 So, in this lecture, we are going to see how to obtain response of a single degree of freedom system subject to harmonic loading, we will first start with undamped system and then find out the solution of the differential equation and then we are going to look into damped system. So, let us get started. So, till now, basically what we have studied on this response of single degree of freedom system, let me draw that chart. So, let us say we have a single degree of freedom system. Now, in terms of the response of a single degree of freedom system, we could have free vibration or we could have forced vibration. Now, in free vibration, it can be characterized or categorized as undamped free vibration and then there would be a damped free vibration. And for the forced vibration, it can be categorized based on what kind of force that is being applied on the single degree of freedom system. So, we could have for example, harmonic or periodic, harmonic or periodic excitation, alright. And then we could have say arbitrary excitation, for example, it could be a pulse, ok. So, pulse excitation like sine pulse or triangular pulse or step function like that or we could also have random excitation. Example could be seismic excitation like you know and Wendell others, ok. Now, we have already studied the free vibration of damped and undamped systems, ok. So, today what we are going to start, we are going to see how to obtain the response of a single degree of freedom system subject to forced vibration and more specifically in this chapter, we are going to focus on harmonic or periodic excitation, ok. So, let us start with that. Now, in terms of periodic excitation, any excitation can be characterized as periodic if or any function can be characterized as periodic. If I can write as f of t is equal to t plus some integral number times t naught, where t naught is the period of function f, ok. So, and like you know j is in the integer, ok. So, it could be I mean it could be from minus infinity to plus infinity all the integers. So, let us say I write it as minus 3, minus 2, 1, 0, 1, 2, 3 and so on, ok. So, if any system can be written like this, it would be characterized as periodic function, ok. And harmonic functions are, ok, functions which can be written as some constant times sin or cos, ok. So, basically if I have functions like something times sin, ok or something times cos, ok. These are called harmonic functions. Now, as you would have noticed, all harmonic functions are periodic. However, not all the periodic functions are harmonic, ok. Now, we will start with harmonic excitation. So, one might ask a question that what is the uses of studying the response of a single degree of freedom system subject to harmonic excitation. In reality, forces are not always harmonic and in reality, forces are not always single degree of freedom system. But the idea is that we are going to start with simple understanding, ok. And then we are going to build on that platform to understand more complicated systems subject to more complex excitations, ok. So, basically the idea is that if you understand the response of single degree of freedom system to harmonic excitation, it would assist you in understanding or provide you insight how the system would behave to other type of excitation, ok. So, I mean if you know like you know Fourier transformation, if you remember from your mathematics class, you know that any periodic function, ok. So, let us say I have any function f of t that I have written above, ok. If it is a periodic function, it can be written as sum of many harmonic excitation, ok. So, we are going to start with harmonic excitation. And once we understand the behavior subject to harmonic excitation, it would also give us insight into understanding to any any type of periodic functions, ok. So, we know that any periodic function f of t can be expressed as sum of these terms, right. This is from your mathematics class, ok. So, I can write this as simply, ok. So, if you understand the response to harmonic excitation, ok, then there are basically, if the excitation reality is being applied as in harmonic excitation, for example, a few examples could be like in a machine foundation. So, you could have machine foundation where the load is applied as you know harmonic excitation or you could have any unbalanced rotatory load. So, any unbalanced and we are going to model that actually in this chapter, these can be, these can be directly modeled as harmonic excitation, ok. And even if it is a periodic excitation, but not harmonic, I could still obtain the response as sum of several harmonic excitation. So, if we understand the response to simple harmonic excitation, ok, then I would be, I would be equipped with like you know knowledge to interpret basically response to any other periodic function as well. And like you know same goes for let us say earthquake as well. So, earthquake in reality might be a random excitation, ok. So, earthquake like you know it is composed of several frequencies, ok. And like you know you will see that later when you will get into that. If the power spectral density, ok, there is a term called power spectral density which basically defines how much of energy is actually so situated at each of these frequencies. So, let us say this is my ground excitation and I am representing it with some random function which is neither periodic, neither harmonic. Even this has basically distribution of these frequencies here, ok. And depending upon which frequency is predominant, I could have, ok, like this up to infinity, ok. And this is basically power spectral density and this is frequency. So, if I understand the response of a single degree of freedom system to individual excitation frequency, right. And then it would