# Spring pendulum

1. Nov 3, 2011

### qartveli

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. Nov 4, 2011

### BruceW

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..