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Homework Help: Fundamental Frequency of a string

  1. Feb 2, 2009 #1
    1. The problem statement, all variables and given/known data
    The length of a string is 1440 cm. The
    string is held fixed at each end. The string
    vibrates in eight sections; i.e., the string has
    eight antinodes, and the string vibrates at
    150 Hz.

    What is the fundamental frequency? Answer
    in units of Hz.

    2. Relevant equations

    f = nv / 2L frequency = (# antinodes)(velocity) / (2 * length)

    3. The attempt at a solution

    I already found the wavelength which is 3.6 m. I got that by substituting nv/2L for f into the equation v=λf (velocity = wavelength*frequency). Then I followed the above frequency formula:

    v=(3.6 m)(150 Hz) = 540 m/s

    f = nv / 2L
    f = (8)(540 m/s) / (2*14.4 m) = 150 Hz

    My online HW said it was wrong. So I tried again.

    f = (2)(540 m/s) / (2*14.4 m) = 37.5 Hz

    I used two because it said that fundamental frequency is the lowest frequency and you can use the first harmonic or something. Didn't really understand it. But the two is because the 1st harmonic has 2 antinodes.

    I am so confused and I have no clue what I am doing! Please help <:S
    Last edited: Feb 2, 2009
  2. jcsd
  3. Feb 2, 2009 #2


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    If there are 8 antinodes then there are 4 wavelengths represented on the string.

    But I think 8 antinodes means it's the 8'th harmonic of the string.

    So isn't 1/8 of 150 hz then the first harmonic and isn't that the frequency you use with 2*L to determine v?
  4. Feb 2, 2009 #3


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    Or do you not need to find the v?

    Only the f1 fundamental frequency?
  5. Feb 2, 2009 #4
    i already found v. i just dont understand the fundamental frequency part.
  6. Feb 2, 2009 #5


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    8 antinodes means 8th harmonic = 8 times the fundamental frequency (first harmonic).
  7. Feb 2, 2009 #6


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    Here's a link:
    http://www.glenbrook.k12.il.us/GBSSCI/PHYS/Class/sound/u11l4d.html [Broken]
    Last edited by a moderator: May 4, 2017
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