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Wine glass acoustics - wavelength not what expected

  1. Mar 15, 2008 #1
    Hi,
    I have been conducting a lab experiment using a piece of latex glove to stimulate a tone from a wine glass that is rotating on a turntable. I used the equation [tex]\lambda[/tex]=v/f (using an audio spectrometer setup to find f) to find the wavelength of the emitted tone.
    We expected the top part of the glass would be a quarter of the length of the wavelength, as in an organ pipe with one closed end. What I found though, was that it was very close to half.
    Does anyone know why my initial assumption was wrong?

    Thanks

    -Jam
     
  2. jcsd
  3. Mar 15, 2008 #2

    mgb_phys

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    The air in the glass isn't vibrating like a standing wave down the length of the glass as in an organ tube, the rim of the glass is vibrating like a bowstring.
    Imagine that two opposite points on the rim are staionary and the curve between them vibrates.
     
  4. Mar 15, 2008 #3
    omg thanks, that's been annoying me for weeks!
    Thankyou!:biggrin:
     
  5. Mar 15, 2008 #4
    ok, that leaves me with a second question;
    I just did a rough calculation, and it turns out that the wavelength is about the same as the circumference of the glass, which would lead me to expect to observe two nodes and two antinodes as I moved a microphone around the glass, i.e., two 'quietest' points, and two 'loudest' points.
    However, my previous research (and observation) suggests that there is in fact a quadrupole configuration, that is, there are four of each such nodes and antinodes.

    Why would this be?
     
  6. Mar 15, 2008 #5

    mgb_phys

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    Perhaps 2 waves at the same time and this is the most stable configuration?
    (only a guess)
     
  7. Mar 15, 2008 #6
    hmm... that was my initial thought too, but then the wavelength of the wave on the glass rim is then half that of the emitted tone...
    ok, thanks for your help :-)
     
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