 So now that we know a little bit more about degrees of freedom, let's develop some intuition as far as classical dynamics is concerned. Now do this thought experiment. I mean you can beautifully imagine this in your mind, but if you have the facilities go do that yourself. You have a very strong spring and you tie that to the roof and you tie a mass of a known magnitude there. And you notice if it pulls down, if you hang it, it pulls the spring down and you can actually measure it without the load and then with the load and you'll see well it goes down a certain distance. Now you know exactly what the force on the spring is caused by this mass because we know weight which is gravitational force. This is the mass times gravitational acceleration g. So you know exactly what that force is and what you now notice. Well if you double or if you change this downward force by changing this mass g stays constant. You're doing it in the same laboratory. It's going to pull down a bit further and you can plot these. If I put this force on the spring pulls out that distance. If I put another force on it pulls down a different distance and what you eventually notice is the follow. And this is really school physics but what you notice is there's a direct linear relationship between the force and the distance. Now let's just call that distance x in as much as we'll plot this. Not as a y-axis but as a x-axis and the distance that it pulls down. We just call that distance x. So the force in this instance is down and it pulls a certain distance down. So when you do this proportionality or when you do this experiment and you plot force and the distance that it pulls out x. Force x, force x you'll see well there's a direct linear relationship and when there's that we know we can put in a constant of proportionality and we'll call that constant of proportionality x, k I should say. And then from school we also know that that's rook's law for spring. We needn't go there. But within a reason we can suggest that definitely there's a spring constant because we can also then do the same force the same distance and for a different spring we'll get a different k value there. So we can actually work out what k is for each and every spring by just putting x on this side. The simple is that that's not the point though there is this relationship. Now I apply an added force. Well what I'm getting at let me just show you directly what we're getting at. And what we are getting at directly is that force can also be written as mass times acceleration which is x double dot. And the dots refer to the fact that here we are dealing with time derivative. So I can also say that the force here is mass times d squared x dt squared. That's what we're dealing with. And if we say that that equals kx we are dealing with what? We are dealing with a differential equation in as much as we have a variable x and we have one of its derivatives here the second derivative in that. So what can we write? Let's just write this as x prime prime whichever way you want to write this. So I'm going to have that in x prime prime. And now we've just got to be careful with the signs here and what we do with that. We know that this force that the spring pulls up now that was the force down and you can still draw them as also be a force in an opposite direction that's what the spring pulls on that mass so that we do each equilibrium. If you hang that mass this is going to hang there. It's an equilibrium and therefore the forces must balance. So we must just be very careful what we do here. So we know that because it's a restorative force we usually have something like that. If we now have that negative in there we bring it over to the other side plus kx equals 0 and I can also just divide everything by mass and I have x double prime plus k over m x equals 0 and that is a differential equation. That is a differential equation. And how are we going to solve this differential equation? That's why you can't really do proper physics, proper classical dynamics. If you're doing the linear algebra and you haven't had a course in differential equations. Now fortunately these are some, these are one of the easiest differential equations to solve. And as much as, and just remember this is for a bit of revision. I can write something as dx dt. I can write this as x prime. I can also write this as dx. So that's all the different notation. So to think about it I could have written that as d squared of x plus k over m of x equals 0 which I could also have written as d squared plus k over m x equals 0. And you will notice this is a homogeneous differential equation in second order and because it's homogeneous it's equal to 0. Remember I can make use of the auxiliary equation. I can say m squared plus k over m equals 0. Now that m is not that m so probably I should call this something else and usually we call k over m we call that omega squared. And perhaps we'll get to how that, how that is. This is the angular velocity. In angular velocity omega equals 2 pi mu. That's 2 pi times the frequency. The frequency is how many times per second. Now it's very easy to do. This is you had the intuitive thought experiment of the mass just hanging there and how we got to this differential equation to start off with. You can also imagine a wheel turning and on the periphery of that wheel constant motion the periphery of that wheel is a little light and if you now view it edge on exactly like this all you'll see is actually this mass on the spring going up and down. It's exactly the same motion. And we can show that this is such. So usually here we'll have omega squared. Omega squared we'll have omega squared. And we'll show, perhaps I'll show that the spring constant divided by mass gives you the square of the angular velocity. And angular velocity is in radians per second. How many? This up and down motion can be expressed as radians per second. And this is how many radians per second. If I drew a line on this disc that's spinning and it goes certain radians a unit of angles, certain radians per second. Okay, we can express this as this. And this is an easy, very easy equation to solve. In other words, I can have that m squared equals negative omega squared. And in the next video we'll do, maybe I'll put some values in the annual show. I'll still decide how to get the solution to this. This is obviously going to be negative. In other words, we are going to deal with imaginary numbers here.