Next question would be, what if you have a car running over the road, and the car has many vibration excitation source. We would like to know the vibration of this general, I'd say, complicated structure. How to do it? Conclusion. The vibration of this car, if I denote the vibration in y direction, that is, function of three-dimensional space and time, it can be expressed as we did before. Sum of the magnification factor multiply by many many modes. How to get this mode? That's our primary objective, to utilize what we learned in this vibration lecture. How to get this. To get an idea how to get this, let's go back to the simplest case that we learned. Remember, the simplest vibration case we learned is two degrees of freedom, frictionless flow, same mass, same stiffness, this case, and we measure the vibration of this mass using coordinate X_1 and using coordinate X_2. Using Newton's second law, we obtained the equation that represent the vibration of this. Of course, we put the force over here and force over here too, F_1 and F_2. Equation turns out to be, there is a mass matrix, X vector, double dot, K matrix, X vector, and that has to be F vector. This means that intuitively, even though it has two degree of freedom system, it follows same concept as we used for single degree of freedom system. If you look at more detail about this, you'll find that this mass matrix is composed by m, zero, zero, m, X_1, Y plus K matrix. This is the K matrix, K, one, one meaning that the force exerted to mass one when I impose the unit displacement over here. So that is 2K and that is 2K. Because this is a symmetry system, minus K, 2K. X_1, X_2. That is equal to using the notation I wrote over here, F_1, F_2. Remember, how we obtain the mode shape. How we did to get a mode shape or mode shapes? What we did is, as we did before for the string case, the mode shape, in other words, the essential fundamental shape of a mode is not depending on the excitation force. That is physical observation. Mathematically, we say it could be the solution of homogeneous equation. So we try to solve characteristic equation. In this case, that is m equals zero. This is characteristic equation. For string case, you have homogeneous equation, but to get the mode shape, we set up. We make excitation force is zero, then solve the characteristic equation. So for string case, a beam case, for two degree of freedom, simple case, we follow the same route. Exactly the same route. The difference is physical interpretation. So looking at more detail about this equation there, I'm assuming that X double dot is equal to minus Omega square X, in other words, harmonic excitation. Thanks to Fourier mathematician, Fourier civic contribution. Then we can write, this is minus Omega square m. In other words, I am putting this over there, plus 2k, minus k, and minus k, minus Omega square m, plus 2k, X, is equal to 0. Everybody knows the solution of this equation. Trivial solution, this one is zero. That is meaning this, therefore, determinant of this one has to be zero, that gave us two natural frequency Omega one square, and Omega two square that you can get on from a previous lecture. What if I have a two degree of freedom system, but not as simple as this? Because in this case, by using this simple two degree of freedom system is that we can intuitively get or guess the mode shape. Because this is symmetric, therefore, one possible mode is moving two mass in equal phase and equal distance. The other one is moving this mass equal distance, but out of phase. No other possibility. Every vibration or be the summation of this plus that. Now, for more general case, say, there is a mass m_1, and mass m_2, spring, and spring, raised wall, frictionless button, and there is a spring also. Say this is k_1, and this is k_2, and this is k_3 generally. At the same token, I'm measuring displacement of mass one, by x_1 coordinate displacement m_2 by x_2 coordinate, again, assuming that I have excitation over here, F_1, and excitation over here, F_2. Now, of course, the equation follows generally this form, but the detail of a mass matrix, and stiffness matrix will be different. To see the detail, what we have to do is draw a free body diagram, that shows the force acting on this mass, and the excitation force, and apply Newton's second law to this mass, showing all the external force. Then, we can obtain this matrix expression. But is there any other way to do this more systematically? There is. Okay, for us, let's see mass matrix. Mass one has to respond to this coordinate because m_1, x_1 double-dot, is what we call inertia. What about then this term, multiply by x_2? What about this term, multiply by x_1? There's only two mass, and