In the previous lecture, we obtained the response. There's two degree of freedom system can be expressed like that, and determinant of this matrix can be expressed like that. We examine the one, this denominator or this determinant, guess to be zero, make the system five rate, infinitely resonance. Then, if you use the Omega one square, that is k over m, and Omega two square equal to 3k over m over here. Then, this expression can reduce to m square, then Omega one square, minus Omega square, and Omega two square minus Omega square. The reason why I got this, is because that has to be adjoint of this matrix. Adjoint is a minor of the cofactor, so this element has to be this one, this one. Eliminated this row element and a column element that I can, have minus Omega square m plus 2k, that is corresponding to this. Similarly, I got this, the reason why I got plus over here is because this is the cofactor of the minor. Now, so what I have over here is this, and then I have the matrix over here. Then, I have F_1 and F_2 for simplicity. Assume that there is no F_2, which means I am exciting the excitation only applied to the mass one. Okay, now. From this expression, I can get the the displacement X_1, complex is to be equal to, that has to do with the denominator, that is m square, Omega one square minus Omega square, and Omega two square minus Omega square. Then, I have 2k minus m Omega, Omega square multiplied by F, that is X_1. Similarly, I can get X_2, getting this mathematical expression often regarded to be very tedious, but we have to be patient to get over this kind of tedious approach, because it will provide us very interesting, exciting physical insight. So that has to be m square, Omega, one square minus Omega square, and Omega two square minus Omega square. Then, I have this one multiplied by this, so I have simply kF. As case K is equal F_1 and F, there's no F_2. So looking very carefully this expression, and noting that this one has, this one multiply that one over some numerator, and the same multiplication in denominator and a numerator. So recall that A multiply by B over C can be expressed something A, something B plus, maybe I can write a C_1 and C_2. So that means, very interestingly, this expression also be composed by m multiply Omega one square, minus Omega two square, something, plus m multiply by Omega two square minus Omega square over something. What does it mean physically? Okay. Looking at carefully the first to top, which has the denominator is, m multiply Omega one minus Omega square. That system has only one natural frequency, which is Omega one. Second expression has, m multiply by this one. So that's system also have its own natural frequency at Omega two. The amplitude over here and there expresses how much it move corresponding to that natural frequency, for its natural frequency and second natural frequency. Okay. So if I get out those magnitude, then I can write X_1 over here is equal to F_1 over m. I have one over Omega one square minus Omega square, and then plus one over Omega two square minus Omega square. You can obtain this readily. X_2 is equal to F_1 over, in this case interestingly, 2k. Then, I have one over this, and then I have one over, there has to be minus, three minus Omega over Omega one square. This is one minus Omega over Omega one square. The detailed derivation, I will supply to you guys supplementary materials. If I simplify this expression by using the relation between Omega_1 square and k over m, then I can write this is F, and then I divide this time by Omega_1 square that will give me 1 minus Omega over Omega_1 square, and then I have Omega_1, I have to divide Omega_1 square over here. So what I can do is I can multiply Omega_1 square here that will give me k and I have over here one, and then I can divide this one by Omega_1 square, then I have one and here, Omega 2 square divided by Omega_1 square that is three, and I have Omega over Omega_1 square. So essentially, X_1 is expressed in terms of new parameter that is non-dimensional frequency component that is Omega_1 Omega divided by Omega_1 square. Essentially, this one sees the excitation frequency in terms of forced natural frequency. Similarly, we can convert X_2 like that. Now, this mathematical expression possesses many physical insight to get a physical insight or what this mathematical expression implies. One easy way to see this is to look at this mathematical expression in terms of graph. Okay, I draw this and want to see this in terms of Omega over Omega_1. First, I want to draw this one. Then, when Omega_1, this non-dimensional frequency tends to be zero, then this is zero, then this is one, this is zero, 1 over 3. So that has the value of 1 minus 1 over 3, that is 2 over 3 and F over k. So start with some here over here, and then when this approach to one, then this blows up, and this is some scale that is just 1 over 2. So at 1 Omega equal to Omega_1, this will go up. Therefore, the line has to be like that. Another interesting point is when Omega over Omega_1 is a square root to 3. So that is about 1.7. So if here is 2, this is 2 and over here, this gets to be very, very large. So there is another interesting line that blows up the magnitude of this one, and as you can see here over vicinity over here that is plus. So this is like that, and then little bit over 1. So if this is a little bit bigger than one, the term of this one, and term of this, and comparing those things is will be minus, therefore, look like that, and over in this region, you can readily find the curve looks like that. So physically, this curve physically means, one thing is resonant behavior, and another one is over here. There's certain frequency that make the motion of X_1 stationary. In other words, X_1 does not move. Remember, we have m, there is x_1, k, there is x_2, okay, like that. So at certain frequency of excitation, the x_1 does not move at all. This one is moving. Wow, that has a lot of practical application. This we called nodal point. So by properly arranging Omega_1, in other words, k over m, we can find the nodal point where the x_1 is a station. So this could be related with later on vibration absorber, and then let me draw the response of x_2, that is expressed like this which is quite similar with X_1, and that looks like, for I use other, and this will look like that, and then we have something look like that, and we have look like that. All right. Other physical observation we can see from this graph is that. Okay. If I excite the system with a certain frequency over here, for example, then X_1, there is a response of X_1 as well as a response of X_2. Remember, this mode corresponding to certain natural frequency. This mode corresponds to another natural frequency. That is the natural frequency, in other word, it represents two degree of freedom vibratory systems characteristics. Okay. But if I excite the system with the frequency other than natural frequency, then as you can see here, these two different mode is vibrating. As we see, the solution says, the solution says, X is equal to determinant of Z Omega, and at joint matrix of Z Omega, and F_1, F_2. Okay. Any frequency, in other words, if I excite the system with any frequency, there is X, composed by as we saw before, two mode shape. So if I excite the system over here, the mode one has amplitude like that, mode two has amplitude like that. Okay. Excite the system over here, is a very little vibration over here, very little vibration over there. Okay. We start with the vibration system that has mass, m_1, and some stiffness, k_1, mass, m_2. This is related with another stiffness, k_2, has mass list hill that is moving along this road. So we use the coordinate y(t), and coordinate X_2, and coordinate X_1. Actually, they've now solved this problem. This is the model that could mimic the car vibration in this direction. In this case, our objective is to minimize X_1, minimize X_1. Maybe, X_1 square. How to do this? Also, we attempted to understand the rather simple case. Same mass, same stiffness, frictionless roads on frictionless floor. In this case, we have to use coordinate, X_1 and X_2, assuming there is excitation F_1, F_2. We also studied this system. My demonstration has to do with a certain stiffness k_1 and m, and k_2. In this case, we have to use X_1 coordinate, X_2 coordinate, I oscillate this. So you can have y(t) over here. All these three system can be tackled by using governing equation. That comes from Newton's second law. We examined this governing equation normally, graphically. So we look at the governing equation in terms of graph. Okay. Magnitude and phase value. Okay. For two degree of freedom system, this magnitude and phase diagram looks rather differently as you can see over there. But using this illustration, if we attempt to find out the practical application, for example, nodal point, and magnification factor, and natural frequency. So having this model and investigating the behavior of this model by using Newton's second law, produces governing equation. We analyze this mathematical expression in terms of graph, as well as some demonstration, experiment, graph xy as experiment, and then find out practical application. This sort of rule, the sort of rule, mass physical interpretation and practical application. Normally looking at some extreme cases will be essential three parameters to understand vibratory system. Okay.