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Standing Wave Question

  1. Nov 29, 2007 #1
    1. The problem statement, all variables and given/known data
    A particularly beautiful note reaching your ear from a rare Stradivarius violin has a wavelength of 39.1 cm. The room is slightly warm, so the speed of sound is 344 m/s.
    If the string's linear density is 0.620 g/m and the tension is 160 N, how long is the vibrating section of the violin string?

    T = 160N
    Linear Density = 0.62g/m
    Wavelength of violin string = 0.391m
    Speed of sound in warm room = 344m/s
    m = 1?

    2. Relevant equations

    v = [tex]\sqrt{\frac{T}{\mu}}[/tex]

    v = [tex]\lambda[/tex][tex]\times[/tex]f

    3. The attempt at a solution

    I all know is to find the velocity of the string first which is the square root of 250 or about 15.811m/s. I am unsure of what to do next. Also, what is the "vibrating section of the violin string"? I am unsure of what that is. Thanks in advanced.
  2. jcsd
  3. Nov 29, 2007 #2

    The vibrating section of the string is that of the standing wave, i.e. the bit better the bridges of the violin. This length is dependent on the note played. To solve this problem you need to figure out which harmonic the string is vibrating at.
  4. Nov 29, 2007 #3
    I still don't quite get what the "vibrating section" of the violin string is. Are you saying that it is the length of the string; which makes no sense. I think the harmonic for stringed instruments is one or m = 1. Somewhere in between I have to use this formula:

    f1 = [tex]\frac{1}{2L}[/tex][tex]\times[/tex][tex]\sqrt{\frac{T}{\mu}}[/tex].
  5. Nov 30, 2007 #4
    With stringed instruments you have as many harmonics as there are notes, effectively that is the physics behind it. A violinist can place his finger very lightly on a string so that it divides the string into half. He will hear a harmonic (the note an octave higher than the open string). By placing his fingers in other places he can get more harmonics, e.g by touching the string a quarter of the way down he gets the next harmonic.
  6. Dec 5, 2007 #5
    Sorry for the late reply, the way I approached the question was correct. My answer was incorrect because the linear mass density in the question was in g/m while I needed to convert to kg/m. This question took me hours to figure out, I was sure the way I was doing was correct.

    For future references, the answer to this question is 28.9cm.
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