What we are seeing, that this small mode in this part, is a summary of what we learned in the last lecture. The reason why I suggest to you to go somewhat details about this, is because this is a fundamentals. This carries fundamental and essential technique and information related with a two degree of freedom system. Then later on, understandings two degree of freedom system will be power to understand multiple degree of freedom system. This part shows general two degree of freedom system. X1 is the coordinate and measures the displacement of the mass one. X2 is the coordinate and measures displacement of mass two. F1 and F2 are the forces acting on mass one and mass two. Then, we have to find out the stiffness matrix by drawing free body diagram, which means there how much force is acting on mass one and mass two. As you can see over here, the force acting on mass one by having displacement X1 about like this, and the displacement at X2 like this, then, the difference of displacement, that is X1 minus X2, multiply by K2, is the force acting on mass one due to those displacement. K1 X1 is the force acting on mass one due to the displacement at X1. By the same token, we can also see the force acting on mass two as you can see over here. Then, U-turns equation says, this has to be equal to mass one, M1, and the acceleration multiply by the M1 is M1, X1 double dot. Then using the coordinate sign convention, which is positive, then we can obtain the equation that governs the motion of this two mass system. Let me draw another thing that is acting on the acceleration multiply by the M2. That is M2, X2 double dot. Then I can write the governing matrix equation, that is, M1, M2, 00, and then here is X1, double dot, X2 double dot, plus K matrix. As you can see here, the contribution to the displacement X1 and X2 has to be K1 and K2, because there is a K2 X1. Therefore, there should be K1 plus K2. The contribution to the displacement X2, can be found over here, that is minus K2 X2, therefore I put minus K2 over here. What about the k matrix component over here? That can be found over here. The contribution to X2 would be, K3 plus K2 over there, therefore K2 plus K3 has to be here, and then contribution to X1 can be found at this term that is minus K2 X1 minus K2. That has to be equal to excitation, F1 F2. This is governing equation. Then what is the next step? Next step is to find out the mode shape of this two degree of freedom system. How to do it? We go to characteristic equation, which simply assuming that, X1 X2 displacement is equal to displacement multiply exponential j omega t. In this notation, we use the small x1 x2. That is a little bit confusing, but let's keep this is the displacement of the mass one, and mass two. This is displacement multiply by the exponential j omega t. Then, this harmonic assumption will bring the characteristic equation. This is the solution of the characteristic equation that we derived essentially expressing the relation between natural frequency and mass one and mass two and the spring constant K1, K2, K3. This looks rather complicated, but the first thing we can find out is that, first natural frequency is bigger than second natural frequency, which is obviously following our general physical concept. The other thing we can find out is that equation essentially shows as mass one and mass two changes. For example, if mass one getting larger than mass two, it certainly approach to certain value. In other words, physically it means that mass one is very, very large than mass two, meaning that the first natural frequency will approach to the essentially based on mass and the spring constant of K1, because mass one is very big. If mass two is very big compared with the mass two, essentially vibration characteristics will be dominated by mass two that expresses in this equation. This summarizes what we learned before. First, the case when mass one and mass two is the same as mass M, and if all the spring constants are same, then the first natural frequency will be K over M because in this case, the two mass as we demonstrated before, move with in phase, in this case their mass is 2M, and the spring constant over here is K and K, therefore natural frequencies K over M, which is fundamental or first natural frequency. The second natural frequency will approach to 3K over M because in this case, two masses are moving out of phase, therefore the natural frequency room approach to 3K over M. If this mass one and mass two changes, and the spring constant K1, K2 changes, then of course the first natural frequency and the second natural frequency will be changing. So to see the trend, our changing of natural frequency compare with the spring constant, K1, K2, K3, this graph essentially demonstrate what was going on, or what will be changing. So over here, that is the case when K3 and K1 is very similar, then, the stiffness over here and over here is very similar, for example, to K2. Then, that is the case what we explained over here. So the natural frequency, omega two and omega one will be as we demonstrated over here or approach to this one minus this one. But as the difference between K3 and K1 is getting large and large, then of course the difference between omega two and omega one will be getting large and large as we demonstrated in this graph. Similar patterns can be seen with respect to the difference between mass one and mass two.
