 A pendulum has an oscillation period, which is assumed to depend on the pendulum's length, its bob mass, the angle of its swing, and the acceleration of gravity. The first thing I want us to do is write out the relationship in a dimensionless form following the six steps of the Buckingham Pi theorem. Then I want us to use data from Earth where we have a length, a mass, a period, an angle of swing, and an acceleration of gravity to figure out what the period will likely be in a different circumstance. It's essentially a model prototype relationship except with Earth and Moon to be able to change the acceleration of gravity a little bit more meaningfully as well. The first step of the Buckingham Pi theorem is to list all the physical variables involved. For that, I can start with the period of oscillation, T, the pendulum's length, L, the bob mass, M, the angle of swing theta, and the acceleration of gravity, G. Step number two is to write out the dimensions of each of those variables. First up, I have the period, which is measured in time. Then I have length, which is measured in length. Then I have mass, which is measured in the mass dimension. Note that the mass dimension is an uppercase M, not the mass variable, which is a lowercase M. Theta is dimensionless. The acceleration of gravity would be length per times squared. I can write that as length times time to the negative two. Step three is to determine a J value. J here is how many of our variables are going to repeat between the dimensionless parameters. A good general rule of thumb is to guess J equal to the number of dimensions that appear, and if you can't make it work, decrease it by one and try again. Here we have time, length, and mass appearing. Those three dimensions imply to us that we should start with a J value of three. Since we have five total variables, three of them are repeating. That means two will be non-repeating. Since we're generating a pi group for each non-repeating variable, that means we should have two non-repeating variables. Our repeating variables can be selected from these five. I typically avoid parameters that don't have any units, so I will leave out theta. And then I'm going to semi-arbitrarily neglect period of oscillation from my repeating variables because that's our dependent variable. Therefore, our repeating parameters will be length, mass, and gravity. Step five is to generate the pi groups one at a time. Since I have two non-repeating variables, I will have two pi groups, one for each of the non-repeating variables. The first one will use period of oscillation just because I'm working my way from left to right here. So I would write that as period of oscillation times length to the A times mass to the B times gravity to the C. What I'm doing in step five is trying to select exponents that leave pi group one as a dimensionless parameter. I personally like to write this out kind of like balancing a chemical reaction. So I will write my goal on the left and then I'll write all the dimensions of all the parameters here. And then I build an equation for each of the dimensions. For the mass equation, I have zero on the left and I only have B appearing on the right that makes solving for B pretty easy. For length, I have zero on the left. I have A plus C on the right. For time, I have zero on the left. One plus negative two times C on the right. So B is zero. If I pull out the time equation, what I'm doing is writing two times C is equal to one. Therefore, C is equal to one half. And then plugging that into the L equation, if C is one half, that means A will be negative one half, which means pi group one will be period of oscillation times length to the negative one half power times mass to the zero with power, which is just one times gravity to the one half power or period times the square root of gravity divided by length. I can repeat the same process using the other non-repeating parameter theta. For pi group two, I have theta multiplied by length. I need a new exponent this time. So A, B, C, let's start with D times mass. That should be a lowercase mass. The E times gravity to the F. And then I write it out in terms of all the dimensions I want to achieve. Mass to the zero, length to the zero, time to the zero is equal to, there are no dimensions of theta, so I can skip that. Then I have length to the D, mass to the E, length to the F, times time to the negative two F. My equation for M will be zero is equal to E. My equation for L will be zero is equal to D plus F. My equation for time will be zero is equal to negative two times F. Solving these three equations for the three unknowns, I first get E is equal to zero, and then from the T equation I get F is equal to zero, and then from the L equation zero is equal to D plus zero, therefore D is zero. Therefore pi group two is equal to period of oscillation times length to the zero with power, times mass to the zero with power, times gravity to the zero with power. All three of these terms become one, therefore pi group two is just theta. So what I can say from my two pi groups is that pi group one is a function of pi group two, or period of oscillation times square root of gravity over length is a function of theta. With that I can approach part B of the problem. If I have two situations I'm going to call earth one and the moon two. If I have L one, M one, T one, theta one, and I have L two, M two, theta two, along with both accelerations of gravity, what will T two be? So when I'm comparing earth and the moon here I'm saying theta one is equal to theta two, and then because pi group one is a function of theta that means pi group one at one must equal pi group one at two, or rather period times the square root of gravity over length is equal to period times the square root of gravity over length. Note here that the mass doesn't matter. The mass doesn't affect the period of oscillation. Neat. Then I know gravity one, I know length one, I know T one, I'm looking for T two, I know gravity two, I know length two. So T two is going to equal T one times the square root of G one over G two times L two over L one. Period one was 2.04 seconds. G one was acceleration of gravity on earth. So I'm going to use 9.81 divided by the acceleration of gravity on the moon, which is 1.62. The length of the pendulum on the moon, which was 30 centimeters divided by the length of the period on earth, which was one meter. If my length dimension units cancel and my gravitational acceleration units cancel, that means everything under the radical is a unitless proportion, which means that the radical is unitless. Therefore, whatever my units are for the period of oscillation number one will be the same units for the period of oscillation number two. If I jump into my calculator, I can take 2.04 multiplied by 9.81 divided by 1.62 times 30 divided by 100 all within a square root sign. So my period of oscillation on the moon will be 2.74959 about two and three-quarter seconds.