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Wave speed, freq, and period of wave on a string

  1. May 1, 2013 #1
    1. The problem statement, all variables and given/known data

    A musician frets a guitar string of length 1.5 m at x = 0.34 m with one finger, and simultaneously plucks the string at x = 0.17 m with another finger (raising it to a height h = 2.1 mm. Both fingers are simultaneously removed from the string, and it is allowed to vibrate. The string has a tension of 76 N, and a linear mass m = 4.9 g/m.

    2. Relevant equations

    [itex]v=\sqrt{\frac{T}{\mu}}[/itex]
    where
    [itex]\mu=\frac{mass\,in\,kg}{Length\,in\,m}[/itex]

    [itex]v=f\lambda[/itex]



    3. The attempt at a solution

    I started off converting the linear mass:

    [itex]4.9\frac{g}{m} \times\frac{1 kg}{1000 g} = .0049\frac{kg}{m}[/itex]

    Then I plugged that into the equation:

    [itex] v=\sqrt{\frac{76 N}{.0049\frac{kg}{m}}}[/itex]

    and got v=124.54 m/s

    From here I used [itex]v=f\lambda[/itex]
    where [itex]\lambda=.68 m[/itex]

    and got frequency = 183.147 Hz

    Taking the reciprocal of that:

    Period = .00546 s

    These aren't the right answers, but I don't see where I went wrong

    Also, picture attached
     

    Attached Files:

  2. jcsd
  3. May 1, 2013 #2

    haruspex

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    You don't say exactly what is being asked. You have calculated, correctly I believe, the period as though the fret finger is still in place, but is it what was asked for?
     
  4. May 1, 2013 #3
    I included what was asked for in the title, but forgot to include it in the question. It's asking for the wave velocity, frequency, and period.

    I accounted for the finger being lifted off by making the wavelength twice the interval between the finger and the origin. Is that not right?
     
  5. May 1, 2013 #4

    haruspex

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    No, that's what it would be if the finger were left in place. But even then, 1.5m is not twice 0.34m. How do you propose to bring 1.5m into the equation?
     
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