This post is background for another post I’m doing tomorrow or possibly later this week. It involves that old standby of freshman physics, the pendulum. We want to find out the period of a pendulum, the length of time the pendulum spends making one complete back-and-forth cycle. It’s literally one of the oldest calculations in the book, but personally I’d rather have a background that’s more complete than needed instead of possibly leaving important things out. And just for the fun of it we’ll use a very slightly atypical method.

This, straight from Wikipedia, is a classic pendulum diagram boiling the entire thing down to its basic geometrical essence:

Let me rename one of the variables in the diagram just for the sake of convenience. Let’s rename y1 to L, for the length of the pendulum.

Now the only mass involved is the pendulum bob at the end of the rod or string. We consider the string to have no mass, which is actually a very close assumption for even “heavy” strings as long as the bob is much heavier by comparison. As such, we can find the potential energy of the system. We know the gravitational potential energy is the mass m of the pendulum bob times its height times the local gravitational acceleration g. Therefore the potential energy U = mgh. And, for the record, h = L(1 – cos(theta)). Take a moment to confirm this for yourself, and we’ll use that in a bit.

Its kinetic energy is of course K = 1/2 mv^2, but we’d like to think of the pendulum in terms of the angle theta on the diagram. How can we express v in terms of theta? Well, look at the distance along the curve from the lowest point to whatever point the bob happens to be at. Call that distance c. From the basic geometry of a circle – because the pendulum bob is constrained by the string to move in a circular path – we know that c = L*theta. The rate of change of position (c being the position) is the velocity v, and with constant L we know that:

The fraction-looking thing being the calculus way of saying “the rate at which theta changes with time”. Since we have both the kinetic and potential energy in terms of theta, we can write the Lagrangian now:

Plugging that into the Euler-Lagrange equations gives, after some slight simplification: [EDIT: The following two equations should have a minus sign on one side. Sorry for the typo!]

However! sin(theta) is approximately equal to theta, if the angle is small. Here, it is. Therefore we can make the replacement without sacrificing much accuracy at all, and it makes the calculation much simpler:

We’ve solved this differential equation before, it’s just the simple harmonic oscillator. Reusing our knowledge from that problem tells us the period of the pendulum will be:

This is remarkable for the fact that the period is independent of the amplitude of a swing. Swing a pendulum a little or a lot, and so long as our sin(theta) approximation isn’t broken too badly it will still take the pendulum the same time to make each swing. Larger swings mean the bob has more distance to cover, but it’s doing so faster. The effects cancel out. Even for swings as large as 30 degrees, the period will deviate from this equation by less than 2%.

Given this fact, we can measure the length of a pendulum cord by timing the period. That’s what we’ll do shortly.