In the last lecture, we studied the behavior of a two degree of freedom system. To understand the physics behind the two degree of freedom system as well as practical application of a two degree of freedom system, we start with the simplest case that has spring K and mass, same spring K and the mass and the spring. Which is located by some roller that rolls on the frictionless floor. To measure the motion of these two mass, we need to employ coordinate. Also we assume that there is excitation force acting on this mass. I denote this is F1 and this is F2. So, this is the graphical illustration of the system two degree of freedom system. We are going to handle hoping that the system will provide us some physical insight. From this physical insight we would like to get some practical application. So, we actually are handling these two degree of freedom system in three different reasons. To understand the physical behavior of the system we need to convert this physical system into mathematical expression. That gave us in the previous lecture mass matrix X1 double dot, X2 double dot plus stiffness matrix X1, X2 that has to be equal to excitation force. In this case there is F1 and F2. Those are complex exponential J Omega T and Omega is the excitation force. This stiffness matrix has to be here M because M multiply X1, double dot represent inertia force acting on this mass 0, 0, and M. Stiffness matrix is rather interesting over here K11, and here we normally call K12. This has to be K21, and this is K22, meaning that KIJ, meaning the stiffness spring force or stiffness acting on ice area bounded due to the motion of J. So, K11 is actually the stiffness force acting on this mass. I suppose to move this mass by a little bit or unit or X1 in this way. Then you can immediately see that there will be [inaudible] acting on this way and KX1 acting on that way. Therefore we can have over here 2K. This is the stiffness matrix due to the motion of this mass acting open over here. So, I suppose move this mass by unit then this will be elongated. Therefore the force acting on this mass would be K multiplied by m in this direction. That is the direction same as the X1. Therefore rearranging this I have minus K. Now, by the same token I have minus K over here and I have 2K over here. So that is what I obtain physical meaning of this matrix formation. I can see that this stiffness matrix is symmetric because the system is a symmetric. The mass matrix is also symmetric. So this is equivalent if I write this in another matrix rather compact matrix form then I can say generally any degree of freedom system has mass metrics that can be N by N. I have displacement vector double dot. I have K matrix and this placement over here that constructs the stiffness force and that has to be equal to force vector or column matrix. If I consider the damping force over here that is proportional to the linear damping coefficient C. Then I can have immediately a damping matrix that is proportional to velocity vector. That certainly generalize the version that we can think about from this rather simple symmetric two degree of freedom system. Now, solving this would provide us some physical insight. So, how to solve this? We know that the solution of this matrix equation will be the following. Suppose I have excitation with the frequency of Omega. Then I can assume same harmonic response that's mathematically expressed as I have X Beta but I want this one as complex X vector exponential J Omega T. This assumed response over to the governing equation in matrix form. Then I can have the following expression m 000 X matrix double dot column complex amplitude a column vector plus I have 2K minus K minus K 2K, X column vector is equal to F vector exponential J Omega T. Therefore I have to put- Because I need to put minus Omega square M, minus square of M, minus square of M0. Therefore, what I get is simply minus Omega square M plus 2K and minus K, minus K, minus Omega square M plus 2K matrix has to be equal to complex amplitude equal to excitation force. This is the solution. Of course, you can solve it by expressing this matrix form into two algebraic forms. It is well known, however, the solution of this equation is simply X vector has to be equal to determinant. For simplicity, let's call this Z Omega. So, I call this Z Omega, then the solution of this has to be the joint matrix of Z Omega and then F vector. That comes from linear algebra. So, this is a mathematical expression. What does it mean physically? So, we have to investigate the meaning of this solution by looking at some of the extreme value of this expression to find out physical insight, and then, from that physical insight, we would like to find out possible practical applications. As you can see over here, this is the solution of the displacement, actually, how the solution of this displacement has to be related with excitation, and this is the system property. So, determinant of Z Omega, Z Omega as over here, determinant of Z Omega is simply this one, product of this one and the minus of product of these two elements. Therefore, this will be minus Omega square M plus 2K minus K square. That is the determinant. So, one of the extreme case of this determinant