Thermal Expansion — Question 8

Question 8

A warming pendulum clock loses time

A small-angle pendulum clock is calibrated at temperature $T_0$. Its rod expands as $L=L_0(1+\alpha\Delta T)$ with $\alpha=20.0\times10^{-6}\,\mathrm{K^{-1}}$. Use $P=2\pi\sqrt{L/g}$, with constant $g$. The clock counts oscillations and assigns each its original calibrated duration. The operating temperature is $40.0\,\mathrm K$ above $T_0$.

1. Find the period ratio and the indicated time after a true $86400\,\mathrm s$.

2. Find the first-order daily time loss and compare it with the exact modeled result. State the sign of the error.

Question 8 - Solution

A warming pendulum clock loses time

(a) Method and calculation

$$\frac P{P_0}=\sqrt{1+.000020(40)}=\sqrt{1.0008}.$$$$t_{\mathrm{indicated}}=t\frac{P_0}{P}=\frac{86400}{\sqrt{1.0008}}\,\mathrm s\approx86365.461\,\mathrm s.$$The clock loses approximately $34.539$ s.

(b) Interpretation and further calculation

$$(1+\epsilon)^{-1/2}\simeq1-\epsilon/2,\quad\epsilon=\alpha\Delta T.$$$$t-t_{\mathrm{indicated}}\simeq\frac{t\alpha\Delta T}{2}=34.560\,\mathrm s.$$The first-order loss exceeds the exact loss by about $.021$ s; the clock runs slow because each real oscillation takes longer.

Physical check

Clock time scales inversely with the actual period, not proportionally to it.