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I think its a trick question

  1. Jul 21, 2004 #1
    Hey guys, I ran into a problem that I can't figure out:

    In order to decrease the fundamental frequency of a guitar string by 2.0%, by what percentage should you reduce the tension?

    I know it's some kind of trick question, but I have no clue what the relationship between tension and frequency is. Could someone help me out? Thanks.
  2. jcsd
  3. Jul 21, 2004 #2

    f = fundamental frequency
    L = string length between fixed points
    T = string tension
    [itex]\rho[/itex] = string mass per unit length
  4. Jul 21, 2004 #3
    Thanks for that formula!
    Okay, now that I've been given the relationship between tension and frequency, how do I go about finding the percentage to reduce the tension by? I came up with a very complicated equation that lead me nowhere...
    f - f/.02 = (1/2L)(T/u)^1/2 - (1/.0008L)(T/u)^1/2, where T = tension, u = mass density, f = frequency, and L = length. I don't know where to go from here, or even if that's the right direction.
  5. Jul 22, 2004 #4
    Assume that [itex]\rho[/itex] and L are constants.


    The only variable is T so one could say f is proportional to a constant(k) times the root of T:

    f\ \propto\ k\sqrt{T}

    If you let k=1 then you're left with(for your purposes at least):

    f\ \propto\ \sqrt{T}

    You could do the problem leaving k in if you like because it will cancel in the end, but that's your perogative.

    You should be able to solve it now. When looking for things like percent change try and figure out which variables are actually changing. Here, L would not change because you are not moving the points of connection and a 2% change would have nominal effect on the actual length of the string. Rho is mass/length thus it too is unchanged. We can group the constants and say the chage in frequency is dictated by the change in the tension of the string.

    Also, your approach to %change doesn't look entirely correct.

    \% change_T=\frac{T_i-T_f}{T_i}\times 100\%

    Where Ti is initial tension and Tf is final tension.

    f_i\ \propto\ k\sqrt{T_i}\ \propto\ \frac{f_f}{98\%}

    and you know that [itex]f_f[/itex] is equal to [itex](100\%-2\%)f_i[/itex]. Solve the entire problem using the above percent chage formula in terms of [itex]f_f[/itex] or [itex]f_i[/itex]. Either way works but you need to use the same frequency throughout.

    Good luck.

    [edit] fixed my crazy sqrt
    Last edited: Jul 22, 2004
  6. Jul 22, 2004 #5

    Doc Al

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    Staff: Mentor

    You can certainly do this by brute force. But an easier way to approach it is to realize that small changes can be approximated by differentials. A 2% change is small enough.

    Start with this variation of faust9's proportion:
    [tex] T \propto f^2 = k f^2[/tex]
    Then [itex]\Delta T = \frac{dT}{df} \Delta f[/itex]. Figure the derivative and express this equation in terms of percent (fractions) to find the simple answer.
  7. Jul 22, 2004 #6


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    Science Advisor
    Homework Helper

    A slighlty different description of what faust said:

    You could set it up this way:
    [tex]0.98 \times f_1= f_2[/tex]
    [tex]0.98 \times \frac{1}{2L} \sqrt{\frac{T_1}{\rho}}=\frac{1}{2L}\sqrt{\frac{T_2}{\rho}[/tex]
    ...a bunch of stuff drops out
    [tex]0.98 \times \sqrt{T_1} = \sqrt{T_2}[/tex]
    and so on.
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