Calculating Energy of Vibrating String Problem

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Homework Help Overview

The problem involves calculating the energy of vibrations of a string fixed at both ends and pulled up at the center. The context includes concepts from mechanics and wave physics, particularly focusing on tension, work done, and energy associated with vibrating strings.

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Approaches and Questions Raised

  • Participants explore different methods to calculate the energy, including integrals related to work done and energy in terms of frequency and linear density. Questions arise about the validity of the approaches and the relationships between variables.

Discussion Status

The discussion is active, with participants sharing different approaches and questioning their calculations. Some participants express uncertainty about their results and seek clarification on the relationships between the variables involved.

Contextual Notes

There are indications of confusion regarding the calculations and assumptions made, particularly in relation to the integrals and the derived expressions for energy. Participants are also considering the implications of dimensional analysis on their results.

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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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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.
 
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.
 
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...
 
Last edited:

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