Period of Rotation of Plane of Oscillation of Foucault's Pendulum

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Theorem

Let $P$ be a Foucault pendulum.

The plane of oscillation of $P$ rotates through a full circle in time $T$, where:

$T = \dfrac D {\sin \lambda}$

where:

$\lambda$ denotes the latitude on Earth at which $P$ is located
$D$ denotes the length of a sidereal day.


Proof

From Angular Speed of Rotation of Plane of Oscillation of Foucault's Pendulum, the angular speed of rotation $\alpha$ of the plane of oscillation of $P$ is given by:

$\alpha = \omega \sin \lambda$

where:

$\lambda$ denotes the latitude on Earth at which $P$ is located
$\omega$ denotes the angular speed of rotation of Earth.

We have:

$D = \dfrac {2 \pi} \omega$

Hence:

\(\ds T\) \(=\) \(\ds \dfrac {2 \pi} \alpha\)
\(\ds \) \(=\) \(\ds \dfrac {2 \pi} {\omega \sin \lambda}\)
\(\ds \) \(=\) \(\ds \dfrac D {\sin \lambda}\)

$\blacksquare$


Sources

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