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Energy of a Plucked String

  • Thread starter e(ho0n3
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  • #1
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Problem. A string fixed at two ends (which are a length L apart) is pulled up at the center to a height of h. Assuming that the tension T remains constant, calculate the energy of the vibrations of the string when it is released. [Hint: What work does it take to strech the string up?]

The work to pull the string is

[tex]\int_0^h \frac{y}{c} \, T \, dy[/tex]

where

[tex]c = \sqrt{y^2 + (L/2)^2}[/tex]

right? And if I were to calculate the energy directly, I would need to know the frequency of vibration and the linear density of the string right?
 

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  • #2
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Nevermind. One may find the frequency given the linear density using the fact that a standing wave is produced when the string is released so:

[tex]f = \frac{v}{\lambda} = \frac{v}{2L} = \frac{\sqrt{T/\mu}}{2L}[/tex]

The energy is then given by:

[tex]E = \int_0^L 2 \pi^2 f^2 D(x)^2 \mu \, dx[/tex]

where

[tex]D(x) = h \sin (\pi x / L)[/tex]

right? I should, in theory, get the same answer using this method and the method in the first post.
 
  • #3
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I ask because the latter integral is much easier to calculate (at least for me) than the former one. For the latter one, I get [itex]\pi T h / L[/itex] as the answer.
 
  • #4
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It just dawned on me that

[tex]\frac{d}{dy} \sqrt{y^2 + (L/2)^2} = \frac{y}{\sqrt{y^2 + (L/2)^2}}[/tex]

Duh! So the integral in the first post becomes [itex]T(\sqrt{h^2 + L^2/4} - L/2)[/itex]. This doesn't agree with what I posted earlier. (After a quick dimensional analysis, I realize that the energy I calculated in post #3 is wrong.) Hmm...
 
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