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Using Waves to Find Depth of Well

  1. Jun 12, 2015 #1
    1. The problem statement, all variables and given/known data

    A student uses an audio oscillator of adjustable frequency to measure the depth of a dried up well. Two successive resonances are heard at 17.0Hz and 23.8Hz. The speed of sound in air is 343 m/s. How deep is the well?

    2. Relevant equations


    3. The attempt at a solution

    I solved the problem and the answer was 25.22m. However, I got this answer by using the equation for a tube open at both ends. I was confused because I thought that the correct equation would have been the one for a tube open at one end and closed at the other. Can someone explain why f=nv/2L was used instead of f=nv/4L
    Last edited by a moderator: Jun 12, 2015
  2. jcsd
  3. Jun 12, 2015 #2

    Simon Bridge

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    You can and should use the closed ended pipe formula - did you try it?
    Remember that n in the closed pipe formula, can only take odd-number values.
  4. Jun 12, 2015 #3
    Yes, I tried it (f=nv/4L) with n equal to one and the online assignment page said that it was wrong and then when I used the open-open formula (f=nv/2L) with n equal to one, the website said that the answer was correct. I was assuming that it was an error with the answer, but I just wanted to make sure.
  5. Jun 13, 2015 #4

    Simon Bridge

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    The assignment does not appear to be in error - you will get the same result using the correct equation provided you understand what you are doing.

    For a closed tube resonator, if the first frequency is the nth harmonic, then the next freqeuncy will be the (n+2)th harmonic. The difference between them will be given by:
    $$f_{n+2}-f_n = \frac{(n+2)v}{4L} - \frac{nv}{4L} = \frac{v}{2L}$$

    But for an open open tube, the nth harmonic is followed by the (n+1)th harmonic - so the difference becomes:
    $$f_{n+1}-f_n = \frac{(n+1)v}{2L}-\frac{nv}{2L} = \frac{v}{2L}$$
    ... exactly the same value.

    The hint was trying to get you to realize that there was a shortcut.
  6. Jun 13, 2015 #5
    Thank you!

    I can't believe I missed that, it seems so simple now.
  7. Jun 13, 2015 #6

    Simon Bridge

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    This is why it is important to understand the relations instead of just learning to use the equations.
    In general, you should not follow a hint blindly - it should give you that "ah-ha!" moment at some stage.
    If that does not happen, then go back to working without the hint.

    I don't like to give out hints in my coursework because they can obscure important lessons.
    i.e. Without the hint you'd probably have figured out that the question had set you up with two equations with two unknowns (n and L) then you solve by simultaneous equations to find L.
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