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Two Compressed Springs -> Unstable Equilibrium

  1. Nov 17, 2005 #1
    I will make a crude visualization of this system:

    |-------------O--------------|
    <-----a------><------a------>

    Identical springs: k1=k2=k
    Natural Length: l > a

    The problem is to prove that the system is unstable.

    Obviously, a slight movement directed off the horizontal axis will cause the springs to unstretch to a natural position vertically above or below the current position. The setup is arranged on a frictionless horizontal table.
    I know that the second derivative of the potential energy will tell me about the stability, so I am trying to write down the potential energy. My problem is how to write down the 'x' for the two springs, i.e.

    [tex] U(x) = \frac{1}{2} k x^2_1 + \frac{1}{2} k x^2_2 , x_1=x_2[/tex]
    [tex] U(x) = k x^2 [/tex]

    I suppose it is just a geometry question, but I'm not sure to find that compressed length [tex]x[/tex].
     
    Last edited: Nov 17, 2005
  2. jcsd
  3. Nov 18, 2005 #2
    ^bump
    ....................................
     
  4. Nov 18, 2005 #3

    NateTG

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    Science Advisor
    Homework Helper

    Let's say that the 'midpoint' is displaced upwards by some distance [itex]d[/itex]. Can you calculate the length the springs would have then?
     
  5. Nov 18, 2005 #4
    so, if I take my orgin at the far left, with [tex]l_o[/tex] the natural length.

    [tex]x = l_f - l_o [/tex]
    [tex]x = \sqrt{a^2 + d^2} - l_o [/tex]

    so, now I need to replace d, right?

    [tex]d = \sqrt{(l_o + x)^2 - a^2} [/tex]

    but that can't be right, because I would have [tex] x = x(x) [/tex]. I must be missing something.
     
    Last edited: Nov 18, 2005
  6. Nov 19, 2005 #5
    anyone.......................................................................
     
  7. Nov 19, 2005 #6

    Astronuc

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    Staff: Mentor

    Presumably if the net force is greater than zero, in either up or down then the system is unstable.

    What is the net force if O is displaced upward by d?

    If the springs were constrained in the horizontal, then one could establish an equation for SHM with one spring a+x(t) and the other a-x(t).
     
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