Fun with dimensional analysis 1 – Simple pendulum

Given a simple pendulum with all the “usual” assumptions like no friction or air resistance, we have a massless rod of length L connected to a bob of mass m deflected by angle \theta. I assume the only physics we need inject is the gravitational strength of the Earth (or whatever body the pendulum is on), so we include g as well.

The angle \theta is dimensionless so dimensional analysis can’t really tell us anything about what part it plays, but the other parameters are fair game. We wish to find how the period \tau varies as L, m, and g. So we have

\displaystyle \tau \sim L^\alpha m^\beta g^\gamma or

\displaystyle \lbrack\tau\rbrack = \lbrack L\rbrack^\alpha \lbrack m\rbrack^\beta \lbrack g\rbrack^\gamma so

\displaystyle T = L^\alpha M^\beta L^\gamma T^{-2 \gamma} meaning

\displaystyle \beta = 0 \\ \alpha + \gamma = 0 \\ 1 = -2 \gamma \Rightarrow \gamma = - \frac{1}{2} \\ \Rightarrow \alpha = \frac{1}{2}

therefore

\displaystyle \tau \sim \sqrt{\frac{L}{g}}

We find from this simple exercise that the period of the pendulum varies not at all with its mass, which coincides with Galileo’s principle that different masses fall at equal rates. The placement of the remaining terms is expected: the longer the rod the more time we might expect the bob to take to swing through a full period, and the stronger the gravitational acceleration is the faster the bob is pulled through a swing. What the dimensional analysis shows in particular is what the powers of these terms are necessary.

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