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Foucault's Pendulum and Its Time Period

Foucault’s pendulum is a simple pendulum designed to demonstrate the rotation of the Earth. It was first publicly demonstrated by Léon Foucault in 1851 at the Panthéon in Paris.

It consists of a heavy bob (mass) suspended from a long, flexible wire.

The key feature is that it is free to swing in any vertical plane (isotropic suspension).

As the pendulum swings, the plane of oscillation appears to rotate relative to the ground.

Principle of Foucault’s Pendulum

The motion of a pendulum is governed by the principle of inertia. When the pendulum is set into oscillation, its plane of oscillation tends to remain fixed in space.

However, the Earth rotates about its own axis. Therefore, an observer standing on the rotating Earth sees the plane of oscillation slowly change its direction.

Hence, Foucault’s pendulum provides direct experimental evidence of the rotation of the Earth.

Angular Velocity of Foucault’s Pendulum

If the pendulum is situated at latitude \(\lambda\), the angular velocity of rotation of its plane of oscillation relative to the Earth is

\[ \boxed{\omega_F=\omega_E\sin\lambda} \]

Substituting \[ \omega_E=\frac{2\pi}{T_E}, \] we obtain

\[ \boxed{ \omega_F= \frac{2\pi}{T_E}\sin\lambda } \]

where \(\lambda\) represents the latitude of the place.

Time Period of Rotation of the Plane

The time period corresponding to the angular velocity \(\omega_F\) is

\[ T_F=\frac{2\pi}{\omega_F} \]
\[ T_F= \frac{2\pi} {\left(\frac{2\pi}{T_E}\sin\lambda\right)} \]

Therefore,

\[ \boxed{ T_F=\frac{T_E}{\sin\lambda} } \]

Since the period of Earth's rotation is approximately 24 hours,

\[ \boxed{ T_F=\frac{24}{\sin\lambda}\text{ hours} } \]

Case I: At the Pole \(\lambda=90^\circ\)

At the pole,

\[ \sin90^\circ=1 \]

Therefore,

\[ T_F=\frac{T_E}{\sin90^\circ} \] \[ T_F=\frac{T_E}{1}=T_E \]

Since \[ T_E=24\text{ h}, \] we obtain

\[ \boxed{T_F=24\text{ h}} \]
Result: At the poles, the plane of oscillation completes one complete rotation in approximately 24 hours.

Case II: At the Equator \(\lambda=0^\circ\)

At the equator,

\[ \sin0^\circ=0 \]

Therefore,

\[ T_F=\frac{T_E}{\sin0^\circ} \] \[ T_F=\frac{T_E}{0}\rightarrow\infty \]
Result: At the equator, the plane of oscillation does not rotate relative to the Earth.
\[ \boxed{T_F\rightarrow\infty} \]

Direction of Rotation

  • In the Northern Hemisphere, the plane of oscillation appears to rotate clockwise when viewed from above.
  • In the Southern Hemisphere, it appears to rotate anticlockwise.
  • At the Equator, there is no apparent rotation of the plane of oscillation.