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Imposing boundary conditions on a string

  1. Feb 14, 2005 #1
    I need to know how different boundary conditions on the DE representing a string under a force can be physically implemented. For example, if you need y(0)= 0, just tie the string to y=0 at that end. If you need y'(0)=0, attatch it so that it can freely slide up and down a pole at x=0. But what if you need y'(0) = a, some constant? I was thinking you could put the pole at an angle, but then the string wouldn't always be attatched at x=0. How could you do it?
     
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  3. Feb 14, 2005 #2

    dextercioby

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    The end would be oscillating.The amplitude of the point 'x=0' will change with time under the influence of an external factor...And that will send the string into a prescribed oscilating movement...
    So the description of y'(0)=0 means the the point 'x=0' will have a fixed amplitude,not necessarily 0...

    Daniel.
     
  4. Feb 14, 2005 #3
    I'm sorry, I should have been clearer. y(x) is the position of the string, and y'(x) is the slope. The entire string is acted on by a force f(x). I need to do something with the setup to fix a certain slope at the x=0 end, regardless of what the force applied is. There is no time in the problem.

    Now that I think about it, I don't know if my idea for y'(0)=0 would work either. The string would just slide down until the vertical component of tension equaled the f(0). Now I'm completely lost.
     
    Last edited: Feb 14, 2005
  5. Feb 14, 2005 #4

    dextercioby

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    It really doesn't make any sense what you're telling me here.The time dependence HAS TO BE INCLUDED.The concept of force itself requires that the "shape" of the string will vary in time...
    So please set the initial conditions on the amplitude
    [tex] y(x,t) [/tex] and the "velocity" [tex] y'_{t}(x,t) [/tex].

    Daniel.
     
  6. Feb 14, 2005 #5
    The force does not vary in time. It reaches an equilibrium with the tension and the string takes on a fixed shape y(x).
     
  7. Feb 14, 2005 #6

    dextercioby

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    Yes,but that's a "snapshot",y=y(x) is the image of a string at a certain moment of time...But that's it...You cannot make any assumptions on it...

    Daniel.
     
  8. Feb 14, 2005 #7
    If you want to be really precise, with f(x,t) = 0 for t<0 and f(x) for t>0, it is the snapshot at t=infinity, after all transient effects have disappeared. But it isn't necessary to involve time at all, you just need to balance the forces. The DE the string must satisfy is T0 d2y/dx2 = F(x), where T0 is the original tension in the string (it's an approximation). I'm just trying to find out what you physically have to do to ensure y'(0) is some constant 'a', regardless of the force applied.
     
  9. Feb 14, 2005 #8

    dextercioby

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    Solve the PD equation
    [tex] \frac{1}{v^{2}} \frac{\partial^{2} y(x,t)}{\partial t^{2}}=\frac{\partial^{2} y(x,t)}{\partial x^{2}} +F(x) [/tex]

    with the initial conditions that u want:
    [tex]y(x,0)=blah,blah,... ;y'_{t}(x,0)=a [/tex]
    and the boundary conditions what u like
    [tex] y(0,t)=y(L,t)=0 [/tex] (fixed ends).

    Daniel.
     
  10. Feb 14, 2005 #9
    It's not a wave, it's a string under a force. If f(x) was, say, g, it would be a string hanging under the influence of gravity, and the solution would be the catenary curve. And only one of the ends is fixed, the other end is supposed to have a fixed slope.
     
  11. Feb 14, 2005 #10

    dextercioby

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    It's an oscillatory motion of string giving rise to mechanical waves...
    This is of course,unless the problem is totally different.I thought you were referring to that...
    In the case u describe,what is the problem...?

    Daniel.
     
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