Built on Facts

Period of a Pendulum

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:

i-533b2e16f10154efec6cd04efd1d7f23-pendulum.png

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:

i-d4bb9fe09d6b78eb832985821d8a307f-1.png

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:

i-88516ca505d84f969b7ab15a965c3fb3-2.png

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!]

i-d18063683dcc0d42b9be45451a84d1e3-3.png

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:

i-bb5c2c6b0452df43a61e3974bd9b473f-5.png

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:

i-8d7070ce28f0af28047c7a5fb8246d53-4.png

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.

Comments

  1. #1 Uncle Al
    January 13, 2010

    Do you believe the length of a one-second pendulum is coincidental, or that somebody measured ideal equator to pole distance good to 10^(-7)? “No” to both. If only the meter had been a factor of 1.00069181 smaller.

  2. #2 andy.s
    January 13, 2010

    Corrected equations are at:
    http://mathurl.com/ydyts86

    and

    http://mathurl.com/yaadex6

    if you need ‘em.

  3. #3 Anonymous
    January 14, 2010

    Say, what is the basis behind the concept of the Lagrangian? It seems really arbitrary, yet it seems to work really well, since you get the same equation the Newtonian way.

  4. #4 Matt Springer
    January 14, 2010

    The Lagrange equations are pretty much just Newton’s laws after a very brilliant (I’d have never thought of it) restatement in terms of energy and in a generalized coordinate system. The Wikipedia article has a pretty good derivation of Lagrange from Newton. As a rough heuristic, the first term is “ma”, the second is “F”, and it’s just Newton’s F = ma restated.

  5. #5 IBY
    January 14, 2010

    @Matt
    Thanks for the response. That is why I find it so strange. How do people come up with something like that? It seems like they pull it out of nowhere like magicians. ^_^

  6. #6 Nia
    September 11, 2010

    What are 3 variables that affect the period of a pendulum?

  7. #7 yas
    September 14, 2010

    HI ,
    Can you please tell me how to write kinetic energy equation for a simple pendulum with a massless support moving horizontally with constant acceleration of a?!!

    Thanx

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