# Can someone please derive $T=2\pi\sqrt{l/g}$ or prove it without using calculus?

I don't know much calculus, but I want to know that how one derives the formula to find time period $T$ of a simple pendulum.

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Do you know differential equations? – Bernhard Mar 16 '13 at 10:00
@Bernhard I very much doubt it. – Michael Brown Mar 16 '13 at 10:19
As a general note: A complete derivation must use the geometrical properties of π, as the latter is a trancendental number and thus can only be obtained by using limits (which are calculus). Such limits are implicitly present in geometry as well, but they are sufficiently hidden that one does usually not consider it to be calculus. – Wrzlprmft Jan 28 at 13:55

## 3 Answers

You can get a proportionality using one of the most basic techniques in science: dimensional analysis. What you do is look at the dimensions of the various quantities in the problem and try to see how they can fit together. Here you want the period, which is a time. I'll write $[T]$ to indicate that it has dimensions of time. What could it depend on? If we're ignoring friction (which is often reasonable) the only things it could depend on are:

• The mass of the pendulum $m$, with dimensions of mass $[M]$
• The length of the pendulum $\ell$, with dimensions of length $[L]$
• The acceleration of gravity $g$, with dimensions $[L/T^2]$

You want to put this together in a way that gives you a time. The most general thing you can write down is

$$m^a \ell^b g^c,$$

for some numbers $a,b,c$. The dimensions of this are

$$[M^a L^{b+c} T^{-2c}],$$

which you can see by putting in the dimensions and multiplying it out. Equating the dimension to $[T^1]$ we find:

$$\begin{array}{lcl} a &=& 0, \\ b+c &=& 0, \\ -2c &=& 1. \end{array}$$

From which you find $a=0,\ b=-c=1/2$. So we know that

$$T = k \sqrt{\frac{\ell}{g}},$$

where $k$ is some pure number (no dimensions) which we can't work out this way. It turns out that $k=2\pi$ but dimensional analysis can only take you so far. I don't know of a way to get $k$ without using at least a little bit of calculus. :)

Note: More generally the period can and does depend on the magnitude of the swing, given by $\theta_{max}$, the largest angle of deflection from the vertical. Angles are dimensionless quantities, so really we can only write

$$T = \sqrt{\frac{\ell}{g}} f(\theta_\textrm{max}),$$

where $f(\theta_\textrm{max})$ is some function we need to work out using Newton's laws and calculus, but it turns out that for reasonably small swings $f(\theta_\textrm{max})\approx 2\pi$ independent of $\theta_\textrm{max}$.

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This answer elaborates on the one given by JKL. The only reason to enter it in a separate answer is that it won't fit within the 500 character limit of a comment.

The pendulum bob can swing in any direction in the horizontal plane. Take the following motion pattern: the bob swings around in a horizontal circle (so that the pendulum wire sweeps out a vertical cone). With such a circular motion pattern the distance of the bob to the equilibrium point remains the same.

Interestingly, this circular motion can be thought of as a combination of two harmonic oscillations, perpendicular to each other. This is a general rule when descibing motion: If the motion is in two dimensions then you can decompose everything in two perpendicular components, and the laws of motion apply perfectly fine for the individual motion components.

When the pendulum is not swinging it's like a plumb line. I will call the point where the bob then hangs the 'equilibrium point'.

In the following expressions I use the letter 'r' for distance of the bob to the equilibrium point.

For small deviation from the equilibrium point the force(-component) on the bob towards the equilibrium point is proportional to the angle of the pendulum wire. To a good approximation the following is valid for the force:

$F = mg\frac{r}{L}$

The factor r/L is there because the ratio of r and L is a measure of the angle.

As mentioned by JKL, for circular motion the required centripetal force is given by:

$F_c = m\omega^2r$

Equating the two:

$mg\frac{r}{L} = m\omega^2r$

m and r drop out:

$\frac{g}{L} = \omega^2$

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+1: Nice work, but you can actually go a little further with this. You can use that fact that for a harmonic oscillator potential the frequency is independent of amplitude (indeed, this is the nontrivial physics), to continuously deform the motion from a circle to a flatter and flatter ellipse until you finally have motion in a single plane. The period remains the same. This completes a demonstration for the simple pendulum problem. The tricky thing to show (without calculus) is that the amplitude doesn't matter. – Michael Brown Mar 16 '13 at 15:22

SIMPLE PHYSICS FOR THE PEDULUM EQUATION: No Calculus, only some algebra.

Forgive me if I assume you only know some good algebra and some basic physics, such as the equations of circular motion and SHM (simple harmonic motion.) With these constraints in mind I will try to show you how to prove the equation you are working on, based only on pure physics arguments, some simplifying assumptions, and very little mathematics. Once you get done with the explanations, it will then look easy!

But first, a couple of things from circular motion and SHM:

The speed of a particle in uniform circular motion around a circle of radius r is

$v=\omega r$

where $\omega=2\pi/T$. The acceleration of the particle is

$a=\omega^2r$

Now, let us see how we can apply all these in the case of the pendulum.

It is very crucial to understand that the equation you are trying to prove holds only for very small deviations of the bob (the mass) from its “middle” equilibrium position.

Now imagine you have the mass $m$ hanging vertically from some fixed point, by a piece of light and inextensible string of length $L$. Give the mass a very small deviation from the equilibrium position. As the mass oscillates it is moving on an arc of a vertical circle of radius $r=L$, but the deviation must be small. The smaller the deviation the better, because otherwise even using calculus will not lead to the simple equation we want to prove. In fact, although the formal mathematical derivation requires a small deviation, it does not set a limit how small it must be. So we shall take it to be so small, that the centripetal force that holds the particle in the circular motion is the weight, $mg$, of the bob. That means the centripetal acceleration is $g$. Therefore, according to the second of the above equations we can write

$g=\omega^2L$

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

Solve the last equation for T to you get: $T=2\pi\sqrt{\frac{L}{g}}$.

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Two problems here: the motion is not uniform circular so the period is not related to the angular velocity by $\omega=2\pi/T$! Second: the force balance is wrong. The tension on the string must be more than $mg$ otherwise the bob would not accelerate upwards! At the bottom of the arc the acceleration is towards the centre of the circle, so there must be a net force upwards. – Michael Brown Mar 16 '13 at 13:19
@MichaelBrown Thanks aagain! I would try to pay more attention to the scope of an answer, which has been made clear in the first paragraph, and the simplifying assumpions made there and elsewhere in the answer. Imagine an infinitesimal deviation around the equilibrium position of the mass. Your own answer is, I will not say wrong, only approximatly correct since $f(\theta_{max}\sim 2\pi$ for very small deviations! The $T$ in your last equation comes up only as an average period. I think the correct answer involves an elliptical integration, but the approximation will do for practical purposes. – JKL Mar 16 '13 at 18:16