.
     
    Last edited: Dec 8, 2012
  6. Dec 8, 2012 #5

    Andrew Mason

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    Ok - maybe this is a closed tube because the ear is very close to one end, effectively closing the tube. If that is the case, then the L (1.4 m) represents an odd number of quarter wavelengths and the difference in the number of quarter wavelengths between consecutive resonances is 2:

    ie: nλ1/4 = L

    (n+2)λ2/4 = L

    where n is an odd number.

    AM
     
  7. Dec 8, 2012 #6

    gneill

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    Question: If the tube is effectively closed, won't the ear which forms the closed end be sitting at a node when there's a standing wave (resonance)? Wouldn't that produce a sound minimum for the observer? ...just a thought...

    Actually, the problem statement as given doesn't explicitly state that the ear forms the closed end. It may be a closed-end tube and the observer places his ear close to the open end while listening (sort of like placing a tuning fork near your ear).
     
  8. Dec 8, 2012 #7

    Andrew Mason

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    No. The intensity does not depend on whether there is a node or antinode at the ear. It depends on whether there is a standing wave in the tube. If the tube is closed at one end the standing wave can occur only if there is an node at the closed end (at the ear) and an antinode at the open end.

    That is the confusing aspect to the question.

    AM
     
  9. Dec 8, 2012 #8
    That ambiguity is my fault: The original wording said "...a cardboard tube of length 1.4m placed close to his ear (the tube hence behaves as open at one end and closed at the other)."

    You're right, there'd be a node at the ear end.
     
  10. Dec 8, 2012 #9
    So, if the closed end is at the observer's ear, and the open end is at the vibrating string, and we use (n+2) where I used (n+1) in my original attempts, then v=665 m/s, which is still double.
     
  11. Dec 8, 2012 #10

    gneill

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    I would have thought that there would be no air pressure changes at the nodes when a standing wave is present, hence minima for the observer's ear at the resonances.

    Edit:
    Of course, this still would not address the high speed issue, since we'd then expect maximas to occur halfway between minimas, and that would still leave us with the same Δf between frequencies of interest, and hence the same velocity value (##v = 2L\Delta f##).
     
    Last edited: Dec 8, 2012
  12. Dec 8, 2012 #11

    gneill

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    Another consideration could be that the 'air' is not standard air... could be a rich helium mix which would drive the velocity up.

    -just doing a bit of lateral thinking here... :smile:
     
  13. Dec 8, 2012 #12

    mfb

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    Closed end <-> no velocity <-> maximal pressure differences
    The eardrum is a good approximation of a closed end, I think. The connection cardboard tube <-> ear is different issue.


    @gneill: I had that problem in a lab course once - I think it was CO2, and I could estimate the fraction of CO2 afterwards and even explain why that fraction might have been present in the tube, but it is annoying to find that in the data analysis.
     
    Last edited: Dec 8, 2012
  14. Dec 8, 2012 #13

    gneill

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    Ah yes, I was confusing air motion envelope with pressure envelope. Thanks.
    Indeed!
     
  15. Dec 8, 2012 #14

    Andrew Mason

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    n has to be an odd number (of quarter wave lengths). There is only one value that n can have with these frequencies. If you don't want to work it out, just do trial and error starting at n=1 until you get the right ratio of frequencies.

    eg. if n = 1 then n+2 = 3 so λ2/λ1 = 3. But we know that λ2/λ1 = f1/f2 = 592.9/356.3 = 1.66. and 1.66≠3

    So try n=3 (it must be odd)..... etc.

    AM
     
  16. Dec 9, 2012 #15

    Andrew Mason

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    I have corrected the above so that λ1 is the longer wavelength: λ1/λ2 = 3

    f2/f1 = (v/λ2)/(v/λ1) = λ1/λ2 = (n+2)L/nL = n+2/n = 1 + 2/n = 1.66; n = 2/.66 = 3
     
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