Let us look at what really happened of this system, low-frequency, resonance, high frequency. To understand what happened in this vibratory system, we have to look at this and then why it oscillate like that by applying Newton's second law upon this mass. So I measure the response of this mass by using coordinate X of t. This is stiffness k. I oscillate this base, say this is y with a certain frequency of f. So I express this like this y is complex amplitude of y, therefore I denote this as complex amplitude X. The reason why this is the same as f over there is because this is linear system. Therefore, if a frequency is 200 hertz, the output frequency has to be 200 hertz. In other words, if I oscillate this system with the base with a frequency of 200 hertz, but I got a frequency other than 200 hertz, that means this system is not linear system. So that is one way to find out non-linear system too. So from this picture, we want to have mathematical equation because what they normally do is mathematical domain. Then we see this in physical domain and also we see this is practical domain. There is three bridges, and more often, you cross the bridge the more you get understanding. So travel around this three domain using this features as far as you can. So get mathematical expression of this vibratory system using what, Newton's second Law. Because Newton's second law governs the force and motion. So we have to apply Newton's second law upon this mass. What kind of force is acting? In this case, as long as I lift up, the force due to the presence of gravitational force field is already balanced. So forget about the force due to gravitational force field. Then because we have a coordinate y(t) in this direction, the force I moved this y of t and I assume that x of t move about this way. Then the stiffness, the force acting due to this stiffness would be k(y minus x). That is a force acting on this mass, that has to be equal to, according to Newton's second law m. Newton's second law says, it has to be equal to m times the acceleration of this mass that will express x double dot. That is a pictorial or graphical expression of Newton's Second Law that determines how this mass will move due to the foundation excitation. So looking at this vector expression according to the coordinate I selected, this is the coordinate and this is positive side. So that has to be k(y minus x) has to be equal to mx double dot. If I rearrange this mathematical expression that will look like mx double dot plus kx equal to ky. That is the governing equation. So this is mathematical expression. So I want to move to practical domain based on this mathematical expression. By exploring this mathematical expression in terms of physical parameters like m, k and y. Okay. We have vibratory system and applying Newton's second law, we obtained mathematical expression that looked like ky. Simply compare with the mathematical expression we obtained before. F of t in this case is ky. This is the case when we have mass, spring and we measure the response from the mass. So, what happen when we have this excitation on the response? We use the same approach we done before. So, say assume y of t has complex amplitude and exponential j2 pi ft and assuming x or similarly has complex amplitude and exponential j2 pi ft. I explained already why there's two frequency has to be same. Plot this assumed solution to the governing equation but it gave me the following result which is similar with what we have on this system. That is minus 2pi f square m plus k and complex amplitude of x has to be equal to complex amplitude of base excitation y. Put k over here. Therefore from this expression we can obtain easily how much amplitude I get for unit base excitation of y. As is simply expressed as over here k and here k minus 2pi f squared m which is a similar but big different. Now, let's see how this expression look like. In this case, the complex response compare with the basic citation. We obtain the various similar results that we obtained before. The numerator has to be k and denominator we obtain k minus two pi f squared m. If I divide numerator and denominator by mass m I obtain k over m and I obtain k over m over here and minus 2pi f square and we know that is natural frequency square. So, that is 2pi f0 square and this is of course the same as what I just wrote. So, rearranging this expression gives me simply 2pi f zero squared divide by 2pi f0 squared minus 2pi f square. So, I want to see the response this vibratory system in terms of the natural frequency then I have to divide this one and that one by 2pi f0 squared and I obtain one, and over here I obtain one and I obtain f over f0 squared. I should write down over here. That is correct with the 2pi has to be disappear. So, this is very compact expression certainly possesses a lot of physical insight that could be applied or used in practical application. So, as we observed before we can see that this expression look very differently when excitation frequency passes through the natural frequency. In other words, when f equal to f0 certainly this one behaves as some singular behavior. So, then f equal to f0, his one tends to go zero, so it blows up. So, if I draw this one, in other words the complex amplitude compare with the basic excitation amplitude over here when f excitation force equal to natural frequency we'll have infinity and over here when excitation force is zero. In other words, the load profile is a flat. Of course we will have zero response. So, the behavior look like that. When excitation force is greater than f0, in other words in this region we observed that this one is bigger than one, therefore we have negative value. So, if we draw as we did before in terms of magnitude and phase over here it looks like that. Over here its magnitude will look like that. Therefore, similarly if you draw the phase. Now, if I express the negative in terms of phase, it would be 180 degree or 2pi. There is a zero. So, over here therefore it goes like that and it goes like that. To see more clearly I will use another color. So, they look like that. If you prefer you can use minus 180 degree instead of 180. So, we call this region low-frequency region especially this region we call the stiffness controlled region because over here the stiffness, control the whole system, over here when we have high frequency region mass control region. So, if I demonstrate this region, a very low frequency, when I excited this, the vibration follows the base displacement like that. Over here and the base excitation is very high. I don't know whether I can excite this system very high but I will try. When I excite the system with high frequency then very little response. So, when you drive a car, when you have this wavy system when drive very fast, the excitation frequency will increase. Then interestingly you feel small vibration. But when you get over here, when I excite the system with the natural frequency the system will getting bigger and bigger and bigger. So, natural frequency is very interesting and vital and essential behavior you have to understand to understand vibration system. You can use this behavior to get more energy or if you understand very well about natural frequency behavior, you can either control your vibration system putting your system into stiffness controlled region or putting your system into mass controlled region so that you can obtain what do you want to get in practical application. So, in summary you can use this system to for example energy harvest. Example, obviously you want to get a resonance behavior of your system so that you get more energy from somewhere else.
