Pendulums — Question 9

Question 9

Finite amplitude makes an ideal pendulum clock run slow

For a simple pendulum, use the supplied small-amplitude correction $T(\theta_A)\approx T_0(1+\theta_A^2/16)$, with angle in radians. A clock is calibrated using $T_0$, but operates at amplitude $\theta_A=.20\,\mathrm{rad}$. It counts each physical cycle as $T_0$ of indicated time.

1. Estimate its indicated time after $3600\,\mathrm{s}$ of true elapsed time and the time lost.

2. If amplitude is reduced to $.10\,\mathrm{rad}$, how does the leading fractional period excess change? State the approximation limits.

Question 9 - Solution

Finite amplitude makes an ideal pendulum clock run slow

(a) Derivation and calculation

Using the supplied approximation,$$T=T_0(1+\epsilon),\qquad\epsilon=\theta_A^2/16=.20^2/16=.00250.$$$$N=t_{\mathrm{true}}/T,\quad t_{\mathrm{indicated}}=NT_0=\frac{t_{\mathrm{true}}}{1+\epsilon}.$$$$t_{\mathrm{indicated}}=3600/1.00250=3591.0224\,\mathrm s.$$

(b) Result and interpretation

$$\boxed{\text{time lost}=3600-3591.0224=8.9776\,\mathrm s}.$$$$\theta_A\mapsto\theta_A/2\Rightarrow\epsilon\mapsto\epsilon/4=.000625.$$$$\text{new loss}=3600\frac{.000625}{1.000625}=2.2486\,\mathrm s.$$The longer actual period makes a clock calibrated to $T_0$ run slow.

Physical check

The leading estimate $3600\epsilon=9.00\,\mathrm s$ agrees with the more precise ratio to the stated approximation order.