Problem — Linearising the pendulum

Problem

Linearising the pendulum. You have a series of measurements $(L_i, T_i)$ for the pendulum. To verify that $T \propto \sqrt{L}$ you have two options: plot $T$ vs $L$ or $T^2$ vs $L$. (a) Which of the two should give a straight line through the origin? (b) Sketch the two qualitative graphs for ideal data. (c) What is the advantage of linearising before performing the regression?

Solution

Step 1 — Identifying the linear relation (a). The pendulum law is $T = 2\pi\sqrt{L/g}$. Squaring both sides:

$T^2 = \frac{4\pi^2}{g}\,L$

This has the form $y = m\,x$ with $y = T^2$, $x = L$ and constant slope $m = 4\pi^2/g$. So it is the graph of $T^2$ vs $L$ that gives a straight line through the origin, not $T$ vs $L$:

$\ev{\text{linear in } L \text{ is } T^2,\ \text{not } T}$

Step 2 — Qualitative graphs (b). For ideal data (see figure):

• $T$ vs $L$: a curve concave downward (a square root), with a vertical tangent at the origin.
• $T^2$ vs $L$: a straight line starting from the origin, with slope $4\pi^2/g$.

Step 3 — Advantage of linearisation (c). Bringing the data onto a straight line before the regression is useful for several reasons:

1. Immediate visual check: the eye instantly recognises whether points lie on a straight line or not; a curve is much harder to judge “by eye”.
2. Simple, robust linear regression: the least-squares method for a straight line is elementary and directly provides slope and intercept with their uncertainties.
3. Extraction of the physical parameter: from the slope $m = 4\pi^2/g$ one obtains $g = 4\pi^2/m$, i.e. $g$ is measured from the simple inclination of the line.
4. Checking the origin: a line that passes (within error) through the origin confirms the absence of systematic offsets; a non-zero intercept flags an error to be corrected.

Linked notes

Topics: Grandezze fisiche e misura Skills: Lettura dei grafici · Linearizzazione dei dati Objects: Pendolo semplice