Interesting question. If we had just a 1d spring, the potential energy would be given by: V = \frac{1}{2} k Q^2 (Where Q is the displacement from equilibrium). And for a 2d pendulum, the potential energy is: V = -mgrcos(\theta). Now if we define the displacement from equilibrium to be r-l (in other words, the change in length of the pendulum, where l is simply a constant), and if we add the two potentials together, we would get a total potential:
V = \frac{1}{2}k(r-l)^2 - mgrcos(\theta)
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:
KE = \frac{1}{2}m(\dot{r}^2 + r \dot{\theta}^2 )
And now, we can use the Euler-Lagrange equations to find out the laws of the system:
-grsin(\theta) = \frac{d(r \dot{\theta})}{dt}
mr \dot{\theta} - k(r-l) + mgcos(\theta) = m \ddot{r}
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 sin(\theta) \rightarrow \theta and cos(\theta) \rightarrow 1 - \frac{1}{2} \theta^2 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..