even help me in basically analyzing or interpreting the behavior of the structure subject to earthquake excitation as well, ok. So, I hope this provides a background that why we are doing like you know response to harmonic excitation first and how to, how it would be useful in subsequent chapters, ok. So, in terms of harmonic excitation, we are first going to start with an undamped system and then we are going to go to a damped system, ok. So, first let us do undamped harmonic excitation. So, if it is undamped, I know that my damping term in the equation of motion would be 0 for the single degree of freedom system. So, we know that our equation of motion is what the general equation of motion is this. Now, if the damping is 0, I can simply delete this term and then I would be left with, ok, k is equal to p t. Now, any harmonic excitation p t can be represented as either sin omega t or cos omega t, ok. So, first we would be doing sin omega t and the response subject to cos omega t would also not differ by much, ok. So, this excitation, which I am representing as, ok, p t equal to p naught times sin omega t, p naught is the force amplitude of the force that is being applied, ok. So, basically it is the, whatever the force that you are applying, this is the peak value of that force, ok. So, this is p naught is the peak value or amplitude of the applied force and note here, now I have an additional frequency omega, ok. And this is different from the natural frequency of the system which was omega n, right. And we saw that my omega n for this system, right, I could simply get it as under root k by m, correct. Now, omega is different from this omega n, omega is basically the applied frequency of the force, ok, of the harmonic force that is there and it is called excitation frequency or the forcing frequency. So, we would be referring it further as either excitation frequency or forcing frequency, ok. And if you plot this function p naught sin omega t, you know that it would look like a simply sinusoidal function, ok, which let us say starts here, ok. So, I can plot it like that. So, this is my applied force, ok. And this is, sorry, this is time here, ok. This is the forcing time period or excitation time period which I can write it as 2 pi by omega, ok. So, now basically my differential equation becomes, ok, my differential equation becomes p naught sin omega t. And now I need to solve this differential equation. So, we are going to employ the same techniques that we have used in previous chapters to solve this second order linear differential equation, ok. So, how did we solve it? Basically, we said that the total response can be represented as sum of particular solution and then sum of complementary solution. And this complementary solution had two unknown constant, ok. So, this particular solution is any unique solution that can satisfy this equation, right. And complementary solution was the general solution to the homogeneous part of this equation. So, if you, in this equation here, if you set the right hand side equal to 0 and then whatever the solution you got, that was basically the complementary homogeneous solution, ok. And it had two unknown constants and those unknown constants were determined using initial conditions, ok. So, let us assume our initial conditions are given in terms of u 0 and u dot 0. So, initial displacement and velocity, ok. Now, let us see how do we solve this. So, basically in terms of particular solution here, I would write this as, let us say I want to get any unique solution that satisfy this equation, alright. Now, what I see here on the left hand side of this differential equation, I have a term which is u of t or up of t and then a double differential of term, right. And then sum of that is actually equal to some sine function times some constant, ok. Now, let us say if I assume this u p of t as some constant times sine omega t, knowing that the sum of term u p of t and then the double differentiation of that term would be again some constant time sine t, right. Keeping that in mind, I have selected this as particular solution. The only thing that need to be found out here that what is the value of c. So, if I substitute this here, I can write it as this m omega square c u p of t or let me just write instead of u p, here it would be sine of omega t, ok. And that should be, sorry, there is another term here k times c sine omega t and that should be equal to p naught of sine omega t. So, if we compare the coefficient of sine omega t, right, we get as the value of c as p naught divided by k minus m omega square and I can further rearrange the terms here to get p naught by k, ok. And this would be omega square divided by k by m, ok. If you arrange it, we will get it like that and I know that k by m is nothing but omega n square, ok. So, that I am going to substitute here to get this as omega by omega n square, ok. So, the particular solution I have obtained as this term here, all right. Now, the second thing that remains is the complementary solution which is nothing but solution to this equation here. And that we saw in previous chapters can be directly written as some constant times cos of omega n t plus another constant times sine of omega n t. So, I can write the total solution as particular solution plus the complementary solution, ok. And this can be written as this is the particular solution and then a cos omega n t plus b sine omega n t. Now, I have two unknown constants here a and b which can be found out by substituting at t equal to 0 initial displacement at the initial velocity. So, you will need to differentiate it and then substitute again t equal to 0. If you do that, you will see after substituting these conditions u equal to u of 0 and then u dot of t equal to u of 0 