# Foucault Pendulum at the North Pole

• Hak
Hak
Homework Statement
A small ball of mass ##m##, and carrying a positive charge ##q##, is suspended
by an insulating string of length ##\ell##. The pendulum so formed is placed, at rest, in a
homogeneous, vertical magnetic field of strength ##B##. Experiment shows that, if the
ball is initially knocked slightly sideways, then it swings back and forth, with the
plane of its swing slowly rotating.

1. Find the final angular frequency ##\omega_0## of the pendulum.
2. How long does it take for the plane to make one complete revolution?
Relevant Equations
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Consider a pendulum swinging at the North Pole in an inertial frame of reference $$xyz$$ with the $$x$$ and $$y$$ axes in the plane tangent to the pole and the suspension placed at elevation $$z = l$$. There is no induction $$\mathbf{B}$$ and in the chosen system the plane of oscillation $$xz$$ of the pendulum is constant. The acting forces are the weight $$m \vec g$$ and the tension$$\vec T$$of the wire. Given $$\theta$$ the angle of oscillation we have $$T_x =-mg \sin \theta$$and $$T_z = mg \cos\theta$$ and the equations of motion result in $$m\frac{ d^2 x}{dt^2} = - mg \sin\theta$$ and $$m \frac{d^2 z}{dt^2}=mg \cos\theta - mg$$. Now we assume that the pendulum oscillation is small, i.e., $$\sin\theta \sim \theta$$ and $$cos\theta \sim 1$$, whereby the second equation (and the related motion) is negligible while the first equation can be posed $$\frac{d^2\theta}{dt^2} = -\frac{g}{l}{\theta}$$ and represents a harmonic motion of pulsation $$\omega_1 = \sqrt{\frac{g}{l}}$$, hence the well-known formula $$\tau_1= 2\pi \sqrt{\frac{l}{g}}$$. Now suppose we enter the vector $$\vec B$$ directed as the $$z$$ axis. The ball comes to be subject to the Lorentz force $$\vec F_L = q \vec v \times \vec B$$, which in the $$xy$$ plane results in a centripetal force relative to a circular trajectory of radius $$R$$. The ball comes to be subjected to the Lorentz force $$\vec F_L = q \vec v \times \vec B$$, which in the $$xy$$ plane results in a centripetal force relative to a circular trajectory of radius $$R$$. The force is perpendicular to the velocity v (thus at zero work on the ball) but tends to move it around the center of the circle. The observer thus observes a rotation of the pendulum-ball's plane of oscillation m. Similar to what the noninertial observer of the Foucault pendulum sees, which participates at the north pole in the rotation of the Earth, due to the Coriolis force $$F_C = -2m \vec \omega_\tau \times \vec v_r$$. If we refer to the center of the circle, the Lorentz force is opposite to the verse of $$\vec R$$ as the Coriolis force. But the thing to note is that it (is easily demonstrated) is much less than the weight force $$mg$$, about $$10^{-5}$$ of the same. This means that the angular frequency $$\omega_0$$, to which it will give rise with the same velocity $$v$$ of pendulum oscillation, will be much less than the frequency $$\omega_1= \sqrt{\frac{g}{l}}$$ of pendulum oscillation. Therefore, the rotation of the pendulum-ball oscillation plane appears very slow to the observer. Also, having made the premise about the signs, we can refer to magnitude. It will be $$m\frac{v^2}{R}=|qv|B$$ to obtain, simplifying and dividing by $$m$$, $$\omega_0 = \frac{|qB|}{m}$$. Then the relative period would be $$\tau = \frac{2\pi m}{|qB|}$$. In reality, only in one half-period does the pendulum swing so as to travel clockwise around the circumference, while in the other half-period the swing of the pendulum-ball is in the opposite direction. Therefore, the actual period, time taken to carry out one revolution around the center in a clockwise direction, will be twice as long as $$\tau$$, that is, $$\tau_0= 2 \tau = \frac{4 \pi m}{|qB|}$$.

