Problem — Required precision on g

Problem

You want to measure $g$ with relative uncertainty $\le 1\%$ using a pendulum of length $L=1$ m and a length measurement with $\Delta L = 1$ cm. Knowing that $\Delta g/g \approx \Delta L/L + 2\,\Delta T/T$, find the maximum acceptable uncertainty on $T$ and comment on whether a hand-operated stopwatch is sufficient (reaction time $\sim 200$ ms).

Solution

Contribution from the length. The relative uncertainty on the length is $\frac{\Delta L}{L} = \frac{0{,}01 \ \text{m}}{1 \ \text{m}} = 0{,}01 = 1\%$

Splitting the error budget. The propagation formula gives $\dfrac{\Delta g}{g} \approx \dfrac{\Delta L}{L} + 2\dfrac{\Delta T}{T}$. To keep the total within $1\%$ we cannot let the time term be large: we require the time contribution to remain small, say $2\,\Delta T/T \le 0{,}5\%$, i.e. $\frac{\Delta T}{T} \le 0{,}25\% = 0{,}0025$

Uncertainty on the period. The period of a $1$ m pendulum is $T = 2\pi\sqrt{\frac{L}{g}} = 2\pi\sqrt{\frac{1}{9{,}8}} \approx 2{,}0 \ \text{s}$ hence $\Delta T \le 0{,}0025 \times 2{,}0 \ \text{s} = 5 \times 10^{-3} \ \text{s} = 5 \ \text{ms}$

Comment. A hand-operated stopwatch has an uncertainty dominated by human reaction time, $\sim 200$ ms, i.e. roughly 40 times larger than the required limit. It is completely insufficient. Solutions: use photogates (automatic measurement of the passage), or time $N$ oscillations and divide: by timing $N=100$ periods the error on a single period is reduced by a factor $N$, bringing the overall $200$ ms down to $2$ ms per period.

$\ev{\Delta T \lesssim 5 \ \text{ms} \;\Rightarrow\; hand\text{-}operated\ stopwatch\ insufficient}$

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Topics: Metodo sperimentale Concepts: Cifre significative e incertezza Skills: Propagazione degli errori