Physics Friday 112

Continuing from last time, we noted that the equation of motion for our driven pendulum,
$\ddot{\theta}+\left(\omega_0^2-\alpha\omega^2\cos\left(\omega{t}\right)\right)\theta=0$
cannot be solved analytically even in the small angle approximation.

Recall that we required that the driving motion of the pivot be much smaller in amplitude than the length of the pendulum, so that $\alpha\equiv\frac{a_0}l\ll1$ . Now, let us assume that the driving frequency ω of pivot is much, much faster than the natural (small-angle) frequency of the pendulum ω₀; so that $\alpha\omega^2\gg\omega_0^2$ ,
and thus the $-\alpha\omega^2\cos\left(\omega{t}\right)$ term in the equation of motion will dominate the $\omega_0^2$ term for all but a tiny fraction of the driving cycle.
Now, we note that, superimposed upon any larger-scale motion of the pendulum will be a “wiggle” of frequency ω. Now, to compare that wiggle to the driving motion, we note that when the pivot has upward acceleration, that means an upward force acts on the pivot, a force whose torque on the pendulum would tend to reduce θ, while a downward acceleration of the pivot will tend to increase θ.

Since the vertical acceleration of our pivot is $-a_0\omega^2\cos(\omega{t})$ , we expect the wiggle in θ to be proportional to $\cos(\omega{t})$ . So, let us look for an approximate solution of the form $\theta(t)\approx{b(t)}\cos(\omega{t})+c(t)$ , 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.
Plugging into the equation of motion;
$\left(-b\omega^2\cos(\omega{t})\right)+\left(\omega_0^2-\alpha\omega^2\cos\left(\omega{t}\right)\right)\left(b\omega^2\cos(\omega{t})+c\right)=0$ ,
Now, neglecting the $\omega_0^2$ ,
$\left(-b\omega^2\cos(\omega{t})\right)-\alpha\omega^2\cos\left(\omega{t}\right)\cdot\left(b\omega^2\cos(\omega{t})+c\right)=0$
$\omega^2\left(-b\cos(\omega{t})-\alpha{c}\cos\left(\omega{t}\right)-\alpha{b}\cos^2\left(\omega{t}\right)\right)=0$ ,
which means that we see that to leading order, we need $b+\alpha{c}=0$ , so that $b=-\alpha{c}$ , and our approximate solution is thus
$\theta(t)\approx{c(t)}\left(1-\alpha\cos(\omega{t})\right)$ .
From our equation of motion, we have acceleration of θ given by
$\begin{eqnarray}\ddot{\theta}&=&\left(\alpha\omega^2\cos\left(\omega{t}\right)-\omega_0^2\right)\theta\\&=&c(t)\left(\alpha\omega^2\cos\left(\omega{t}\right)-\omega_0^2\right)\left(1-\alpha\cos(\omega{t})\right)\end{eqnarray}$ .
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.
Thus, we see:
$\bar{\ddot{\theta}}=-c(t)\left(\frac12\alpha^2\omega^2+\omega_0^2\right)$ .
Now define $\bar{\omega}\equiv\sqrt{\omega_0^2+\frac12\alpha^2\omega^2}$ . 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
$\bar{\omega}=\sqrt{\omega_0^2+\frac12\alpha^2\omega^2}=\sqrt{\frac{g}{L}+\frac{a_0^2\omega^2}{2L^2}}$ .

Thus, in this approximation, our pendulum swings with a frequency $\bar\omega$ , faster than the natural frequency $\omega_0$ ; 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 $\alpha=\frac{a_0}{L}$ .

Plot of the approximate solution (blue) with L=1 m, a₀=0.01 m, and ω=100 s^-1. The undriven oscillation of similar overall amplitude (gray) is shown for comparison. (Note: Click on graph for link to full size image, as it antialiases weirdly when scaled down).

[Note—Major edit on 3-15: Corrected a sign error that had greatly complicated the analysis and caused an error in comparing driven vs. undriven frequencies].

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Tags: Approximation, Friday Physics, Pendulum, physics

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One Response to “Physics Friday 112”

Physics Friday 113 « Twisted One 151's Weblog Says:
March 19, 2010 at 12:09 am | Reply
[...] Friday 113 By twistedone151 We return again to our driven pendulum. This time, we consider what happens when the pendulum is near the upright [...]

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