has to do with when this is approached to zero. When this is approached to zero, determinant actually is a scalar value. Therefore, the response tends to be increasing very abruptly, meaning resonance. So, that's the physical meaning of the determinant tends to be zero. So, we can find that there are two frequencies that make determinants as zero, meaning make resonances. Let's calculate this. From last lecture, we obtained the one frequency is equal to K over M, the other resonant frequency, Omega 2 square, is 3K over N. So, there are two resonance frequencies that are corresponding to two degree of freedom system because essentially, two degree of freedom system, meaning that we need two independent coordinates that completely describe the motion of this vibratory system. We will extend that understanding to modal analysis concept. There are two natural frequencies. One is K over M, one is proportional to 3K over M. What's the physical meaning of this natural frequency? In [inaudible] , we can envisage that this rather simple but symmetric system can move in two ways. One is these two mass move co-linearly, move same pattern, then this stiffness does not have a chance to contribute to this type of vibration. So, in this case, this mass M only feels the stiffness by this one, and this mass M is also influenced by the stiffness over here, this stiffness. Therefore, the natural frequency of the whole system to be 2K over 2M. Therefore, that is proportional to K over M. That corresponds to the natural frequency of this motion or mode later on. Another possible way, of moving this to mass is this way, opposite direction. Then as you can see here, this stiffness, will contribute the motion or vibration of this system, very significantly compared with other cases. In this case, and if I want move this, the stiffness of this and stiffness of that, and the stiffness of that, contribute equally likely. Therefore, the Omega two squared has to be 3k over m. Okay, let's go back to what you did for single degree of freedom system. For single degree of freedom system, we have a mass, and we have stiffness of k here, and we have coordinates. Then measure displacement over here, and we impose, there is an excitation force. We got mathematical expression of this system, that is mx double plus kx equal to F of t. By the same token as we used before, we assume that this is harmonic excitation. Then, harmony response will be induced, that gives us minus Omega square m plus k, and x that is complex, amplitude that is complex force. Of course, Ft is complex force multiply by exponential j Omega t. The solution of this response is very simple. X is equal to F over minus Omega square m, plus k. When this denominator get to be zero, that corresponding to the resonance as very similar with what we have over here. Okay. Let me demonstrate a little bit. The force to degree, I mean single degree of freedom system. Okay, when I get a resonance, this we're getting bigger and bigger, or low-frequency moves like that, high-frequency move like that. But two degree of freedom system, there's two different resonances. I don't know whether I can demonstrate successfully. I hope you can see it. Okay, this is forced natural frequency. That is one-third of a second natural frequency, that has to be small. So, this is like that. This is not quite same as this system, but I only try to give you a physical insight. Okay. Second natural frequency. Sorry, not easy to demonstrate. Try, we're here. Okay, I can do that. Okay, this is second natural frequency, which is different with that. Okay. Forced this and we observe from this experiment, or it is very likely the natural frequency expressed in this expression happens. Okay. Also, remember that we use the graph, which shows this behavior. This is when natural frequency coincides with the excitation force. We also draw the phase diagram. All right. The second thing we observed. It's not what we observed in single degree of freetest system. When we excited with a frequency of Omega one, the two degree of freedom system moves like that, and when we excite the system with the different frequency, the motion of this two degree of freedom system selected, and that is different. So, each frequency corresponds to each vibration shape, which we call mode shape. Okay. So, what we can intuitively draw, the mode shape of this force natural frequency is the case when this two masses moving in the same direction. So, moving same direction. So, we can write the modal shape is like that. Okay this is x_1 and this is x_2. The second mode shape is like and this go up, this will go down. Okay. So, mode shape corresponding to the second natural frequency, looks like that. All right. Back to mathematical operation. How to get this mode shape from this mathematical expression? It's this simple. Pluck this Omega one square over here, will give me the solution of x like that, and pluck this Omega two square over here will give me the solution of x like that. Okay, let me write down again. This expression over here. Determinant has to be minus Omega square m, plus 2k minus case, that is determinant. I have all the joints of Z Omega. That is simply like that, and I have F_1 and F_2 over here.