at t equal to 0, you obtain the final solution as u of t should be equal to u of 0 cos omega n t plus u dot of 0 and then this term here sine omega n t and then I have the particular solution which is sine omega t, ok. So, this is the final solution for undamped vibration subject to harmonic excitation which is P naught sine omega t, ok. Now, what you see here that one thing I am important thing to note here we have in there is one not two frequencies, right. Let me just write it here again. We have two frequencies. First frequency is basically the natural frequency of the system and then the second frequency is actually the applied frequency of the force or the excitation frequency, ok. So, one question might arise if you try to plot this system at what frequency would the system vibrate? Would it vibrate at omega n t or would it vibrate at omega or the applied frequency? You know and then the question also becomes that what do these frequency actually mean? What is the what is the physical significance of this frequency? So, let us draw each term that we have here, ok and then see how do they look like, ok. So, let me say I am trying to draw here the displacement. I am not going to say if it is the total displacement of if it is like one of these displacement. Let me first just say that it is a displacement quantity, ok, right. Now, first I am going to draw this term here, ok. Then I am going to plot this term here, ok. So, this is by let me just call it u 1. Let me call it as u 2, ok. So, if you plot u 1 what you are going to see? It vibrates at its own frequency omega n which is the natural frequency of the system and let us say it looks like something like this with some initial conditions. Remember that I do not have any damping. So, amplitude is going to remain constant here, ok and it will keep on vibrating like this with time period equal to 2 pi by omega n. Now, I have another frequency, another term the sin omega theta, ok and let us see let us plot that and see what do we get, ok. So, let us say it looks like something like this. Now, which one would look like which plot for example, you know you might argue that why it is why the first u 1 looks like this and why not the other way? Well, it depends on the relative ratio of the frequency omega by omega n, ok. So, for just this case just for demonstration I am assuming that let us say it looks like this. So, let us say this is my u 1 and this is my u 2. So, these are the two components, ok of the response that we get subject to the harmonic excitation of single degree of freedom system, ok. It is important to note here this u 1 is basically. So, we can say that two components of the vibration response the component 1 is basically vibrates at omega n of the natural frequency of the system, ok and u 2 basically it vibrates with the whatever the applied frequency is, ok. So, it vibrates at the applied frequency, ok. This response is called force response or steady state response and we are going to come back to that why do we call it like that or steady state response and this is called transient response, ok. So, for this let us say I try to so I have drawn like you know u 1 and I have drawn u 2. So, if I try to again draw the total response which is u equal to u 1 plus u 2, ok it would look like something like this, ok. It is going to look like or let me just first draw it then we are going to. So, it is going to look like something like this, this is not to scale of course, ok. You can perhaps plot this kind of graph using MATLAB. So, this is the total response, ok. So, this is u 1 plus u 2 state transient response plus steady state response which is basically some of these two response. So, what it is basically doing? It is taking the transient response which is u 1 and then it is vibrating along this steady state response. So, the sum of two function would look like this. This is basically the total response of the system, ok. So, there are certain characteristics of each of these response, ok. If you look at the transient response or let me just write down the expression for ut again the total force or sorry the total displacement, ok. We had this expression u of 0 cos of omega n t and then we had sin omega n t, ok and then we had this term here which was sin omega t, ok. So, this is sin of omega t. So, as I said this is the transient response and this is the forced or more commonly it is called steady state response. And these terms are better explained when we will get into like you know damped forced vibration. However, note here the transient response depends on initial conditions, right. It depends on u of 0 it depends of u dot of 0. However, my force or steady state response is it does not depend on initial condition, ok. And that you can easily imagine. For example, consider let us say or not like that let us draw the spring mass representation, ok. If you apply a force which is harmonic force, right, whether there is initial condition or not, whether there is no initial velocity, ok and no initial displacement is still as long as you have applied force, ok. The body would have some response, ok and that is the steady state response, ok due to this force, all right. Now, for the transient response it depends on the initial condition. However, even if the initial conditions are 0 this force P naught would still provide some response to the system that would contribute or that would oscillate at the frequency omega n. And how much of that response would be the total response contribute to the total response? Well, it depends it would depend on the frequency omega by omega n which we would see later, ok. Now, in reality what happens? So, if you look at here you know the transient response it seems that it would remain constant, ok and it would