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Spring pendulum

  1. Nov 3, 2011 #1
    Hello everybody :) I need some help in an experiment simulation :) If you will make a pendulum with a spring instead of a string or rope (nonelastic), then bend it over on θ angle and then let it go - it will start to oscillate around with different manner than a simple pendulum :) with centripetal force spring will get longer and longer, after T/4 spring will start to get shorter and etc... This difference in length will change the period of oscillation of a pendulum and the trajectory too. So can anybody make simulate this experiment in the computer and post the picture of trajectory???
  2. jcsd
  3. Nov 4, 2011 #2


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    Interesting question. If we had just a 1d spring, the potential energy would be given by: [itex]V = \frac{1}{2} k Q^2[/itex] (Where Q is the displacement from equilibrium). And for a 2d pendulum, the potential energy is: [itex]V = -mgrcos(\theta)[/itex]. Now if we define the displacement from equilibrium to be [itex]r-l[/itex] (in other words, the change in length of the pendulum, where [itex]l[/itex] is simply a constant), and if we add the two potentials together, we would get a total potential:
    [tex]V = \frac{1}{2}k(r-l)^2 - mgrcos(\theta)[/tex]
    Now I'm going to talk about the 2d case, because the equations are easier. So the kinetic energy of the object is given by:
    [tex] KE = \frac{1}{2}m(\dot{r}^2 + r \dot{\theta}^2 )[/tex]
    And now, we can use the Euler-Lagrange equations to find out the laws of the system:
    [tex]-grsin(\theta) = \frac{d(r \dot{\theta})}{dt} [/tex]
    [tex]mr \dot{\theta} - k(r-l) + mgcos(\theta) = m \ddot{r} [/tex]
    And there is also the equation for the conservation of energy, which simply says that the kinetic energy plus the potential energy is conserved.

    So, the equations are a bit complicated. We could also make the small angle approximation, which would make [itex]sin(\theta) \rightarrow \theta[/itex] and [itex]cos(\theta) \rightarrow 1 - \frac{1}{2} \theta^2[/itex] But it would still look quite complicated.

    You could use these equations for a simulation on computer, and that would show the kind of trajectory to expect. And maybe there is a way to do stability analysis, which would show that certain trajectories are more stable than others, I'm not sure..
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