Problem — Ranking pendulums

Problem

Four simple pendulums, with respective lengths and masses: (A) $L = 0{,}25\;\text{m}$, $m = 0{,}50\;\text{kg}$; (B) $L = 1{,}0\;\text{m}$, $m = 0{,}10\;\text{kg}$; (C) $L = 1{,}0\;\text{m}$, $m = 5{,}0\;\text{kg}$; (D) $L = 4{,}0\;\text{m}$, $m = 1{,}0\;\text{kg}$. All at $g = 9{,}81\;\text{m/s}^2$, small oscillations. Rank them from the shortest to the longest period.

Solution

The period of the simple pendulum is $T = 2\pi\sqrt{\frac{L}{g}}$ and does not depend on the mass: so $m$ is irrelevant and only $\sqrt{L}$ matters. In units of $2\pi/\sqrt{g}$: $T_A \propto \sqrt{0{,}25} = 0{,}5, \quad T_B \propto \sqrt{1{,}0} = 1, \quad T_C \propto \sqrt{1{,}0} = 1, \quad T_D \propto \sqrt{4{,}0} = 2$ B and C have the same length (the different masses do not matter) and therefore the same period. Order from shortest to longest period: $\ev{A < B = C < D}$

Links

Topics: Oscillazioni e moto armonico Concepts: Pendolo semplice · Periodo e frequenza Skills: Ragionamento di ranking Objects: Pendolo semplice