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Change in Tension & Fundamental Frequency of a String

  1. Apr 1, 2007 #1
    [SOLVED] Change in Tension & Fundamental Frequency of a String

    Problem. Show that if the tension in a streched string is change by a small amount [itex]\Delta F_T[/itex], the frequency of the fundamental is changed by a small amount [itex]\Delta f = 1/2 (\Delta F_T / F_T) f[/itex].

    Let T be the intial tension and h the change in tension. The velocity of a transverse wave on the string is [itex]v = \sqrt{T/\mu}[/itex]. The initial frequency is

    [tex]f = \frac{v}{\lambda} = \frac{\sqrt{T}}{\lambda \sqrt{\mu}}[/tex]

    The new frequency f' is

    [tex]f' = \frac{v'}{\lambda} = \frac{\sqrt{T + h}}{\lambda \sqrt{\mu}}[/tex]

    The difference is:

    [tex]f' - f = \frac{1}{\lambda \sqrt{\mu}} \, (\sqrt{T + h} - \sqrt{T})[/tex]

    That looks nothing like what I'm trying to show. Now, if I multiply the RHS by [itex]\sqrt{T} / \sqrt{T}[/itex], I get

    [tex]f' - f = \frac{\sqrt{T + h} - \sqrt{T}}{\sqrt{T}} \, f[/tex]

    and if I do it again, I get

    [tex]f' - f = \frac{\sqrt{T(T + h)} - T}{T} \, f[/tex]

    which is as close as I could get to what needs to be shown.
     
    Last edited: Apr 1, 2007
  2. jcsd
  3. Apr 1, 2007 #2

    Doc Al

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    Staff: Mentor

    binomial expansion

    [tex]\sqrt{T + h} = \sqrt{T}(1 + h/T)^{1/2}[/tex]

    Hint: Approximate that expression by taking a binomial expansion to first order in h/T. (Note that h/T << 1)
     
  4. Apr 2, 2007 #3
    Great hint! I never considered it. The approximation is given below:

    [tex]\sqrt{T(T + h)} = T + 1/2 \, h[/tex]

    and so

    [tex]f' - f = \frac{T + 1/2 \, h - T}{T} \, f = \frac{h}{2T} \, f[/tex]

    Thanks.
     
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