Okay, let's try our third application. This is the pendulum. We have a mass m connected by a mass-less rod of length l to a pivot point. The pendulum then is able to swing on an arc. Gravity is acting on it with a force mg down. There is a friction that's opposing the motion damping the motion. Then there's an external force that is driving the pendulum. So, one way of thinking about it is a dad pushing a child on a swing. The dad is providing the external force. The child is the pendulum on the swing. How do we write the governing equations for this? This is again Newton's equation. So, F equals ma. What is our coordinate system here, the easiest way to write a coordinate system is to use the fact that this mass is constrained to move on an arc. So, we can just measure distance along the arc. So, we can define our coordinate system along the arc and call that s, with origin s equals 0 corresponds to theta equals 0. So, s the coordinate system will be s which is equal to l theta. So then the acceleration of this mass our a is going to be l. Then theta double dot. So that's the mass times acceleration will be ml theta double dot. What are the forces? There's the gravitational force and F sub g, that's along the arc. So if you do the projection of the downward vertical force onto the arc, you get minus mg sine theta. There's the frictional force. So, the frictional force is going to be opposing the velocity. So if it's moving this way the frictional force will be negative. So it will be minus C, ds, dt with s equals l theta will be minus C l theta.d theta dt. So that's the frictional force. Then finally the external force. We just model as F naught times cosine omega t. Putting that together, we get our differential equation. So, mass times acceleration is going to be ml theta double dot and then on the right-hand side we have the frictional force so we bring it to the left-hand side. So, plus Cl theta dot and then we have gravity on the right-hand side minus mg sine theta. We bring to the left plus mg sine theta. That's equal to the external force which is F naught cosine omega t. So that's our second order differential equation. But this is different than the previous two cases. The previous two cases the LRC circuit and the mass on a spring. The modeling resulted in a linear differential equation. So every term was proportional to theta or its derivatives. Here we have a sine theta term, not linear and theta. So this is a non-linear equation. Nevertheless, we can go ahead and non-dimensionalize it just like we non-dimensionalized previously we can use the idea that if this oscillation is of small amplitude here so theta is small then sine theta is approximately equal to theta. If we divide through by ml this term will have a g over l in it. That's supposed to be our omega naught squared. So we can define our omega naught equals the square root of g over l. Just like we did in the previous equations, if we do that then we can define a dimensionless time. So tau is omega naught times t. That's actually all we can do here. Theta is already a dimensionless variable. The angle is a dimensionless variable and because it enters as sine theta, theta has no units. So, if we define it like this, we can get our dimensionless equation. Our dimensionless equation then becomes d squared theta, d tau squared plus alpha d theta d tau, plus sine theta equals another dimensionless parameter gamma times cosine beta tau. So, now we have three dimensionless parameters. We have alpha equals c over m omega naught, we have beta equals omega over omega naught. Then we have gamma which is an additional non-dimensional parameter equals f naught divided by ml omega naught squared. Three dimensionless parameters and that's because we have a non-linear equation. But for small amplitude we can linearize. So, if we assume small amplitude, so we can make the approximation that sine theta is approximately equal to theta. Then we can replace sine theta by theta. We can scale theta at that point, we can define a new theta capital theta which is theta scaled by gamma. So we can eliminate gamma from this equation and we can end up with the same dimensionless equation that we've always considered which would be d squared theta d tau squared plus alpha d theta d tau plus theta, sorry, these are the capital theta the scaled value of theta. Then that's the gamma goes away and you just have cosine beta tau. So it's the same dimensionless equation as in the other two examples which I think is quite remarkable. So, all three examples we've done if you define the variables, the dimensionless variables appropriately you end up with exactly the same dimensionless equation. So, let me review this example: This is the example, the application to the pendulum, we have a mass on a mass less rod of length l gravity acting on the mass. Frictional force opposing the motion of the mass and an external driving force such as the dad pushing the child on the swing. If we construct Newton's equation F equals ma and he had all the forces, we get this equation which is non-linear equation because instead of the usual linear term here theta is actually a sine theta because it's a non-linear equation you just can non-dimensionalize time. You end up with this dimensionless equation then that has three parameters; the damping, the forcing amplitude and the forcing frequency alpha, gamma and beta, or alpha, beta and gamma. On the other hand, if we want to make the connection to the previous two applications we can consider a small amplitude of the oscillation so we can make the linear approximation sine theta approximately theta. Then we can re-scale the angle theta by gamma introduce a capital theta, write the equation for capital theta and amazingly, we get exactly the same equation that we had in the previous two applications. So, in some sense the LRC circuit, the mass on the spring, the pendulum at small amplitudes of oscillation are all the same problems. Physically they're all different, but they all satisfy exactly the same second order linear inhomogeneous differential equation with constant coefficients with one coefficient alpha here in the derivative term and then a beta and as the frequency and the forcing term. So, we need to solve this equation and that's what I plan to do. I'm Jeff Chasnov thanks for watching and I'll see you in the next video.