Problem — Law from pendulum data

Problem

The period $T$ of a pendulum is measured as the length $L$ varies. Data: $L=0{,}25$ m → $T=1{,}00$ s; $L=0{,}50$ m → $T=1{,}42$ s; $L=1{,}00$ m → $T=2{,}01$ s; $L=2{,}00$ m → $T=2{,}84$ s. Verify graphically, by plotting $T$ as a function of $\sqrt{L}$, that the relation is of the form $T=k\sqrt{L}$ and determine $k$. Compare with the theoretical formula $T=2\pi\sqrt{L/g}$.

Solution

Linearisation. If the law is $T=k\sqrt{L}$, then plotting $T$ as a function of the new variable $x=\sqrt{L}$ gives a straight line through the origin, with slope $k$. Let’s compute $\sqrt{L}$ for each data point: $\begin{aligned} L=0{,}25 \Rightarrow \sqrt{L}=0{,}50, \quad T=1{,}00 \\ L=0{,}50 \Rightarrow \sqrt{L}=0{,}707, \quad T=1{,}42 \\ L=1{,}00 \Rightarrow \sqrt{L}=1{,}00, \quad T=2{,}01 \\ L=2{,}00 \Rightarrow \sqrt{L}=1{,}414, \quad T=2{,}84 \end{aligned}$ The points $(\sqrt{L},\,T)$ lie on a line through the origin: the law $T=k\sqrt{L}$ is confirmed.

Slope. Let’s estimate $k=T/\sqrt{L}$ from each point: $\begin{aligned} \frac{1{,}00}{0{,}50}=2{,}00 \\ \frac{2{,}01}{1{,}00}=2{,}01 \\ \frac{2{,}84}{1{,}414}=2{,}01 \end{aligned}$ The values agree within the uncertainty: $k \approx 2{,}01 \ \text{s}/\sqrt{\text{m}}$.

Comparison with theory. The theoretical formula $T=2\pi\sqrt{L/g}$ can be rewritten as $T = \dfrac{2\pi}{\sqrt{g}}\sqrt{L}$, so the theoretical slope is $k_\text{teo} = \frac{2\pi}{\sqrt{g}} = \frac{2\pi}{\sqrt{9{,}8}} \approx 2{,}006 \ \text{s}/\sqrt{\text{m}}$ in excellent agreement with the experimental value.

$\ev{k \approx 2{,}01 \ \text{s}/\sqrt{\text{m}} \;\approx\; \frac{2\pi}{\sqrt{g}}}$

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Topics: Metodo sperimentale Skills: Linearizzazione dei dati · Lettura dei grafici