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Standing waves on a fixed string

  1. Feb 17, 2014 #1
    Hello,

    1. The problem statement, all variables and given/known data

    Two wires, each of length 1.8 m, are stretched between two fixed supports. On wire
    A there is a second-harmonic standing wave whose frequency is 645 Hz. However,
    the same frequency of 645 Hz is the third harmonic on wire B. Find the speed at
    which the individual waves travel on each wire.

    2. Relevant equations

    [itex]L = \frac{nv}{2f_{n}}[/itex]

    3. The attempt at a solution

    I don't know if I understand the idea of natural frequencies correctly and it's relation to n (an integer value in the above equation).

    If I imagine a string fixed at both ends there are a number of different standing waves that can be made, ie different harmonics.

    The first harmonic has 1 antinode, the second has two etc.

    When working out the velocity of the wave on a string, does the 'n' refer to the harmonic? I assume that the different harmonics can be considered to be the natural frequencies of the string.

    I'm fairly sure I have this wrong, because I get slower speeds for higher harmonics and intuition tells me that this is wrong. I remember having to shake the rope up and down much harder to reach the next standing wave in an 'experiment' that was done in school.

    I'd really appreciate a helping hand :)

    BOAS
     
    Last edited: Feb 17, 2014
  2. jcsd
  3. Feb 17, 2014 #2
    The speed is the same for all harmonics (and fundamental). At least in the first approximation.
    So in your formula, for a given string, L and v are fixed. Only frequency and n change for various harmonics.
     
  4. Feb 17, 2014 #3

    AlephZero

    User Avatar
    Science Advisor
    Homework Helper

    In your experiment, the rope was the same length and tension, and the frequency changed.

    In this question, you have two different wires, with different tensions and/or diameters, and the frequency stays the same.

    Your "relevant equation" is correct.
     
  5. Feb 17, 2014 #4
    So rearranging for v gives [itex]v = \frac{2fl}{n}[/itex] and the n refers to the particular harmonic, hence the standing wave on the second string is slower.

    Thanks for the help - it hadn't twigged in my brain that the two strings are not said to be the same.

    BOAS
     
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