there's no chance to have this mass term, cross products term, to act upon other coordinate. There's no chance this mass can have inertia expressed by x_1. No chance this mass expressed by the x_2 inertia in terms of x_2 coordinate. That's a physical understanding, probably, put this to cross-product from zero, and plus K matrix. This term, what does it mean by this term? Say, this is K_11 in matrix rotation, this is K_12, and this is K_21, this is K_22. What this means K_11? If you see carefully, K_11 multiply Kx_1 is what? Force exerted on coordinate one, by having unit displacement in x_1 direction. So K_11 is a force acting on element one, by having unit displacement on x_1 coordinate. So what is this? So holding this, and the moving this unit displacement, then I will get K_1 plus K_2. Why this is positive? That's good question. Holding this, pushing that, then force exerted in this direction and that direction, and remember, when we write the equation MX double dot, and K multiplied by coordinate. So if you you apply Newton's second law, minus K_1, minus K2x equal to m_1x_1 double dot. So if I move this term to m_1x_1 double dot case, that turns out to be plus. What about this one? This one is force acting on one, by moving unit display to Mx_2. So holding this, and then I'm moving this. Then, how much forces acting on mass one? K_2 multiplied by x_2, because this is acting over there that it has to be minus k_2. What about this one? Moving one by one, and acting force on mass two that is again minus K_2. This is similarly, I can obtain K_2 plus K_3. Wow. That has to be equal to F_1, and F_2. What is the next step? Next step, as I mentioned earlier, we have to solve the characteristic equation. Lets do it. How to do it? Minus Omega square m_1 plus K_1 plus K_2 minus K_2 minus K_2 minus, sorry, it has to be Omega square. Minus Omega square m_2 plus K_2 plus K_3. Therefore determinant of this has to be zero. Are we doing okay? Lets check it. If K1 and K2 is equal, then this is same as this, this is the same as that, this is same as this, and this is the same as this one. So for intermediate chalk, this mathematical expression looks okay. Another one. If K2 is very small, in other words, the link between mass one mass two is very small, then this term is very small. Then the solution pretty much dominated by this diagonal term. That's another physical observation we can find, and the later on what does it mean by practically? Now, let's solve this characteristic. Okay, determinant has to be zero. Therefore I can write minus omega squared m1 plus K1, K2 multiplied by minus omega squared m2 by plus, I think K2 plus K3, minus k2 squared. Let's this be zero. This will give us the natural frequency omega one and omega two square. What modal analysis has to do is, we plot this omega one square and omega two square over here, and then this matrix will provide us the mode shape. Why? Okay, that two simpler case m symmetric case. The first natural frequency of this system was k over m. Second natural frequency of this system is, right? Say if this mass moving at the same time, at the same phase, the natural frequency is obviously. I've got this. This will go up, therefore this stiffness will not act upon the system, therefore natural frequency is k over m. In other words, if I excited this system with natural frequency, first natural frequency it will move like that. Second natural frequency it will move like that. Therefore, if I put this omega one over here, that means I'm exciting this physical system as I show you over here like this. That will provide us mode shape. Okay, so first I had to get omega one square, omega two square from this rather complicated equation. Okay, let's do it. Omega to the fourth m1, m2, I'm multiplying this with that, and minus omega squared m1, k2, k3. Then I am multiplying this too that, that give me minus omega square and 2k1, k2. Then multiplied this one to that, I have k1 plus k2, k2 plus k3. That has to be equal. That is characteristic equation. So I have similar term over here that has minus omega square. But m1, k2, k3, and then m2, k1, k2, okay. So one is omega to the fourth, one is omega to the square, and one is omega to the generals. So we can get a solution about omega square. Omega square has to be, I say this is term B for simplicity. This is term B. This is term A. This is terms C. Then omega square has to be, what is it? Minus B, plus minus. Let me rearrange this. So omega to the 4A. Here I will say minus B omega square, and this is a C. This has to be plus. Then the solution has to be minus B plus minus square root to B squared minus 4AC over 2A. That is correct. Okay, then, rather not very straightforward, we got a solution.