Problem — Length of a pendulum that beats the second

Problem

We want to build a pendulum that “beats the second”, i.e. with period $T = 1{,}0\;\text{s}$. What must its length be? Use $g = 9{,}8\;\text{m/s}^2$.

Solution

We start from the period formula of the simple pendulum $T = 2\pi\sqrt{\ell/g}$ and isolate the length:

$\ell = \frac{g\, T^2}{4\pi^2} = \frac{9{,}8 \cdot (1{,}0)^2}{4\pi^2}\;\text{m}$

Substituting the numbers:

$\ell = \frac{9{,}8}{39{,}48}\;\text{m} \approx 0{,}248\;\text{m}$

$\ev{\ell \approx 0{,}248\;\text{m}}$

Note: here “beating the second” means period $T = 1{,}0\;\text{s}$. This is different from the classic pendulum clock, which has period $T = 2\;\text{s}$ (one second per half-oscillation) and a length close to $1\;\text{m}$.

Links

Topics: Oscillations and harmonic motion Concepts: Simple pendulum · Period and frequency Objects: Simple pendulum