Work done by a distributed force on a string

  • Thread starter Eidos
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  • #1
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Main Question or Discussion Point

Hi All

I'd like to know how I could calculate the work done by a distributed force on a string.

Let's say the force at a point [tex]x[/tex] at a time [tex]t[/tex] is given by

[tex]F(x,t)[/tex].

Now the instantaneous amplitude of the string is given by [tex]y(x,t)[/tex], say

I think that the work done by the force in changing the configuration of the string from some [tex]y(x_0,t_0)[/tex] to [tex]y(x,t)[/tex] should be something like
[tex]\int_{y(x_0,t_0)}^{y(x,t)} F(x,t) dy[/tex]

I'll use the total derivative on [tex]y(x,t)[/tex] which gives

[tex]dy=y_t dt+y_x dx[/tex]

where

[tex]\frac{\partial y}{\partial x}=y_x[/tex]

Now the integral becomes something like

[tex]\int F(x,t) y_t dt + \int F(x,t) y_x dx[/tex]

My concerns here are the limits I need to put in each integral.

Any help would be greatly appreciated :D
 
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Answers and Replies

  • #2
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I would like to ask some questions to understand this better.

How does the force [tex]\vec{F}(x,t)[/tex] act on the string? I'm guessing it's perpendicular to the [tex]x[/tex] direction and along the [tex]y[/tex] direction?

If it is perpendicular, why are you integrating from [tex]y(x_{0},t_{0})[/tex] to [tex]y(x,t)[/tex] ? Shouldn't it be integrated from [tex]y(x,t_{0})[/tex] to [tex]y(x,t)[/tex] ? (i.e. instead of [tex]\int_{y(x_{0},t_{0})}^{y(x,t)} F(x,t) dy[/tex] , I think it should be [tex]\int_{y(x,t_{0})}^{y(x,t)} F(x,t) dy[/tex] ). I mean, why does the [tex]x[/tex] coordinate change?
 
  • #3
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I've attached the picture of how the force enters the system.

I agree with your assertion that the limits should include changes in [tex]t[/tex] only.
How will this effect my change of variable then?

That is when I take the total derivative of [tex]dy[/tex],
how do I exclude [tex]y_x dx[/tex] from the total derivative?
 

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  • #4
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If the tension in the string is completely neglected, my approach would be

  • Find the infinitesimal work [tex]dW(x)[/tex] done between [tex]x[/tex] and [tex]x + dx[/tex] by [tex]dW(x) = \int_{y(x,t_{0})}^{y(x,t)} F(x,t) dy[/tex]. If [tex]F[/tex] doesn't cahnge with time, just take it as [tex]F(x)[/tex]. If it does, then find [tex]y(t)[/tex] using Newton's second law for the part of string between [tex]x[/tex] and [tex]x+dx[/tex] with mass [tex]m[/tex] by [tex]F(x,t) = m \frac{ \partial^{2} y }{ \partial t^{2} }[/tex]
  • Find the total work by [tex]W = \int_{x=x_{1}}^{x_{2}} dW(x) dx[/tex]

But a string in real life would have elasticity and tension and that analysis would be different. Then there would also be the extra work done to stretch the string.
 

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