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Analyze the movement after displacing the mass xo horizontally

  1. Feb 20, 2005 #1
    Dear friends,

    I need some help with this problem. As you can see in the picture we have this mass attached to two springs of proper lenght l and I'm asked to analyze the movement after displacing the mass xo horizontally (being xo very small).

    I don't want you to solve the problem for me. Just give me a hint how I should begin. I know the oscillator is anharmonic and I believe the best thing to do is analyze its potential and draw the phase plane. But I don't know how to do that.

    Thanks for your help
     

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  3. Feb 20, 2005 #2

    s_a

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    What I'd do is determine the potential energy (U) stored in the springs for any value of q (using Hooke's Law), and then use the fact that the force F = -dU/dq. From this you can determine the acceleration in terms of the position q and hence form a simple 2nd order DE to solve for the equation of the motion of the mass.
     
  4. Feb 20, 2005 #3

    Andrew Mason

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    Are you sure xo is a horizontal displacement? If so, wouldn't we have to know the distance q when the mass displaced horizontally?

    AM
     
  5. Feb 20, 2005 #4

    Curious3141

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    Perhaps it's a plan view ? Anyway, the problem would be even more complicated if we had to consider gravity.

    Assuming that that is a plan view and that all the displacements are in one horizontal plane, the simplest approximate differential equation I could come up with to describe the motion was :

    [tex]\ddot{x} = - \frac{k}{ml^2}(x^3)[/tex]

    which I don't know how to solve. It certainly isn't simple harmonic motion.

    The exact d.e. is a lot more complicated and almost certainly unsolvable exactly.

    k, BTW, is the spring constant of one of the identical springs.
     
    Last edited: Feb 20, 2005
  6. Feb 20, 2005 #5

    Curious3141

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    But if q is indeed a vertical displacement and x is a horizontal one, then it becomes an oscillation in two planes. I think that one would call for solution with a Lagrangian. It's been a long time since I did anything like that, and it isn't elementary.
     
  7. Feb 20, 2005 #6

    Andrew Mason

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    Right. I get the same equation:

    The restoring force is:

    (1)[tex]F = -2k dL sin(\theta) \approx -2kdL(x/L)[/tex]

    Since:

    [tex]L^2 + x^2 = (L + dl)^2 = L^2 + 2Ldl + dl^2[/tex]

    ignoring the dl^2 term for small x,

    [tex]dl = x^2/2L[/tex]

    substituting in (1),

    [tex]F = -2k(x^2/2L)(x/L) = -kx^3/L^2 = m\ddot x[/tex]

    Perhaps someone with access to Maple or Mathematica will be able to tackle this. I found this general solution to third order differential equations.

    http://virtual.cvut.cz/dyn/examples/examples/equations/eqs7/

    AM
     
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