Conserved Quantities in Dynamical Systems
You only need to differentiate using common derivative rules and then susbtitute the expressions from the given system into the result. See:
$$ \frac{dE}{dt} = \frac{d}{dt}\left(\frac{1}{2}\omega^2-\cos\theta\right)=\omega\frac{d\omega}{dt}+(\sin\theta)\frac{d\theta}{dt} = \omega(-\sin\theta)+(\sin\theta)\omega=0.$$
Related videos on Youtube
Math Student
Updated on August 07, 2020Comments
-
Math Student over 3 years
A simple pendulum is modeled by the (nondimensionalized) differential equation $$\frac{d^{2}\theta}{dt^2} = -\sin(\theta)$$
With the auxiliary variable $w$, this is equivalent to the first-order system $$\begin{align*}\frac{d\theta}{dt}&= w\\ \frac{dw}{dt}&= -\sin(\theta)\end{align*}\qquad (3)$$
We know that this system has equilibrium at $(\theta, w) = (2n\pi,0)$ and $(\theta,w) = ((2n+1)\pi,0)$. The second set of equilibria were saddle points, and therefore unstable, but the first set were centers whose stability we were able to determine. In this problem, we'll use a conserved quantity of this system to prove that these fixed points are in fact stable.
The quantity $E = (1/2)w^2 - \cos(\theta)$ is proportional to the total energy (kinetic plus potential) of the pendulum. We expect this to be constant on physical grounds, Use $(3)$ to prove that $dE/dt = 0$
I think I need to solve the first differential because I tried to solve the problem without it and got nowhere. I'm not really sure how to solve it though. Any ideas on how to proceed?
-
rcollyer about 12 yearsAs a point of note, this site operates on accumulated reputations which is acquired via others voting on the questions and answers you have posted and possible having your answers accepted as the best/correct one. For this system to work, you should vote on the answers you find useful, and accept those answers which best answer your questions. To do otherwise, in the long term you will eventually find your questions being ignored.
-