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Standing waves and variable frequency

  1. Oct 6, 2011 #1
    •• Working for a small gold mining company, you stumble across an
    abandoned mine shaft that, because of decaying wood shoring, looks too
    dangerous to explore in person. To measure its depth, you employ an audio
    oscillator of variable frequency. You determine that successive resonances are
    produced at frequencies of 63.58 and 89.25 Hz. Estimate the depth of the shaft.

    I pictured the mine or cave as a one open end pipe or string tied at one end. To estimate the depth I only need to find L.
    In other words:

    maximum Lambda or Lambda(sub1)=4L

    Lambda-sub1= velocity of sound/fundamental frequency=13.36969m.
    Fundamental frequency is equal to the subtraction of the consecutive frequencies given.

    Then,
    (maximum Lambda/4)=L

    should be equal to the maximum depth of the mine. My answer is 3.34m

    Why is this answer wrong?

    Thank you.
     
  2. jcsd
  3. Oct 6, 2011 #2

    berkeman

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    Staff: Mentor

    It's a tricky question. They don't say that the first resonance given is the fundamental resonance. They only say that the two resonances given are successive....

    Does that fix you up?
     
  4. Oct 6, 2011 #3
    I am aware that the first resonance given is not the first harmonic.

    The difference of two consecutive frequencies for an open ended pipe or a string attached on one end is the fundamental harmonic or first harmonic. So, first harmonic is equal to 25.67 Hz.
     
  5. Oct 6, 2011 #4

    berkeman

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    Staff: Mentor

    Ah, I see the shortcut you used. Yeah, when working the numbers, I get pretty much what you got. What are you using for the speed of sound? I get the depth as 3.31m with the speed of sound at 340m/s.
     
  6. Oct 6, 2011 #5
    That's still not the answer. I used 343.2 m/s for v-sound.
     
  7. Oct 6, 2011 #6

    berkeman

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    Staff: Mentor

    Okay, then back to my original line of thinking. The two resonances fit into the same length L. Each resonance occurs when that frequency fits some odd number of quarter wavelengths into the length L, right? That seems like the right way to approach it...
     
  8. Oct 6, 2011 #7
    Yes that is the right way to approach it, but why's that the case? And, how do I know, when to use one approach or the other?
     
  9. Oct 6, 2011 #8

    berkeman

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    Staff: Mentor

    I think the problem with the first approach is that it misses the fact that both resonances happen in the same length L. If the problem gave two resonances and asked what the fundamental resonance would be, then using the quarter wavelength trick would have been correct. But we were given two adjacent resonant frequencies that fit into the length L, so we needed to figure out what length L would work for some resonance numbers for both of those two frequencies.
     
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