$$\require{cancel}$$

# 12.13: Foucault pendulum

$$\newcommand{\vecs}{\overset { \rightharpoonup} {\mathbf{#1}} }$$ $$\newcommand{\vecd}{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}{\| #1 \|}$$ $$\newcommand{\inner}{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}{\| #1 \|}$$ $$\newcommand{\inner}{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$

A classic example of motion in non-inertial frames is the rotation of the Foucault pendulum on the surface of the earth. The Foucault pendulum is a spherical pendulum with a long suspension that oscillates in the $$x-y$$ plane with sufficiently small amplitude that the vertical velocity $$\dot{z}$$ is negligible. Figure $$\PageIndex{1}$$: Foucault pendulum.

Assume that the pendulum is a simple pendulum of length $$l$$ and mass $$m$$ as shown in Figure $$\PageIndex{1}$$. The equation of motion is given by

$\mathbf{\ddot{r}} = \mathbf{g} + \frac{\mathbf{T}}{m} − 2\boldsymbol{\Omega} \times \mathbf{\dot{r}}$

where $$\frac{T}{m}$$ is the acceleration produced by the tension in the pendulum suspension and the rotation vector of the earth is designated by $$\boldsymbol{\Omega}$$ to avoid confusion with the oscillation frequency of the pendulum $$\omega$$. The effective gravitational acceleration $$\mathbf{g}$$ is given by

$\mathbf{g} = \mathbf{g}_0 − \boldsymbol{\Omega} \times [\boldsymbol{\Omega} \times (\mathbf{r} + \mathbf{R})] \label{12.78}$

that is, the true gravitational field $$\mathbf{g}_0$$ corrected for the centrifugal force.

Assume the small angle approximation for the pendulum deflection angle $$\beta$$, then $$T_z = T \cos \beta \simeq T$$ and $$T_z = mg$$, thus $$T \simeq mg$$. Then has shown in Figure $$\PageIndex{1}$$, the horizontal components of the restoring force are

$T_z = −mg \frac{x}{l}$

$T_y = −mg \frac{y}{l}$

Since $$\mathbf{g}$$ is vertical, and neglecting terms involving $$\dot{z}$$, then evaluating the cross product in Equation \ref{12.78} simplifies to

$\ddot{x} = −g\frac{x}{l} + 2\dot{y}\Omega \cos \theta \label{12.81}$

$\ddot{y} = −g\frac{y}{l} + 2\dot{x}\Omega \cos \theta \label{12.82}$

where $$\theta$$ is the colatitude which is related to the latitude $$\lambda$$ by

$\cos \theta = \sin \lambda$

The natural angular frequency of the simple pendulum is

$\omega_0 = \sqrt{\frac{g}{l}}$

while the $$z$$ component of the earth’s angular velocity is

$\Omega_z = \Omega \ cos \theta$

Thus equations \ref{12.81} and \ref{12.82} can be written as

\begin{align} \notag \ddot{x} - 2\Omega_z \dot{y} + \omega^2_0 x = 0 \\ \ddot{y} - 2\Omega_z \dot{x} + \omega^2_0 y = 0 \label{12.86} \end{align}

These are two coupled equations that can be solved by making a coordinate transformation.

Define a new coordinate that is a complex number

$\eta = x + iy$

Multiply the second of the coupled equations \ref{12.86} by $$i$$ and add to the first equation gives

$(\ddot{x} + i \ddot{y})+2i\Omega_z (\dot{x} + i\dot{y}) + \omega^2_0 (x + iy)=0 \notag$

which can be written as a differential equation for $$\eta$$

$\ddot{\eta} + 2i\Omega_z \dot{\eta} + \omega^2_0 \eta = 0 \label{12.88}$

Note that the complex number $$\eta$$ contains the same information regarding the position in the $$x−y$$ plane as equations \ref{12.86}. The plot of $$\eta$$ in the complex plane, the Argand diagram, is a birds-eye view of the position coordinates $$(x,y)$$ of the pendulum. This second-order homogeneous differential equation has two independent solutions that can be derived by guessing a solution of the form

$\eta (t) = A_e^{−i\alpha t} \label{12.89}$

Substituting Equation \ref{12.89} into \ref{12.88} gives that

$\alpha^2 − 2\Omega_z \alpha − \omega^2_o = 0 \notag$

That is

$\alpha = \Omega_z \pm\sqrt{\Omega^2_z + \omega^2_0}$

If the angular velocity of the pendulum $$\omega_0 \gg \Omega$$, then

$\alpha \simeq \Omega_Z \pm \omega_0$

Thus the solution is of the form

$\eta (t) = e^{−i\Omega_zt} (A_+ e^{i\omega_0 t} + A_- e^{i\omega_0 t} )$

This can be written as

$\eta (t) = Ae^{−i\Omega_z t} \cos(\omega_0 t + \delta)$

where the phase $$\delta$$ and amplitude $$A$$ depend on the initial conditions. Thus the plane of oscillation of the pendulum is defined by the ratio of the $$x$$ and $$y$$ coordinates, that is the phase angle $$i\Omega_z t$$. This phase angle rotates with angular velocity $$\Omega_z$$ where

$\Omega_z = \Omega \cos \theta = \Omega \sin \lambda$

At the north pole the earth rotates under the pendulum with angular velocity $$\Omega$$ and the axis of the pendulum is fixed in an inertial frame of reference. At lower latitudes, the pendulum precesses at the lower angular frequency $$\Omega_z = \Omega \sin \lambda$$ that goes to zero at the equator. For example, in Rochester, NY, $$\lambda = 43^{\circ} N$$, and therefore a Foucault pendulum precesses at $$\Omega_Z = 0.682\Omega$$. That is, the pendulum precesses $$245.5^{\circ}$$/day.

This page titled 12.13: Foucault pendulum is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Douglas Cline via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.