Below, instead, is the official solution:

The ball, at position $$\mathbf{r}$$ relative to its rest position, and moving with velocity $$\mathbf{v}$$, is acted upon by the tension force in the string, the gravitational force and the Lorentz force caused by the magnetic field. Using the usual approximations for small-amplitude motions of pendulums, the horizontal projection of the tension inthe string is $$-m \omega^2 \mathbf{r}$$, where $$\omega = \sqrt {\frac{g}{l}}$$ is the angular frequency of the pendulum without any magnetic field. The ball’s equation of motion, in the presence of the magnetic field, is $$m \dot v = q \mathbf{v} \times \mathbf {B} - m \omega^2 \mathbf{r}$$.
This quite sophisticated (vector) differential equation can be solved without using calculus, if we notice that it is very similar to the equation of motion of a simple pendulum in a rotating frame of reference (a Foucault pendulum at the North Pole). To describe the small-amplitude swinging of a planar pendulum with angular frequency $$\omega_0$$, in a frame of reference rotating with angular velocity $$\mathbf{\Omega}$$ relative to the inertial reference frame ($$\mathbf{\Omega}$$ is a vertical vector), we need an equation of motionthat contains the Coriolis force as well as the centrifugal one: $$m \dot v = 2m \mathbf{v} \times \mathbf {\Omega} - m \omega_0^2 \mathbf{r} + m \Omega^2 \mathbf{r}$$.
The solution to this equation is well known. Since the oscillation plane is fixed in the inertial frame of reference, in a reference frame rotating with angular velocity $$\mathbf{\Omega}$$, the plane rotates with angular velocity $$- \mathbf{\Omega}$$; the period of a whole revolution is $$T = \frac{2 \pi}{\Omega}$$.
Comparing equations (1) and (2), it can be seen that the observed rotation rate is $$\mathbf{\Omega} = \frac{qB}{2m}$$ and that the angular velocity vector of the rotating plane has a direction opposed to that of $$B$$. The actual angular frequency $$\omega_0$$ of the pendulum can be expressed entirely in terms of the other parameters of the problem:
$$\omega_0 = \sqrt{\frac{g}{\ell} + \left(\frac{qB}{2m}\right)^2}$$,
but this has no bearing on the rotation rate of the plane of oscillation.
So, the time needed for the plane containing the pendulum’s motion to complete one revolution is: $$T = \frac{4\pi m}{qB}$$.
Notes. 1. It is interesting that the rotational period of the pendulum’s plane depends only on the strength of the magnetic field and the data associated withthe small ball, and that it does not depend on the natural period of the pendulum (i.e. it is independent of $$\ell$$ and $$g$$).

2. The quantity $$qB/m$$ is called the cyclotron angular frequency associated with the given magnetic field; it is the angular velocity of a particle, with charge-to-mass ratio $$q/m$$, that will orbit in a homogeneous magnetic field of strength $$B$$. The plane of the pendulum in the problem rotates with one-half of the cyclotron angular frequency.

Where am I mistaken?

I have found this article, related to the problem presented in this thread, but I cannot understand how the actual angular frequency is calculated. Could someone help me out, including checking my errors, given the official solutions?

I'm having trouble following your analysis of the pendulum swinging in the presence of the B field.
Hak said:
Now suppose we enter the vector $$\vec B$$ directed as the $$z$$ axis. The ball comes to be subject to the Lorentz force $$\vec F_L = q \vec v \times \vec B$$, which in the $$xy$$ plane results in a centripetal force relative to a circular trajectory of radius $$R$$. The force is perpendicular to the velocity v (thus at zero work on the ball) but tends to move it around the center of the circle. The observer thus observes a rotation of the pendulum-ball's plane of oscillation m.
I'm uncertain as to what particular circular motion you are referencing. It appears to me that you might be referring to a circular motion of a charged particle that moves subject only to the Lorentz force. Thus, later you write