not decrease with time. So, if you look at here let me go back to this graph here. So, if you look at u 1 the amplitude actually remains constant, ok and it does not change with time. However, in reality all the system would have certain amount of damping and what that damping actually does although not evident from this undamped the equation from the undamped of an undamped vibration. The damping would reduce this transient response to 0 after sufficient amount of time. However, my force response of the steady response due to applied force would still remain and that is why it is called transient. Transient means something that is between a state or a changing state, ok. So, initially system was let us say it was like you know with some initial condition, ok and then finally it achieves a steady state which is the force response, ok. So, in between whatever the response that system has is called the transient response which actually dyes out in a damped system, but you cannot evidently see here. So, when we will do the damped system we will see that this response actually dyes down. Now, in dynamics we are mostly concerned in like in most of the cases whatever is the force steady state response, force response or the steady state response. So, what we have here because in reality all the system would have some damping, ok and the effect of applied force is measured in terms of steady state response. So, we are going to focus right now we are going to neglect this not neglect this, but we are going to turn our focus now to steady state response, ok because as we will see for the damped system from there that that their mathematical equation for the ut this transient response actually dyes down, ok. So, let us look at the steady state response, ok and I am going to write is this although as ut, but remember that I am only considering the steady state response part of this, ok because this is what would be important. However, you have to keep in mind initially the system would have some transient response and steady state response, ok, alright. So, look let us look at what is the steady state response I can write my steady state response as this term times sin of omega t, ok. Now, if you remember from this equation of motion let us say this was the original equation of motion correct. Let us say I ignore the dynamic effect in the system. So, I basically ignore this term here, ok. If I ignore this term here can I say the static displacement subject to a force p which is varying with time, ok would be u of t equal to p of t by k which is nothing but p naught by k times psi omega t, ok. Now, you have to understand what do I mean by this, ok. So, instead of saying u of t let me just write it as u of s t saying this is the static displacement, ok. It simply means that there is no mass in the system. So, that is why there is no dynamic effect and if I am applying a sinusoidal force, ok, the deformation is basically whatever the force is applied divided by k, ok. Now, it was very simple if simply a force let us say if a spring is given to you and set like you know force p naught is applied you would simply find out that the displacement is p naught by the spring constant k here. But if the force is varying with time then you at each and every instant your displacement would be simply p t by k if no dynamic effect is considered, ok. So, this is static displacement here is basically, ok whatever the force you are applying the sinusoidal force here if that is applied without any dynamic effect of the mass then what is the displacement, ok. So, it is different from the ramp loading in which we said like you know we are applying a force p naught and the final displacement was basically p naught by k if this was large enough with respect to the time period of the system, ok. In this sinusoidal variation of the force we are saying that mass term is 0. So, the resultant displacement is and basically force divided by k. Now, the significance of this is that if there was no mass in the system and if I have applied a sinusoidal load the peak displacement of that is u of s t naught equal to p naught by k and it is the same p naught by k that I have here. So, I am going to write my u of t or the dynamic displacement as static displacement times this term that I have here times sin omega t. So, you need to understand what is this u of s t times naught means, ok. u s t naught is basically the maximum value of static displacement, ok, for a time varying load, ok. Otherwise you would argue like you know what do you mean by the maximum displacement of static load because if it is a static load whatever the load does is apply that is the divided by k is the maximum load. However, I am saying that if you apply a load which is varying with time and there is no mass then at each instant of time you can divide that load by the displays or the stiffness k to get the displacement at time t but that is still the static displacement because there is a mass in the system and this u s t of naught represents that peak static displacement subject to the sinusoidal or the harmonic excitation, ok. So, I hope that is clear to you, alright. Now, if I have this term here, ok, let us see what does it mean. For example, you see here that the term omega by omega n or the frequency ratio, ok, of the applied frequency divided by the frequency the natural frequency of the system, ok. Depending upon the value of omega by omega n by value of u t would change, ok. For example, let us consider two extreme cases, ok or the two cases let us say when omega or the applied frequency is smaller than omega n then what will happen this denominator term would be positive and u t would have same sign as sign of t. So, u t vary accordingly to whatever the variation of the right