Problem — Pendulum clock and temperature

Problem

A pendulum clock is calibrated at a temperature of $20\;^\circ$C, with a rod of length $\ell_0 = 0{,}500\;\text{m}$ and period $T_0 = 1{,}419\;\text{s}$. If the temperature rises to $30\;^\circ$C, the rod expands (expansion coefficient $\alpha = 1{,}2\cdot 10^{-5}\;\text{K}^{-1}$). Does the period increase or decrease? By how much, in percentage?

Solution

The period of the pendulum depends on the square root of the length: $T = 2\pi\sqrt{\ell/g}$, i.e. $T \propto \sqrt{\ell}$.

The thermal expansion of the rod for $\Delta\theta = 30 - 20 = 10\;\text{K}$ gives a relative change in length:

$\frac{\Delta\ell}{\ell} = \alpha\, \Delta\theta = 1{,}2\cdot 10^{-5} \cdot 10 = 1{,}2\cdot 10^{-4}$

Since $T \propto \sqrt{\ell}$, the relative change in the period is half that of the length:

$\frac{\Delta T}{T} = \frac{1}{2}\,\frac{\Delta\ell}{\ell} = \frac{1}{2}\cdot 1{,}2\cdot 10^{-4} = 6\cdot 10^{-5}$

The rod lengthens, so the pendulum becomes slower and the period increases:

$\ev{\frac{\Delta T}{T} \approx +6\cdot 10^{-5} \approx +0{,}006\%}$

Note: a longer period means the clock counts seconds too slowly: in the heat, the pendulum clock runs slow.

Links

Topics: Oscillations and harmonic motion Concepts: Simple pendulum · Thermal expansion Skills: Limiting-case and thought-experiment analysis Objects: Simple pendulum