Hak said:
It will be $$m\frac{v^2}{R}=|qv|B$$ to obtain, simplifying and dividing by $$m$$, $$\omega_0 = \frac{|qB|}{m}$$. Then the relative period would be $$\tau = \frac{2\pi m}{|qB|}$$.
The equation ##m\frac{v^2}{R} = |qv|B## is for the case where the particle is moving in circular motion and is subject only to the magnetic force. The angular speed of the circular motion in this case is ##\frac{|qB|}{m}##, as you stated. But then, it appears to me, you take a leap of faith and assume that this is the angular speed of precession of the pendulum swinging in the presence of the B field. We know that the correct precession frequency is actually ##\frac{|qB|}{2m}##.

Hak said:
In reality, only in one half-period does the pendulum swing so as to travel clockwise around the circumference, while in the other half-period the swing of the pendulum-ball is in the opposite direction. Therefore, the actual period, time taken to carry out one revolution around the center in a clockwise direction, will be twice as long as $$\tau$$, that is, $$\tau_0= 2 \tau = \frac{4 \pi m}{|qB|}$$.

The pendulum will precess clockwise for both half-periods. For example, suppose we're looking down onto the xy-plane and the magnetic field points out of the page. For the first half period, let the particle swing from point a toward point b, as shown on the left below. You can check that the magnetic force deflects the particle upward so that it arrives at point b' instead of b. This corresponds to a small clockwise deflection. Likewise, if the particle swings from b back toward a, for the second half period, then the magnetic force deflects the particle downward. This again corresponds to clockwise deflection.

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TSny said:
I'm having trouble following your analysis of the pendulum swinging in the presence of the B field.

I'm uncertain as to what particular circular motion you are referencing. It appears to me that you might be referring to a circular motion of a charged particle that moves subject only to the Lorentz force. Thus, later you writeThe equation ##m\frac{v^2}{R} = |qv|B## is for the case where the particle is moving in circular motion and is subject only to the magnetic force. The angular speed of the circular motion in this case is ##\frac{|qB|}{m}##, as you stated. But then, it appears to me, you make a leap of faith and assume that this is the angular speed of precession of the pendulum swinging in the presence of the B field. We know that the correct precession frequency is actually ##\frac{|qB|}{2m}##.
The pendulum will precess clockwise for both half-periods. For example, suppose we're looking down onto the xy-plane and the magnetic field points out of the page. For the first half period, let the particle swing from point a toward point b, as shown on the left below. You can check that the magnetic force deflects the particle upward so that it arrives at point b' instead of b. This corresponds to a small clockwise deflection. Likewise, if the particle swings from b back toward a, for the second half period, then the magnetic force deflects the particle downward. This again corresponds to clockwise deflection.

View attachment 332441
Thank you very much. So, as I understand it, what is wrong is the final angular frequency. If you read the official solution to the problem and the article attached at the bottom of the page, you can see that they arrive at two different solutions of that frequency. How to explain this? Where do I go wrong in this respect?

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Hak said:
Thank you very much. So, as I understand it, what is wrong is the final angular frequency. If you read the official solution to the problem is the article attached at the bottom of the page, you can see that they arrive at two different solutions of that frequency.
What are the two different solutions for ##\Omega## in the official solution? I see only one result.

TSny said:
What are the two different solutions for ##\Omega## in the official solution? I see only one result.
I've modified my previous message. However, I am talking about the final angular frequency of the pendulum ##\omega_0##, not ##\Omega##. The final angular frequency results written in the official solution in post #1 and in the article attached to post #2 look different to me. Thank you.

Hak said:
I've modified my previous message. However, I am talking about the final angular frequency of the pendulum ##\omega_0##, not ##\Omega##.
Ok, sorry for the misinterpretation on my part.

