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

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 $$2 \pi$$.

• Hint: The motion of a simple pendulum is a type of sinusoidal type motion. – John Omielan Sep 16 at 4:24
• Na that's an easy cop out. There must be a better argument. – KennyB Sep 16 at 5:00
• @HansLundmark As evidenced by the range of different types of answers and by OP's comments to those answers, this question lacks way too many details to be properly answerable. I have no idea what OP actually is asking. For example, at first it just sounds like a question of how to derive the period of a pendulum, but apparently OP is also confused about how $2\pi$ comes into the period of $\sin(\omega t)$ – and one answer doesn't discuss the derivation at all but OP seems to be quite happy about it. – JiK Sep 16 at 14:02
• @HansLundmark I thought it was an interesting question. I'm lecturing dimensional analysis and I derived the period for the students. One student (who studies physics) asked how we can solve for the constant without knowing anything about the problem. I told him that you need to look at the DE and the behavior of the pendulum to know its 2pi periodic. While this was a satisfying answer to the student im not happy with it. Dimensional analysis is powerful because you don't need insight into the problem, only the dimensions and variables. Similarly, can we find the constant in the same manner. – KennyB Sep 18 at 9:34
• @KennyB: I think that if you had added that information to the question, it would have been received much more positively! Context matters. – Hans Lundmark Sep 18 at 9:49

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.

• I prefer this reasoning to the others presented. However, it's more like a good guess to make that substitution. I was hoping there might be a derivation of sorts. Maybe such a derivation doesn't exist? – KennyB Sep 16 at 5:03
• @KennyB This is a standard method for solving a differential equation. No guessing is involved. – Jyrki Lahtonen Sep 16 at 5:11
• No no, what I mean is, the reasoning to substitute that form of T. It's not explicit if you don't know the behavior of the pendulum. – KennyB Sep 16 at 5:25
• @KennyB it's explicit if you know that both sine and cosine have a periodicity of $2\pi$ and then look for time displacements that give you a displacement of $2\pi$. I have a simple demonstration for this but it is too big for this comment. – bracco23 Sep 16 at 14:27

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 $$\sqrt{\tfrac{l}{g}}$$.

• thanks for this answer. I think it's too advanced for the students I'm lecturing, but it's definitely useful and gives some insight into the problem. This also answers my question in the way that the 2pi term is only for the simple pendulum. This is perhaps why dimensional analysis just gives you a constant because it's independent of the small angle approximation. – KennyB Sep 18 at 9:43

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 $$2\pi \sqrt{\frac{l}{g}}$$.

• Unfortunately, you're using the fact that the period is 2pi/omega where omega is the angular frequency. So you have really derived the 2pi – KennyB Sep 16 at 5:06
• @KennyB The fact that the period is $2\pi/\omega$ comes from the fact that the solution is the linear combination of $\sin(\omega t)$ and $\cos(\omega t)$, and from the basic properties of the trigonometric functions. – Botond Sep 16 at 5:45

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.

• This is also one of the better reasonings for the 2pi term as compared to some of the other answers. – KennyB Sep 16 at 5:07
• Why is it radians/sec and not cycles/sec? Both radians and cycles has dimension $1$, so I can't really see the reason. – Botond Sep 16 at 5:47
• The dimension of $\omega$ is $T^{-1}$. The same goes for cycles/sec, in my opinion. – Botond Sep 16 at 7:07
• Yes, since I've not seen every physics book :) – Botond Sep 16 at 15:20
• @JohnDouma In what sense are cycles not unitless? What is the unit? – JiK Sep 16 at 18:38