hand term, ok. So, u t vary are same as whatever the value of applied force in the same direction. So, if this is then then u t and p t are of same sign, ok and basically it means that they are in phase. However, let us consider the second case when omega is greater than omega n then you will see that denominator is negative and then u t, ok, varies as negative of sign omega t which is again negative of p of t. So, in that case omega n omega t are of opposite sign, ok and it is said out of phase, ok. Physically what it means basically that you apply a force in first case if you apply a force to the right, ok, then your system also moves to the right, ok. And if in the second case it means that if you apply the force to the right the system moves to the left and like you know you might have a doubt well how is that physically possible remember that we are not considering here monotonic load, ok. We are considering here a sinusoidal load, ok. So, depending upon the frequency of omega by omega n once the motion is started and like you know providing this vibrational motion here, ok. It might not be in the same direction at any time instant as the applied force, ok. It might move to so if my force is positive it might be moving to the negative direction and that depends on the ratio of omega by omega n, alright. So, let us see if we can formulate some mathematical expression to actually represent my dynamic displacement. So, remember ut here is my dynamic steady state displacement and although in like you know in further lectures or like in a further when we discuss we are going to not I am not going to always refer it as a steady state displacement, but it should be understood, ok. You should understand that I am talking about a steady state displacement but just looking at the expression of ut, ok. So, this is the dynamic steady state displacement, ok. Now can you imagine if I again have a dynamic steady state displacement which is varying as a function of sin of omega t it would again have some peak value. So, it would have some basically amplitude. Can I write my ut as u naught which is the dynamic amplitude or the dynamic displacement amplitude, ok times sin omega t, ok where I write it as u of s t times r d times sin omega t, ok. Now what is the r d here? If you compare this equation here and this equation here, basically r d is this expression 1 minus omega by omega n square, ok. Now if you plot this expression, ok this 1 minus omega by omega n square how does it look like, ok. We talked about the phase. Let us see how does it look like, ok. So, I know that when omega by omega n or the horizontal let me just again draw here, first let me draw here, ok. This is my frequency ratio. So, horizontal direction is basically omega by omega n and the vertical axis is 1 minus omega by omega n square. Now I know that at a very small value of omega by omega n this is equal to 1. So, let us say this is 1 here, this is 2, this is 0 minus 1 minus 2 and so on. So, it starts from 1, ok. I do not know what happens to it as I increase the value of omega by omega n, ok. So, what happens? The value of the ratio r d actually increases because as you increase the omega by omega n, the denominator decreases. So, the overall r d which is 1 inverse of that value would increase. So, it increases something like that. I do not know how it would increase. Other extreme is that by omega by omega n is a very large value. So, it is infinity in that case r d would be equal to 0, ok. So, it will start somewhere from here and as you decrease the value of omega by omega n, then your denominator would actually increase, but in the negative direction, ok. And can I say when omega by omega n is equal to 1, then the value becomes unbounded, ok. So, it becomes unbounded here and unbounded here on the negative side. So, the graph would look like something like this. This is of this specific quantity, ok. And the value actually depends on the frequency ratio, ok. Let us say this is 0, 1, 2 and so on, ok. So, it depends on the frequency ratio omega by omega n, ok. And whether it is positive, this expression or whether it is negative, that also depends on omega by omega. So, what I would do? I would further write this term here or let me just actually delete it here, ok. I do not need that to plot this, ok. I am just trying to plot here u naught, u s t of naught times this expression right here, ok. So, now what I am going to do here? I am going to write my u of t as u s t times r d. Now, I am going to use that factor r d times sin omega t and I am adding a new term here called phi, ok. Now, this phi is called the phase angle, ok. And r d is nothing but the absolute magnitude. So, I have originally this term, but r d is actually the absolute magnitude of this term here. So, r d would always be positive. This is the mod of this function right here, ok. If that is there, then you would ask me, well, then how does the sin of sin t or omega sin of omega t is actually taken into consideration? Because initially it was coming due to this term. Precisely for that I have this angle phi here. Remember we said that when omega by omega n is less than 1, then it is where it varies, the u t varies as a negative of, sorry, it varies as the same positive as sin of omega t and if it is greater than omega by omega, if it is greater than 1, then u varies as negative of that. So, take to take that sin into consideration, I am writing this term here and r d I have taken magnitude of that and all the sin is now being taken care of this phi or the phase angle, ok. So, this is this term here and the phase angle now I am defining phi is equal to 0 for omega less than omega n and it is equal to omega or it is equal to 180 degree for omega greater than omega n, which makes sense. If