Hak said:
The final angular frequency results written in the official solution in post #1 and in the article attached to post #2 look different to me. Thank you.
The official solution in post #1 gets the following result for the angular frequency ##\tilde \omega## of the pendulum swinging in the magnetic field: $$\tilde \omega = \sqrt{\frac g l + \left(\frac{qB}{2m}\right)^2}.$$ In the reference in post # 2, this is given to be $$\tilde \omega = \sqrt{\frac g l }.$$ (I'm avoiding the notation ##\omega_0## since the two references use ##\omega_0## for different quantities.)

In the reference in post #2, they made the approximation that ##\large \left(\frac{qB}{2m}\right)^2## can be neglected compared to ##\large \frac g l##. With this approximation, the two results are consistent.

TSny said:
Ok, sorry for the misinterpretation on my part.The official solution in post #1 gets the following result for the angular frequency ##\tilde \omega## of the pendulum swinging in the magnetic field: $$\tilde \omega = \sqrt{\frac g l + \left(\frac{qB}{2m}\right)^2}.$$ In the reference in post # 2, this is given to be $$\tilde \omega = \sqrt{\frac g l }.$$ (I'm avoiding the notation ##\omega_0## since the two references use ##\omega_0## for different quantities.)

In the reference in post #2, they made the approximation that ##\large \left(\frac{qB}{2m}\right)^2## can be neglected compared to ##\large \frac g l##. With this approximation, the two results are consistent.
Thank you for answering. It's not your fault, I'm the one who didn't explain well. In post #2, they arrive at a solution of the type ##\omega ^2 = \omega_0 ^2 - \omega_B \omega##, which, as you can see, returns two roots similar-but not equal-to the solution of ##\omega_0## in post #1. This is what I do not understand.

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Note that in the second reference, we have

There is a misprint here. The fraction just after the first equal sign should be ##\large -\frac{\omega_B}{2}## instead of ##\large -\frac{\omega_0}{2}##. But the final expression on the right is correct under the approximation ##\omega_B^2/4 \ll \omega_0^2##.

##\omega_{1,2}## do not represent angular frequencies of oscillation of the pendulum. They are frequencies that correspond to circular-motion solutions of the equations of motion

These circular-motion frequencies (##\omega_{1,2}##) correspond to "conical pendulum" motion in counterclockwise and clockwise directions (when the magnetic field is present):

The angular frequency for counterclockwise (##\omega_1##) differs from the angular frequency for clockwise (##\omega_2##).

The oscillating pendulum solution is obtained by adding the two circular motion solutions. This is sort of like obtaining linear polarization of light from the superposition of left and right circular polarization.

They get the following final result for the pendulum motion:

Analysis of this result shows that it represents a precession of pendulum motion with oscillation frequency ##\omega_0 = \sqrt{g/l}## and with clockwise precession frequency ##\frac {\omega_B} 2 = \frac {qB}{2m}##.

TSny said:
Note that in the second reference, we have
View attachment 332444
There is a misprint here. The fraction just after the first equal sign should be ##\large -\frac{\omega_B}{2}## instead of ##\large -\frac{\omega_0}{2}##. But the final expression on the right is correct under the approximation ##\omega_B^2/4 \ll \omega_0^2##.

##\omega_{1,2}## do not represent angular frequencies of oscillation of the pendulum. They are frequencies that correspond to circular-motion solutions of the equations of motion

View attachment 332448
These circular-motion frequencies (##\omega_{1,2}##) correspond to "conical pendulum" motion in counterclockwise and clockwise directions (when the magnetic field is present):
View attachment 332450
The angular frequency for counterclockwise (##\omega_1##) differs from the angular frequency for clockwise (##\omega_2##).

The oscillating pendulum solution is obtained by adding the two circular motion solutions. This is sort of like obtaining linear polarization of light from the superposition of left and right circular polarization.

They get the following final result for the pendulum motion:
View attachment 332449
Analysis of this result shows that it represents a precession of pendulum motion with oscillation frequency ##\omega_0 = \sqrt{g/l}## and with clockwise precession frequency ##\frac {\omega_B} 2 = \frac {qB}{2m}##.
Thank you so much.

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