Derivation of 2$\pi$ in the period of a simple pendulum

Question

Most of us know that the period of a simple pendulum is given by

$$2 \pi \sqrt{\dfrac{\ell}{g}}.$$

But how did the $2\pi$ term get into that argument. From dimensional analysis, we can find the period but not the constant.

Is there a calculus argument as some limit is taken? Is it based on an energy equation? Surely there is a way to derive that $2 \pi$.

Answer 1

The pendulum movement equation is given by \begin{align*} \frac{\mathrm{d}^{2}\theta}{\mathrm{d}t^{2}} + \frac{g}{\ell}\sin(\theta) = 0 \end{align*}

For small values of $\theta$, we can make the approximation $\sin(\theta) \approx \theta$, from whence we obtain the equation \begin{align*} \ddot{\theta} + \frac{g}{\ell}\theta = 0 \end{align*}

whose associated characteristic equation is \begin{align*} x^{2} + \frac{g}{\ell} = 0 \Longrightarrow x = \pm i\sqrt{\frac{g}{\ell}} \end{align*}

Thus the solutions are $\theta(t) = c_{1}\sin\left(\displaystyle t\sqrt{\frac{g}{\ell}}\right) + c_{2}\cos\left(\displaystyle t\sqrt{\frac{g}{\ell}}\right)$. Finally, for $T = 2\pi\sqrt{\displaystyle\frac{\ell}{g}}$, one has \begin{align*} \theta\left(t + 2\pi\sqrt{\frac{\ell}{g}}\right) = c_{1}\sin\left(t\sqrt{\frac{g}{\ell}} + 2\pi\right) + c_{2}\cos\left(t\sqrt{\frac{g}{\ell}} + 2\pi\right) = \theta(t) \end{align*} Therefore we conclude that $T$ is the period indeed.

Answer 2

The Lagrangian of a simple pendulum is $$L=\frac{1}{2}ml^2\dot{\varphi}^2+mgl\cos(\varphi)$$ Using harmonic approximation: $$L=\frac{1}{2} ml^2 \dot{\varphi}^2+mgl\left(1-\frac{\varphi^2}{2}\right)$$ Leaving the constant out: $$L=\frac{1}{2}ml^2 \dot{\varphi}^2-\frac{1}{2}mgl\varphi^2$$ Which will have the following equation of motion: $$\frac{\partial L}{\partial \varphi}-\frac{\mathrm{d}}{\mathrm{d}\varphi} \frac{\partial L}{\partial \dot{\varphi}}=0$$ I.e. $$-mgl\varphi +ml^2 \ddot{\varphi}=0$$ $$\ddot{\varphi}=\frac{g}{l}\varphi$$ Which is a simple harmonic oscillator, with frequency $\sqrt{\frac{g}{l}}$, so the period time will be $2\pi \sqrt{\frac{l}{g}}$.

Answer 3

The expression $\sqrt{\frac{g}{l}}$ represents the frequency in radians per second. Divide by $2\pi$ radians per cycle to get $\frac{1}{2\pi}\sqrt{\frac{g}{l}}$ cycles per second.

Now take the reciprocal to get $2\pi\sqrt{\frac{l}{g}}$ seconds per cycle which, by definition, is the period.

Answer 4

The conserved energy is $$ E = \tfrac12 m l^2 \dot \theta^2 + \tfrac12 mgl \theta^2 $$ so that the system moves on ellipses in the $(\theta,\dot\theta)$ plane.

The period is the time required to go once around an ellipse, which is twice the time it takes to go half a lap, say between $(\theta,\dot\theta)=(-\theta_0,0)$ and $(\theta_0,0)$ with $\dot\theta>0$, where $\theta_0$ is the amplitude of the oscillation, determined by the equation $E=0+ \tfrac12 mgl \theta_0^2$, that is, $$ \theta_0 = \sqrt{\frac{2E}{mgl}} . $$ This gives $$ T = 2 \int_{0}^{T/2} dt = 2 \int_{-\theta_0}^{\theta_0} \frac{d\theta}{d\theta/dt} = 2 \int_{-\theta_0}^{\theta_0} \frac{d\theta}{\sqrt{\frac{E- \tfrac12 mgl \theta^2}{\tfrac12 ml^2}}} = 2 \int_{-\theta_0}^{\theta_0} \frac{d\theta}{\sqrt{\tfrac{g}{l}} \sqrt{\theta_0^2 - \theta^2}} . $$ Now let $\theta = \theta_0 \sin u$ to get $$ T = 2 \sqrt{\tfrac{l}{g}} \int_{-\pi/2}^{\pi/2} \frac{1}{\sqrt{\theta_0^2 (1-\sin^2 u)}} \, \theta_0 \cos u \, du = 2 \sqrt{\tfrac{l}{g}} \int_{-\pi/2}^{\pi/2} du = 2 \sqrt{\tfrac{l}{g}} \cdot \pi , $$ as desired.

You can do the same thing for the full pendulum equation, without the small-angle approximation $\sin \theta \approx \theta$, where the energy is $$ E = \tfrac12 m l^2 \dot \theta^2 - mgl \cos\theta , $$ but then you get a non-elementary integral (a so-called complete elliptic integral) which depends on the amplitude $\theta_0$ (or, equivalently, on the energy $E$), so that the period is some amplitude-dependent constant (greater than $2\pi$) times $\sqrt{\tfrac{l}{g}}$.