9
$\begingroup$

The equation of motion of a pendulum with a bob of mass $m$, and hanging by means of a massless thread of length $T$ is given by $$\ddot{\theta}+\frac{g}{l}\sin\theta=0,$$ and that of a damped one-dimensional oscillator moving along $x$-axis is $$\ddot{x}+\gamma\dot{x}+\omega^2x=0$$ where $\gamma$ is the friction constant

Is there a way to guess that the first equation represents periodic motion with constant time-period while the second is not without solving the equations?

$\endgroup$
1
  • $\begingroup$ -1. You can (and did) guess from the physical process that the motion was or was not periodic. Probably in the same way as Chris did, using conservation of energy. The presence of friction was a bit of a give-away. The equations were not necessary. Asking if periodic solutions exist for a particular ODE is mathematics, not physics. eg math.stackexchange.com/questions/266158 $\endgroup$ – sammy gerbil Oct 19 '17 at 10:48
8
$\begingroup$

A good way to look at these sorts of problems is to talk about conserved quantities. For instance, with some manipulation, it is easy to show that something like energy is conserved in the first formula: $$\ddot{\theta}+\frac{g}{l}\sin\theta=0=\ddot{\theta}\dot{\theta}+\frac{g}{l}\dot{\theta}\sin\theta=\frac{d}{dt}\left(\frac{\dot{\theta}^2}{2}-\frac{g}{l}\cos\theta\right)$$

And so the quantity $\frac{\dot{\theta}^2}{2}-\frac{g}{l}\cos\theta$ is conserved. Since the angular velocity is only a function of position, we are left with three possibilities:

  1. The motion is unbounded: goes off to infinity. This is not possible here because $\theta$ is periodic: if the pendulum flips over, we are back to where we started and this is still periodic motion.
  2. The motion comes to a stop. This is not the case here because $\ddot{\theta}$ and $\dot{\theta}$ do not have the same zeroes, except in the cases where it starts at $\theta=0$ with no velocity or has exactly the right velocity to come to a stop at $\theta=\pi$.
  3. The motion is periodic.

So, except for the special cases where it comes to a stop, this describes periodic motion.

On the other hand we have: $$\ddot{x}+\gamma\dot{x}+\omega^2x=0$$

First off, it is clear that our solution will not have velocity be just position dependent, because the acceleration is velocity dependent. To investigate this, it is helpful to look at a similar quantity to the conserved quantity from the last part:

$$\ddot{x}+\gamma\dot{x}+\omega^2x=0$$ $$\ddot{x}\dot{x}+\omega^2x\dot{x}=-\gamma\dot{x}^2$$

$$\frac{d}{dt}\left(\frac{\dot{x}^2}{2}+\frac{\omega^2x^2}{2}\right)=-\gamma\dot{x}^2$$

Notice that, if $\gamma>0$, the term on the right is always negative, and so the "energy" of the system is always decreasing!

Now if $\gamma$ is small, we can imagine that we get mostly periodic motion just as before for the same reasons (note that the "potential energy" part is unbounded, so it can't go to infinity), but the energy drains out of the system over time (if $\gamma>0$, otherwise energy is being pumped into the system), so the amplitude of each successive oscillation is smaller. If $\gamma$ is large, we don't even have periodic motion initially- the energy is quickly dissipated.

$\endgroup$
6
  • $\begingroup$ In the end of your first equation the derivative of theta should be squared. $\endgroup$ – eranreches Oct 19 '17 at 6:00
  • 1
    $\begingroup$ @eranreches Right you are! Proofreading is not my strong point... $\endgroup$ – Chris Oct 19 '17 at 6:05
  • $\begingroup$ Oh, and now I see that you're also missing 1/2 and a power of 2 in your last equation. Apart from that nice answer! $\endgroup$ – eranreches Oct 19 '17 at 6:31
  • $\begingroup$ I'm quite sure that a double pendulum also has conserved quantities (energy), but its motion is surely aperiodic? $\endgroup$ – LLlAMnYP Oct 19 '17 at 7:20
  • $\begingroup$ @LLlAMnYP It's periodicity relies on the fact that the motion is 1-dimensional. A double pendulum has two different velocities that can vary independently- even if it returns to exactly the same spot, the single conserved quantity is not enough to constrain both velocities. $\endgroup$ – Chris Oct 19 '17 at 18:11
2
$\begingroup$

I would do it by noting that the first system is conservative while the second is not.

Thus, for the pendulum, $$ E=T+V=\frac{1}{2}m\ell^2\dot{\theta}^2+mg\ell(1-\cos\theta)\, . $$ Recognizing the $E$ is constant and can thus be evaluated at the turning points of the (bounded) motion where $(\dot\theta,\theta)=(0,\pm \theta_0)$, we have $E=mg\ell(1-\cos\theta_0)$. Thus, reorganizing and keeping in mind that $\dot\theta=d\theta/dt$, we reach \begin{align} dt&=\pm \frac{d\theta} {\sqrt{\frac{2}{m\ell^2}mg\ell(\cos\theta-\cos\theta_0)}}\, ,\\ T&=4\sqrt{\frac{\ell}{2g}}\int_0^{\theta_0} \frac{d\theta}{ \sqrt{\cos\theta-\cos\theta_0}}= \sqrt{\frac{\ell}{g}}\,f(\theta_0)\, . \end{align} Of course for the second system you cannot do any of this as energy is not conserved.

$\endgroup$
0
$\begingroup$

To add the the other answers, in the general case of more complex systems, you can have conservative systems in which there is no periodic orbit. This is the subject of chaos theory. There is no general rule that can say if a system is chaotic or not. There are many theorems that can help you in some cases, but the general case is not solved.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.