Is the time for a complete rotation of the plane of swing of a Foucault pendulum only a function of latitude?

Question

The context is explained in this post: What is the reason for the orbital movement of the Foucault pendulum?

But now my question is not about the orbital behaviour of the pendulum.

I wonder if the amazing fact that the plane of swing (PS) doesn't rotate with the Earth depends on the amplitude of oscillation.

I think it is reasonable to assume that if the pendulum is not oscillating at all, the mass at the bottom will rotate with the earth as any other object around.

On the other hand, for an amplitude of 2m, PS doesn't rotate along the Earth, and therefore moves for an local observer.

And for 1m, 0.5m, 0.25m ...?

If we suppose that the time for a complete rotation of PS depends only on the latitude, (what results in 31.8 h in Paris for example), and not on the amplitude, is there a sudden discontinuous point at zero?

But if that time is a function of latitude and amplitude, which function is that?

Answer 1

William Tobin writes in his book about the life and science of Foucault that Foucault reported the following about the pendulum in the Pantheon: on very rare occasions there was opportunity to let the pendulum swing for many hours uninterrupted. Foucault reported that on several of those occasions he had kept observing the plane of swing of the pendulum down to an amplitude of swing of about 10 centimeters.

(The pendulum in the Pantheon was undriven, so over the course of hours the amplitude of the swing decayed. As the pendulum in the Pantheon was for the purpose of compelling visual demonstration the normal practice was to restart the pendulum every few hours.)

Foucault reported that down to that very small amplitude the plane of swing had continued to turn at the same rate as the wide swing.

This report by Foucault illustrates that there is no lower bound for the amplitude of swing required to produce the Foucault effect. It's just that as the amplitude becomes smaller and smaller it becomes increasingly difficult to ascertain what the current plane of swing is. Additionally I expect that with smaller and smaller amplitude of swing the effects of otherwise insignificant disturbances (such as a gush of wind) will become major sources of error.

(As to your question about latitude dependency: how the rotation of the plane of swing is a function of latitude is of course described in every single article (webpage, section in physics textbook) about Foucault pendulum physics, that's why I'm not repeating that here.)

Answer 2

Your assumption that

if the pendulum is not oscillating at all, the mass at the bottom will rotate with the earth as any other object

is incorrect. The whole idea of the Focault pendulum is that it swings freely with very little friction, making it very difficult to exert any torque on the system through its anchor point. That's why the pendulum swings in a consistent plane as the earth rotates underneath it - the pendulum is effectively decoupled from the rotation of the earth. If the pendulum is not swinging, none of that has changed - the pendulum is still decoupled from the earth's rotation, so it will not rotate along with the earth. Rather, the pendulum will spin in place, having a rotation entirely determined by its latitude. The behavior is exactly the same as before - the rotation effect is identical, the only difference is that the amplitude of the swing is zero.

It's a better demonstration if you swing the pendulum, since the plane of the swing is very easy to see, but there's no reason you couldn't simply mark a spot on a stationary hanging mass and watch it complete an apparent rotation in exactly one Focault period. In practice, no anchor point is perfectly frictionless, and since static friction is always greater than or equal to kinetic friction, a swinging pendulum will have less torque applied to it from the spinning earth than a stationary one. In addition to being a more dramatic visual display, the swinging pendulum is closer to the theoretical ideal of a frictionless anchor, but both are identical in terms of their ideal behavior. The Focault pendulum's rotational period is determined only by latitude and not by amplitude, even if that amplitude is zero.

Answer 3

I had an idea about my question, so I decided to write an answer.

When the pendulum is oscillating, even for the best lubricated joint that attach the wire at the ceiling, the rotating earth will induce a torsion in the wire. That will produce a small torque, tending to rotate the plane of swing. The air around, that rotates with the earth, will add also a small torque in the same direction.

But except at the extreme points of the oscillation, the pendulum has an angular momentum. It is a vector keeping its horizontal direction and changing modulus all the time (and sign after reaching the extreme points).

The torque vector has a vertical direction.

As $\mathbf T = \frac{d\mathbf L}{dt}, \mathbf L$ acquires a vertical component. That vertical component is a small orbital angular momentum.

So, the pendulum gradually adds an orbital component to its original radial trajectory, what can indeed be observed in the Paris Pantheon for example about half an hour after the start of the oscillations.

If the amplitude of the oscillations is too small, so it is the angular momentum. But the torque is the same because it depends only on the friction conditions, and the angular velocity of the Earth.

For that conditions the orbital component becomes quickly dominant, and the Foucault effect is completely overshadow by the almost circular path of the pendulum.