Worked exercise — Pendulum on Mars

Problem

The same one-metre pendulum ($\ell = 1$ m), left to oscillate on the surface of Mars (where $g_M = 3{,}71$ m/s$^2$), what period does it have?

Solution

Applying the formula $T = 2\pi\sqrt{\ell/g}$ with Martian gravity: $T_M = 2\pi\sqrt{\frac{1}{3{,}71}} \approx \ev{3{,}26 \ \text{s}}$

Almost double that on Earth (where it would be $\approx 2{,}0$ s): the weaker gravity restores the mass more lazily.

Interpretation. The pendulum is a gravimeter: the length $\ell$ stays constant, so measuring the period gives $g$ at that location. Carrying it around the Solar System, each planet “signs” itself with its own period.

Links

Topics: Oscillations and simple harmonic motion Concepts: Simple pendulum · Period and frequency Skills: Analysis of limiting cases and thought experiments Objects: Simple pendulum