Spring pendulum

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..