Problem — Dimensional analysis of the pendulum

Problem

Dimensional analysis of the pendulum. Galileo conjectured that the period $T$ of a pendulum depended on the mass $m$, the length $L$ and the acceleration of gravity $g$. (a) Show, by dimensional analysis, that the only possible combination with dimensions of time is $T = k\sqrt{L/g}$, where $k$ is a pure number. (b) Explain why the mass cannot appear. (c) Knowing that the experimental value of $k$ is $2\pi$ for small oscillations, calculate $T$ for $L = 1\;\text{m}$.

Solution

Step 1 — Setting up the general form (a). Suppose the period is a product of powers of the quantities involved:

$T = k\, m^{\alpha}\, L^{\beta}\, g^{\gamma}$

with $k$ a pure (dimensionless) number. The dimensions of the quantities are $[m]=\text{M}$, $[L]=\text{L}$, $[g]=\text{L}\,\text{T}^{-2}$, while $[T]=\text{T}$.

Step 2 — Equating dimensions. We impose that the right-hand side has the dimensions of a time:

$\text{T} = \text{M}^{\alpha}\,\text{L}^{\beta}\,(\text{L}\,\text{T}^{-2})^{\gamma} = \text{M}^{\alpha}\,\text{L}^{\beta+\gamma}\,\text{T}^{-2\gamma}$

Comparing the exponents of each fundamental dimension:

$\begin{aligned} \text{M}:\ &\alpha = 0 \\ \text{T}:\ &-2\gamma = 1 \ \Rightarrow\ \gamma = -\tfrac{1}{2} \\ \text{L}:\ &\beta+\gamma = 0 \ \Rightarrow\ \beta = +\tfrac{1}{2} \end{aligned}$

Step 3 — Reconstructing the formula. Substituting $\alpha=0$, $\beta=\tfrac12$, $\gamma=-\tfrac12$:

$T = k\, L^{1/2}\, g^{-1/2} = k\sqrt{\frac{L}{g}}$

which is exactly the required form.

Step 4 — Why the mass does not appear (b). The equation for the exponent of M forces $\alpha=0$: mass is the only quantity containing the dimension M, and no combination of $L$ and $g$ can cancel it out. A term $m^{\alpha}$ with $\alpha\neq 0$ would leave an “orphan” M on the right-hand side, incompatible with a pure time. Physically this reflects the fact that in motion under gravity the inertial mass and the gravitational mass cancel out: all bodies fall with the same acceleration, regardless of mass.

Step 5 — Numerical calculation (c). With $k=2\pi$, $L=1\;\text{m}$ and $g=9{,}81\;\text{m/s}^2$:

$T = 2\pi\sqrt{\frac{1}{9{,}81}} = 2\pi\cdot 0{,}319\;\text{s} \approx 2{,}0\;\text{s}$

$\ev{T = 2\pi\sqrt{L/g} \approx 2{,}0\;\text{s}}$

Linked notes

Topics: Grandezze fisiche e misura Skills: Analisi dimensionale Objects: Pendolo semplice