1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    BruceW

    User Avatar
    Homework Helper

    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..
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Spring pendulum
  1. The Pendulum Problem (Replies: 1)

  2. Spherical Pendulum (Replies: 3)

  3. A Pendulum Experiment (Replies: 13)

  4. Spring Pendulum (Replies: 1)

  5. Asimov pendulum (Replies: 1)

Loading...