We've learnt about the vibration system, single degree of freedom system. That mathematically, it's expressed like mx double dot plus kx equal to F(t). We have tried to understand this vibratory system in terms of its transfer function. Previously, we learned this transfer function expression can be understood in terms of complex domain. This is kx, and this would be minus omega square mx, and that has to be F. But, what if we have a damping? What if you have a damping? For example, I made the damping like this. When we have a damping, low-frequency is almost the same as without damping. As you can see over here, without damping case, for low frequency it moves like that. If I put a damping over here, a low-frequency, it also moves like that. It doesn't look like this damp to system has a significant effect on the vibration at low frequency. For high frequency, no damping case. For high frequency excitation, as you can remember, when I excite this with a high frequency, the mass does not really move even though I put a lot of energy. For example, for damped system, and I have high-frequency excitation. So there's not very much move. That means damping does not play a significant contribution in the region of low frequency and high frequency region. What about in resonance frequency region? Resonance simply means that I excite the system with the resonance frequency. Okay, I excite the system with the resonance frequency. It's getting larger and larger. What if I do this? Will the damp persist? I excited this. It does not vividly show you how the damping really contribute the response of vibration system at resonance frequency. But, you can anticipate, oh there is some significant contribution due to damping. That's what physically we observe. To get more understanding about the issue that is demonstrated by this experiment can be demonstrated by more mathematical, look up. For example, I have now mx double dot, I have a damping and I have kx. kx equal to F of t. How to express a damping. Again, I suggest to you, start with a very simple case first. The simple model of damping is, that is proportional to the velocity of the system. So say, this is a cx dot. Or, in some of the book, it is written as rx dot. Because in acoustic, the c is used to express the speed of propagation. But it doesn't matter. So our mathematical model that include the damping look like this. One more term. If we do not have continuous excitation, then there will be a transient to response due to initial displacement or initial velocity. In that case, the response for initial displacement case, it'll look like this. For initial velocity case, it'll look like, this exponential decay term is related with the damping coefficient of C of course. Now, let's look at again this mathematical expression in terms of transfer function. Let x(t) has complex amplitude and exponential j omega t. Also, F of t has complex amplitude exponential j omega. Then, this equation will be expressed like minus omega square mx plus, interestingly j omega c. Because I am differentiating this once, so I have j omega coming out, and the c exponential j omega t, plus kx equal to F. Because we have the same exponential j omega t, we discarded it. Now, this mathematical expression shows that, okay, there is relation between complex amplitude and F. That is one over minus omega square m plus j omega c plus kx. If you take the magnitude of this, you will get the transfer function magnitude. If you see the face of this, we will see the transfer function phase of the transfer function as we saw before. We attempt to understand single degree of pivotary system which is expressed by this mathematical expression. In terms of transfer function, physically seeing this equation, assuming that excitation is steady-state. Expressing F of t is complex amplitude exponential j omega t, resulting this expression as minus omega square m plus jc omega plus kx exponential j omega t. Give me the transfer function as we talked to you before. Of course, transfer functions expresses the magnitude in phase relation between this complex amplitude and complex excitation. Another way to look at this relation is to use some of geometrical expression using the complex domain that has real and imaginary part. Expressing kx like this, because x has a real and imaginary part. Then, we have a jc Omega. So, j in complex domain, that is rotating with respect to a rotating 90 degree because j is, real part is zero and imaginary part is j. So, imaginary part is one, one imaginary part, that has to be in this case. The real part is cosine, some angle, imaginary part is assigned some angle. So, that has to be j sine pi over two. This one has to be cosine pi over two or if you like cosine 90 degree. So, j means, I'm rotating this 90 degree. So, rotating this 90 degree. So, I'm rotating this 90 degree and that magnitude would be c Omega, so that is a jc Omega. Then, what about this? This is a minus Omega square m, minus meaning that the phase is 180 degree, so that has to be, if I use different color, minus Omega square m, multiply by x, it has to be x. That sum, all of those sum has to be equal to F. So, if I redraw this, so that we can see more easily. I convert this to, I have real part and imaginary part, and I have kx and I have jc Omega x. So, if I draw here jc Omega x, because it doesn't matter whether I start with here or there. Then, I have minus Omega square mx, that has to be equal to F. That's what essentially this equation says. So, I converted this equation into rather easier graphical expression. So, in this graphical expression the interesting part we can see is that, this is the phase angle, say five between force and motion response. Interestingly, if Omega is increasing that, then this magnitude is increasing and this pi move like that, approach to 180 degree. So, we could say when this term is dominating in other words, when mass is dominating the phase difference between excitation force and the response turns out to be 180 degree. That is the same as when Omega is increasing. The phase difference between excitation and the force is 180 degree. So that, is this the situation. Or when you ride a car and the car pass over the rough surface, then the excitation frequency Omega is increasing and what you feel inside of a car is like this. The phase difference is 180 degree. Then, Omega getting smaller and smaller, and this will decrease linearly. But this one decrease proportional to Omega scale, so very rapidly decreasing. Therefore, the phase difference between excitation and the stiffness approach to zero. So, when low frequency, the phase difference is zero. So, back to the case when you ride a vehicle passing through the, for example bumper. Then, you move very slowly, then you feel that you are following the shape of the bumper, is like no phase difference. But if you pass the bumper very rapidly, you have a lot of high frequency components therefore you feel a lot of different frequency feeling. That certainly explains the phase relation between the excitation and response. Another interesting point that you can observe from this example, is that looking at the relation between this triangle. Then, you can say immediately the magnitude of square of this and the square of magnitude of square of this, and the magnitude square of this has to be the same because this is right angle. So, if you write that relation you can obtain the magnitude relation of the vibratory system. So, say F square has to be equal to, this one is a c Omega x magnitude square, plus the difference with this and that is KX. The magnitude of this is Omega square m. So, the difference is minus Omega square mx magnitude square. That will give you the transfer function immediately. The physical interpretation has to do with the phase. Magnitude of x and F will look like this. The phase relation between excitation force will look like, between [inaudible] and pi. You can either using this graph or this graph to understand magnitude and phase relation of single degree of freedom vibratory system.