Physics Friday 113

We return again to our driven pendulum. This time, we consider what happens when the pendulum is near the upright vertical position, θ=π, the unstable equilibrium of the undriven pendulum. To do this, let us denote a new angle measure $\phi\equiv\pi-\theta$ , so that $\phi$ measures the angle from the upright vertical. Then $\dot{\phi}=-\dot{\theta}$ , $\ddot{\phi}=-\ddot{\theta}$ , and in terms of $\phi$ , the equation of motion
$l\ddot{\theta}=-\left(g-a_0\omega^2\cos\left(\omega{t}\right)\right)\sin\theta$
becomes
$\begin{eqnarray}-l\ddot{\phi}&=&-\left(g-a_0\omega^2\cos\left(\omega{t}\right)\right)\sin\left(\pi-\phi\right)\\\ddot{\phi}&=&\left(\frac{g}{l}-\frac{a_0}l\omega^2\cos\left(\omega{t}\right)\right)\sin(\phi)\\\ddot{\phi}&=&\left(\omega_0^2-\alpha\omega^2\cos\left(\omega{t}\right)\right)\sin(\phi)\end{eqnarray}$ ;
and for small $\phi$ , this is approximated by
$\ddot{\phi}=\left(\omega_0^2-\alpha\omega^2\cos\left(\omega{t}\right)\right)\phi$ .

Now, one might expect that the pendulum will fall over, as it does in the undriven case. After all, the tangential acceleration, $\left(\omega_0^2-\alpha\omega^2\cos\left(\omega{t}\right)\right)\phi$ , appears like it should average over a cycle to $\omega_0^2\phi$ , a positive quantity. However, this reasoning is incorrect in one important way: recall the “wiggle” in θ, and thus in $\phi$ , due to the driving, and also proportional to a $\cos\left(\omega{t}\right)$ . By examination, we see that when the pivot is accelerating upward, $\phi$ will be driven to increase, and a downward acceleration will produce the part of the wiggle decreasing (or slowing the increase in) $\phi$ .

Thus, the $\phi$ in the driving term of the acceleration $-\alpha\omega^2\phi\cos\left(\omega{t}\right)$ will contain a cosine term, so the above will have a cosine squared term, which, like in our previous rapid ω, small α approximation, will have a nonzero average. In fact, we see that due to the wiggle, $\phi$ will be slightly larger when the driving acceleration is towards lower $\phi$ , and thus this will have a greater effect than when the driving is in the opposite direction, and $\phi$ is slightly smaller. Thus, if ω is large enough, then the net effect of the driving is a restoring force stronger than gravity, and one gets the pendulum oscillating about the upright vertical, with a small wiggle superimposed!

Working as we did in our approximation here, we try for an approximate solution of the form $\phi(t)\approx{b(t)}\cos(\omega{t})+c(t)$ , where b(t) and c(t) are functions whose variation over time is much slower that the “wiggle”, so that on a time-scale of $\omega^{-1}$ , with $\alpha\ll1$ , we can ignore their time dependence and treat them like constants; we also expect b to be much smaller than c (as it represents the “wiggle”).
As before, we substitute into our equation of motion:
$\left(-b\omega^2\cos(\omega{t})\right)-\left(\omega_0^2-\alpha\omega^2\cos\left(\omega{t}\right)\right)\left(b\cos(\omega{t})+c\right)={0}$ .
Now, with $\alpha\ll1$ , and rapid enough driving that $\alpha\omega^2\gg\omega_0^2$ , we can, as before, ignore the $\omega_0^2$ term to get
$\left(-b\omega^2\cos(\omega{t})\right)+\alpha\omega^2\cos\left(\omega{t}\right)\cdot\left(b\cos(\omega{t})+c\right)={0}$ .
Which, to leading order, tels us that $b-\alpha{c}=0$ , so $b=\alpha{c}$ , and so our approximate solution is
$\phi(t)\approx{c(t)}\left(1+\alpha\cos(\omega{t})\right)$ .
From our equation of motion, we have acceleration of φ given by
$\begin{eqnarray}\ddot{\phi}&=&-\left(\alpha\omega^2\cos\left(\omega{t}\right)-\omega_0^2\right)\phi\\&=&-c(t)\left(\alpha\omega^2\cos\left(\omega{t}\right)-\omega_0^2\right)\left(1+\alpha\cos(\omega{t})\right)\end{eqnarray}$ .
Now, taking the average over a period of the driving ( $T=\frac{2\pi}{\omega}$ ), we see that in the product $\left(\alpha\omega^2\cos\left(\omega{t}\right)-\omega_0^2\right)\left(1+\alpha\cos(\omega{t})\right)$ , we will have four terms; the constant term $-\omega_0^2$ will obviously average to itself, while the two cosine terms will each average to zero, and since the average of $\cos^2{x}$ over one of its periods is 1/2, the fourth term, $\alpha\omega^2\cos(\omega{t})\cdot\alpha\cos(\omega{t})$ will have average $\frac{1}{2}\alpha^2\omega^2$ . Lastly, c(t) is approximately constant on this timescale.
Therefore,
$\bar{\ddot{\phi}}=-c(t)\left(\frac12\alpha^2\omega^2-\omega_0^2\right)$ .
Suppose that $\alpha^2\omega^2\gt2\omega_0^2$ , so that the quantity in brackets is positive. Assuming this is true, then define $\bar{\omega}\equiv\sqrt{\frac12\alpha^2\omega^2-\omega_0^2}$ . Then
$\bar{\ddot{\theta}}=-\bar{\omega}^2c$ .
But from $\theta(t)\approx{c(t)}\left(1+\alpha\cos(\omega{t})\right)$ , we see that on the timescale we are averaging, $\bar{\ddot{\theta}}\approx\ddot{c}$ , and so we have
$\ddot{c}+\bar{\omega}^2c(t)\approx{0}$ ,
which is (in the approximation) simple harmonic motion of frequency
$\bar{\omega}=\sqrt{\frac12\alpha^2\omega^2-\omega_0^2}=\sqrt{\frac{a_0^2\omega^2}{2L^2}-\frac{g}{L}}$ .
Now, for this to happen, we noted that $\bar{\omega}$ must be real. With $\alpha\ll1$ , we see then that to have $\alpha^2\omega^2=\alpha\left(\alpha\omega^2\right)\gt2\omega_0^2$ , we need $\alpha\omega^2\gg\omega_0^2$ , as was assumed earlier.
Note that if $\alpha\omega\gg\omega_0$ (instead of just $\alpha\omega\gt\sqrt2\omega_0$ ), then $\bar{\omega}\approx\frac{\alpha\omega}{\sqrt2}$ .

Thus, in this approximation, our pendulum swings with a frequency $\bar\omega$ ; with a small wiggle of frequency $\omega$ superimposed. We also note that the amplitude of the “wiggle” is proportional to the current angle about which it is wiggling, being smaller by a factor of $\alpha=\frac{a_0}{L}$ , creating behavior similar to that here, except with a distinctly different $\bar{\omega}$ , and with a wiggle of opposite phase.

Like this:

Be the first to like this post.

Tags: Approximation, Friday Physics, Pendulum, physics

This entry was posted on March 19, 2010 at 12:03 am and is filed under Math/Science. You can follow any responses to this entry through the RSS 2.0 feed. You can leave a response, or trackback from your own site.

Twisted One 151's Weblog

Physics Friday 113

Like this:

Leave a Reply Cancel reply

Follow “Twisted One 151's Weblog”