you look at it, right, if you look at it here, I would again have the same expression, ok. So, let me first write this as dynamic displacement times sin omega t minus phi. Now, remember that now this is always positive because r d I am considering as magnitude, ok. Now, if omega by omega n is less than 1, which is basically this condition here, my phi is equal to 0. So, u t would become, ok, u naught of sin omega t minus 0, which is sin omega t. However, if omega is greater than omega, then u t would become u naught times sin omega t minus phi, which basically becomes minus of sin omega t. So, I will have a negative sin here, ok. So, this is in phase and this is out of phase and when I say in phase and out of phase is basically the displacement u t with respect to the applied force P of t, ok. Once I have this information, let us plot, ok, let us plot r d and let us plot basically the phi, the phase angle and this then see how do they look like, ok. I know that both of them depends on the frequency ratio. So, r d is nothing but now have to take the magnitude of or the absolute modulus of this function. So, that would mean that this function here would simply invert to the positive side of the x axis, ok. So, let us let us try to draw that. So, I have this r d, ok. So, let me just write it here. So, r d is here, ok. I am plotting r d here and then below this I am also plotting by phase angle, which is phi and both these axis are basically omega by omega n, ok. Value of 0 by 1, let us say 2, 3, 4 and so on, ok. Similarly, 0, 1, 2, 3, 4 and so on and the vertical axis in this case for the phi this is 0 degree, 90 degree, 180 degree and here it is 0, this is the value of 1, then 2, then 3 and so on, ok. So, if I simply take the mod of this function, I again it starts at 1 for a very small value of omega by omega n, ok. So, r d let me just write down the expression anyway for your convenience. So, omega by omega n when it is very small it is equal to 1 and it is unbounded at value of omega by omega n equal to 1, ok. So, it will start at 1 and then it will go like this. And the second case, ok, what happens as you increase the value of omega by omega n it starts from here it will start at value of omega by omega 1 equal to 1 approaching to infinity and then it would decrease as you start and at some point it is going to cross the x axis and then go to 0 for a very large value of omega by omega n, ok. Now, compared to that I have the phase angle, ok. The phase angle which is if omega by omega n is less than equal to 1 it is 0 and if it is greater than 1 it is 180 degree, ok. So, we have talked a lot about r d and phi. Let us see what is the physical significance of these two quantities, ok. Now, as we said r d is nothing but remember we said that my u naught is us t of 0 times r d, ok. r d is nothing but the dynamic amplitude divided by the static amplitude, ok. So, if the force had been applied statically whatever the amplitude of that function is let us say p naught by k and if the system had mass and then the force is being applied, ok whatever the dynamic displacement that you get the ratio of both two is defined as r d and it is called response or displacement or deformation response factor, ok. So, it is called deformation response factor, alright and we will see later that there are different type of response factors, ok. Right now we are only dealing with deformation response factor which represents what is the amplification in the response, amplification or reduction whatever happens with respect to the static system, ok. r d is a measure of that amplification or reduction, alright and phi represents that if the force p t is applied, if the force p t is applied, ok. Then with respect to the applied force p t whether my response or the u is in with in phase or whether it is out of phase. So, it represents phase of u t with respect to applied force. So, if my force p t is moving right whether my displacement is also in the right direction or whether it is opposite to the applied force in the left direction. So, the phi represents that, ok. Now this plot of r d or the displacement response factor here it gives me several useful information in terms of dynamic response because that is what we are trying to study here, right. So, basically if I have a system and a harmonic load is applied on that based on the frequency ratio omega by omega n I would see what happens to my r d, ok which is the displacement response factor. So, let us come to this graph here, ok. If the value of omega by omega n is very very small I do not see any amplification, right. My r d is equal to 1 that means u naught is equal to u s t of naught. So, means dynamic is equal to starting. So, let us consider first case that omega by omega n is very small. It means that omega is very small the applied frequency is very small. When the applied frequency is very small it means that it is a slowly varying load, ok. It is a slowly varying load and in that case r d is equal to 1 or dynamic displacement is equal to static displacement, ok. So, if the load is applied very slowly we do not see any amplification in the system with respect to the static displacement, ok and that that should be intuitive to you, is not it. I mean if you are applying load at very very small frequency then the your system would move, ok in tandem with whatever the force you are applying without having the much dynamic effect of the mass and that is why your dynamic displacement is equal to the static displacement, ok. Now, in the second case when omega by omega n is very very large that means a rapidly applying force or rapidly varying force. In that case r d is equal to 0 or that means your displacement dynamic displacement is equal to 0 it means it implies that for a