Editing Foucault pendulum (section)

== Equation formulation for the Foucault pendulum ==
To model the Foucault pendulum, we consider a pendulum of length ''L'' and mass ''m'', oscillating with small amplitudes. In a reference frame rotating with Earth at angular velocity Ω, the Coriolis force must be included. The equations of motion in the horizontal plane (''x'', ''y'') are:

:<math>
\begin{aligned}
\ddot{x} + \omega_0^2 x &= 2\Omega \sin(\varphi) \dot{y}, \\
\ddot{y} + \omega_0^2 y &= -2\Omega \sin(\varphi) \dot{x},
\end{aligned}
</math>

where:
* <math>\omega_0 = \sqrt{\frac{g}{L}}</math> is the natural angular frequency of the pendulum,
* <math>\varphi</math> is the latitude,
* <math>g</math> is the acceleration due to gravity.

These coupled differential equations describe the pendulum's motion, incorporating the Coriolis effect due to Earth's rotation.<ref>{{cite web |title=Foucault Pendulum Details |url=https://www.phys.unsw.edu.au/~jw/pendulumdetails.html |publisher=UNSW Physics |access-date=2025-01-11}}</ref>

=== Precession rate calculation ===
The precession rate of the pendulum’s oscillation plane depends on latitude. The angular precession rate <math>\Omega_p</math> is given by:

:<math>\Omega_p = \Omega \sin(\varphi),</math>

where <math>\Omega</math> is Earth's angular rotation rate (approximately <math>7.2921 \times 10^{-5}</math> radians per second).<ref>{{cite web |title=Mathematical Derivations of the Foucault Pendulum |url=https://www.idc-online.com/technical_references/pdfs/mechanical_engineering/Mathematical_derivations_of_the_Foucault_pendulum.pdf |publisher=IDC Online |access-date=2025-01-11}}</ref>

=== Examples of precession periods ===

The time <math>T_p</math> for a full rotation of the pendulum’s plane is:

:<math>T_p = \frac{2\pi}{\Omega_p} = \frac{2\pi}{\Omega \sin(\varphi)}.</math>

Calculations for specific locations:

* '''Paris, France''' (latitude <math>\varphi \approx 48.8566^\circ</math>):
:<math>
\begin{aligned}
\Omega_p &= \Omega \sin(48.8566^\circ) \approx 7.2921 \times 10^{-5} \times 0.7547 \\
&\approx 5.506 \times 10^{-5} \, \text{radians/second}, \\
T_p &= \frac{2\pi}{5.506 \times 10^{-5}} \approx 114,105 \, \text{seconds} \\
&\approx 31.7 \, \text{hours}.
\end{aligned}
</math><ref>{{cite web |title=Foucault Pendulum Derivation |url=https://warwick.ac.uk/fac/sci/physics/intranet/pendulum/derivation/ |publisher=Warwick University |access-date=2025-01-11}}</ref>

* '''New York City, USA''' (latitude <math>\varphi \approx 40.7128^\circ</math>):
:<math>
\begin{aligned}
\Omega_p &= \Omega \sin(40.7128^\circ) \approx 7.2921 \times 10^{-5} \times 0.6523 \\
&\approx 4.757 \times 10^{-5} \, \text{radians/second}, \\
T_p &= \frac{2\pi}{4.757 \times 10^{-5}} \approx 132,000 \, \text{seconds} \\
&\approx 36.7 \, \text{hours}.
\end{aligned}
</math><ref>{{cite web |title=Foucault Pendulum Details |url=https://www.phys.unsw.edu.au/~jw/pendulumdetails.html |publisher=UNSW Physics |access-date=2025-01-11}}</ref>

These calculations show that the pendulum's precession period varies with latitude, completing a full rotation more quickly at higher latitudes.