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Fundamental frequency of a guitar string

  1. Nov 11, 2007 #1
    1. The problem statement, all variables and given/known data

    In order to decrease the fundamental frequency of a guitar string by 4%, by what percentage should you reduce the tension?

    2. Relevant equations

    f = sqrt [T/(m/L)] / 2L

    I believe that is the equation that relates frequency to tension...

    3. The attempt at a solution

    I plugged in some theoretical values and got 76.96% which seemed wrong, and it is :)
    How can I go about solving this problem?
     
  2. jcsd
  3. Nov 11, 2007 #2
    [tex] f_1 = (1/(2L)) * \sqrt(F/\mu)[/tex]

    Your formula is correct.

    First, do f and F vary directly or inversely with each other?
     
  4. Nov 11, 2007 #3
    as L gets bigger, F gets bigger, right?
     
  5. Nov 11, 2007 #4
    Yes, except in this case, they will be getting smaller.

    What is the ratio between them?
     
  6. Nov 11, 2007 #5
    if L doubles, than T is quadrupled, right?
     
  7. Nov 11, 2007 #6
    but this has nothing to do with the frequency, since L stays the same...
     
  8. Nov 11, 2007 #7
    Whoops, I misread what you typed. We don't care whether the tension and length vary directly or inversely. We want how frequency and tension are related. The rest of the equation isn't important, since this is just asking for a relative number.

    [tex]f \alpha \sqrt(F)[/tex]
     
  9. Nov 11, 2007 #8
    so the frequency is proportional to the square of the tension?
     
  10. Nov 11, 2007 #9
    Yep. So, putting what we know together;
    [tex] 0.96f \alpha \sqrt(xF)[/tex]
    You need to find x.
     
  11. Nov 11, 2007 #10
    no idea. this is kinda where i got stuck...
     
  12. Nov 11, 2007 #11
    No problem. As [tex]f \alpha \sqrt(F)[/tex], it makes sense that [tex]0.96 \alpha \sqrt(x)[/tex], right? Use [tex] 0.96 = \sqrt(x)[/tex] to solve for x. This will give you a decimal value, which you multiply by 100 to turn into a percent. This is the percent of the original length needed to change the frequency by 4%, so to get the answer, you subtract it from 100%.

    % to decrease = 100% - (100x)
     
  13. Nov 11, 2007 #12
    okay, that's what i was thinking, but i didn't know if i could use an equals sign since we were working with a proportion :) thank you!
     
  14. Nov 11, 2007 #13
    Glad to be of help :)
     
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