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Frequency and length

  1. Sep 23, 2015 #1
    1. The problem statement, all variables and given/known data
    upload_2015-9-23_20-54-40.png

    2. Relevant equations


    3. The attempt at a solution
    for the first harmonic, the wavelength is twice the length of string.
    so 0.296m x 2 = 0.592m is the wavelength

    with the wavelength and frequency known, we solve for v in v = ƒλ

    v = 0.592m x 440 Hz = 260 m/s

    then to solve for the length of new string , we just use the same v (260 m/s) and use the new frequency (523 Hz) and solve for the L?

    if so , I got 24.9 cm
     
  2. jcsd
  3. Sep 24, 2015 #2

    haruspex

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    With regard to the parts of the string, what length have you calculated as 24.9cm? What length does it ask for?
     
  4. Sep 24, 2015 #3
    i calculated the length starting from beneath the finger to the nut to be 24.9 cm
     
  5. Sep 24, 2015 #4

    haruspex

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    What part of the string is vibrating at 523Hz?
     
  6. Sep 24, 2015 #5
    just the 24.9 cm of string between the finger and nut I suppose since the rest isn't vibrating
     
  7. Sep 24, 2015 #6

    haruspex

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    What is being used to make the string vibrate? What part of the string is it acting on?
     
  8. Sep 24, 2015 #7
    to make the string vibrate, you would have to pluck it, and if you held the string down with your thumb, only the part of the string between your thumb and the nut, will vibrate.
     
  9. Sep 24, 2015 #8

    haruspex

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    You could do that, but you would get hardly any sound out of the violin. Whether played with a bow (as in the diagram) or plucked, it is the part between the bridge and the finger that is plucked or bowed. (The vibration is transferred through the bridge to the broad wooden lamina, and thence to the air inside.)
     
  10. Sep 24, 2015 #9
    ah, I had to subtract the two, 29.6 cm - 24.9 cm = 4.7 cm.
     
  11. Sep 24, 2015 #10

    haruspex

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    Yes.
     
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