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 , so that measures the angle from the upright vertical. Then , , and in terms of , the equation of motion

and for small , this is approximated by

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

Thus, the in the driving term of the acceleration 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, will be slightly larger when the driving acceleration is towards lower , and thus this will have a greater effect than when the driving is in the opposite direction, and 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 , 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 , with , 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:
Now, with , and rapid enough driving that , we can, as before, ignore the term to get
Which, to leading order, tels us that , so , and so our approximate solution is
From our equation of motion, we have acceleration of φ given by
Now, taking the average over a period of the driving (), we see that in the product , we will have four terms; the constant term will obviously average to itself, while the two cosine terms will each average to zero, and since the average of over one of its periods is 1/2, the fourth term, will have average . Lastly, c(t) is approximately constant on this timescale.
Suppose that , so that the quantity in brackets is positive. Assuming this is true, then define . Then
But from , we see that on the timescale we are averaging, , and so we have
which is (in the approximation) simple harmonic motion of frequency
Now, for this to happen, we noted that must be real. With , we see then that to have , we need , as was assumed earlier.
Note that if (instead of just ), then .

Thus, in this approximation, our pendulum swings with a frequency ; with a small wiggle of frequency 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 , creating behavior similar to that here, except with a distinctly different , and with a wiggle of opposite phase.


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