Problem: Pendulum that measures g

Problem

A pendulum of length $\ell = 1{,}50\;\text{m}$ completes $20$ full oscillations in $49{,}2\;\text{s}$. Estimate the local value of $g$ from the period measurement.

Solution

The period is obtained by dividing the total time by the number of oscillations:

$T = \frac{49{,}2\;\text{s}}{20} = 2{,}46\;\text{s}$

For the simple pendulum at small oscillations, $T = 2\pi\sqrt{\ell/g}$. Isolating $g$:

$g = \frac{4\pi^2 \ell}{T^2} = \frac{4\pi^2 \cdot 1{,}50}{(2{,}46)^2}\;\text{m/s}^2$

Substituting the numbers:

$g = \frac{59{,}22}{6{,}05}\;\text{m/s}^2 \approx 9{,}79\;\text{m/s}^2$

$\ev{g \approx 9{,}79\;\text{m/s}^2}$

The pendulum thus works as a gravimeter: by measuring length and period one can recover the local value of $g$, which depends on latitude and altitude.

Collegamenti

Argomenti: Oscillations and simple harmonic motion Concetti: Simple pendulum · Period and frequency Competenze: Small-oscillation approximation Oggetti: Simple pendulum