So, let's do our second example for second-order equations. This is a mass on a spring connected to a wall. I guess you can view this as a top view and the mass is sliding and oscillating. There is a external force. So, in addition to the spring there's some forcing acting on it. Okay. How do we write down a differential equation for this? This is basically F equals ma problem. So, this is Newton's law, F equals ma, and acceleration is d squared x dt squared. Right. So, the main point here then is to figure out what are the forces on this mass. So, there's Hooke's law which is saying that the force due to the spring is equal to minus kx. So, linear law for the Hooke's law, meaning that if you move the mass a distance x to the right, the restoring force will be to the left, the spring gets stretched and the force will be proportional to the distance that you move the mass from equilibrium. So, x is measured from equilibrium. So, you move it a distance x, the force is in the negative x-direction and is equal to minus kx. On top of that, you need to model friction. So, we have the frictional force. We're going to model friction as opposing the motion of the mass. So, the mass is moving with some velocity in the positive x-direction. The friction is going to be modeled as proportional to the velocity but in the opposite direction, so opposing the motion. So, we're going to model the friction by minus C times the velocity which will be dx dt. Then finally, we have one more force which is the external force. So, the external force will be driving this mass. So, the external force or model by F_e is equal to f naught times cosine Omega t, as a sinusoidal external force. Okay. So this is the Physics. F equals ma. Hooke's Law for the spring modeling the force due to the spring. Friction, the forces opposing the motion proportional to the velocity but opposing the velocity, causing the mass to slow down, and an external force. So, those are our F's and that's equal to mass times acceleration. So, the governing equation here then is going to be m d squared x dt squared, that's the ma, and that's pull the forces onto the left-hand side. We have the frictional force, was minus C dx dt on the right, we pull it to the left so becomes plus C dx dt, minus the restoring force so it becomes plus kx equals the driving force which is F naught cosine Omega t. Okay. That's our differential equation for a mass on a spring with friction and with a driving force. Again, a second order linear in homogeneous differential equation with constant coefficients given by these parameters; mass, frictional coefficient, spring constant, and the amplitude on the driving force. Again, 1, 2, 3, 4, 5 parameters, we can non-dimensionalise the variables in this equation and come up with a dimensionless equation. Let's do that again. Remember that we divide through by m and whatever is multiplying x becomes our Omega naught squared. So, Omega naught squared is k over m. So here, we will define Omega naught equal to root k over m. Now, we define the dimensionless time Tau equal to Omega naught times t. We can define a dimensionless position variable. Let me call that capital X. If you work this out so that you want to get rid of the term in front of the cosine Omega t, the correct scaling here is m Omega naught squared divided by F naught times X. When you substitute these dimensionless variables into this equation, you'll end up with d squared capital X d Tau squared plus Alpha dX capital X d Tau plus X equals cosine Beta Tau, which is exactly the equation we got for the LRC circuit. So, the non-dimensionalization returns exactly the same equation as for the LRC circuit, except now the definitions are different. This Alpha here which plays the role of a damping coefficient is C over m Omega naught, and this Beta here is still Omega over Omega naught, but Omega naught is different. Omega naught here is square root of k over m. Okay. So, let me review what I did. We have the application of a mass on a spring. We are using Newton's law, F equals ma. The acceleration is the second derivative of x with respect to time. We need to figure out how are we going to model the forces on this mass. We're going to model the spring force using Hooke's law. We're going to model the frictional force by saying it opposes the velocity and is proportional to the velocity, and then we're going to model an external force. So, someone is driving this mass. F equals ma gives us a second-order differential equation in homogeneous because of the external force but with constant coefficients. It looks different than our equation for the LRC circuit, but if we define the dimensionless variables using this Omega naught squared which multiplies X after we divide through by m, if we define dimensionless variables, we obtain exactly the same equation as the LRC circuit but with different physical meanings of the parameters. That's the beauty of non-dimensionalization. Okay. One more application. We'll do that in the next video. I'm Jeff Jasnoff. See you in the next video.