rapidly varying force, ok you do not see any response in the system, ok. So, if you try to imagine this basically what do you have system on which you are applying a very like you know fast moving load this system would not even actually respond to this, ok that is what basically it means the displacement would be equal to 0. Now, if third case omega by omega n, ok so these were the two extreme cases if it is greater than equal to root 2, ok which basically corresponds to let us say this point here root 2, ok. Then my r d starts to become smaller than 1, ok my r d is so that I do not get any amplification in the response, ok. So, the dynamic behavior actually does not lead to amplification but actually leads to reduction in the response and as we will see later this is very useful in some design of some systems. For example, let us say I have a system, ok a static system a building just sitting there would I not be happy that if I apply some force or the frequency on my structure the response become actually less than the response it would have if the load had been applied statically I would be very happy, right. So, that depends on the applied frequency and this actually property is utilized in a vibration isolation, ok we will come back at later. So, my dynamic for r d omega by omega n greater than 1 dynamic is actually less than static, ok and of course then the case where omega by omega n is equal to 1 then r d approaches to infinity, ok. So, the response becomes unbounded, ok the response becomes unbounded, all right. So, these are the four typical cases which you might experience in real life based on how the frequency of the applied force varies with respect to frequency of the system and that decides what is the response of the system, ok. Now, we talked about this, this case here when omega by omega n is equal to 1 the response becomes unbounded we are not going to define the frequency or the applied frequency which is called resonance frequency, ok or the frequency is actually resonant frequency this phenomena, ok. So, this would be the forcing frequency or excitation frequency, forcing or excitation frequency at which the r d or the displacement response factor, ok becomes maximum, ok. So, whatever the value of applied frequency at which it becomes maximum is called the resonant frequency and this phenomena is actually called resonance. Many of you might be aware with this, ok this phenomena here. Now, in this case what happens so the resonance is basically maximum happens when omega by omega n as we saw here is actually equal to 1. So, applied frequency is actually equal to the natural frequency of the system so my r d become maximum. Now, when this happens at this point actually the solution that we have obtained for ut is not valid anymore, ok and we actually obtain a new solution, ok. So, basically what I am saying in this case the solution that we had obtained for our equation of motion this is not valid anymore, ok only at this point, ok it is valid it was valid at all other point. So, we developed a new solution, ok when this happens, ok for this case what do we do basically remember our particular solution initially we had assumed c times sin omega t. Now, omega is equal to omega n so this is c sin omega n t, ok and our homogeneous solution or the complementary solution was a sin omega n t plus b sin sorry first term should be cos here. So, let me just write it again, ok. Now, if you look at this carefully this is not a unique solution anymore because the sin omega n t term because your omega is now equal to omega n is a part of this term here initially because they were different omega was not equal to omega n, ok this was a unique solution but now because I have an omega same omega n term here sin omega n term and then it is part of this complementary solution. So, for these type of cases, ok what we do the solution is obtained by assuming u of p t equal to c t the time constant t, ok as this and if you substitute this in this equation you will obtain as u of particular solution as and you can like you know do that calculation yourself can obtain that as cos omega n t, ok and the total solution you can write it as a cos omega n t plus b sin omega n t p naught omega n t cos omega n t. So, you can see with omega is equal to omega n the solution does not suddenly become or the response does not suddenly become unbounded, ok. If you solve this or if you obtain the response for the initial condition u of 0 and u dot of 0 you will get some expression, ok and what I want to do let us say for both these parameters equal to 0 you can get the response u t as minus p naught by k cos omega n t and sin omega n t, right and if you try to plot this function and you can use any of the softwares like in order you can use MATLAB to plot it it would look like something like this, ok it would look like some let me just plot it here it would be easier to do so. So, my response does become unbounded but I cannot represent my solution using the previous equation I have to derive the new solution for omega by omega n equal to 1, ok because of the reason that we just discussed here now that is not a particular solution is not a unique solution anymore and what happens over the time the amplitude keep on increasing if you plot it the amplitude will keep on increasing and it would become unbounded it does not become unbounded suddenly. Now try to imagine this scenario in reality, ok in reality assuming that it is an undamped system which is in self like you know assumption that might not be true all the system would have some amount of damping but let us say I do have a system which has very small value of damping that can be neglected, ok even given that fact what will happen as the real system will start to vibrate and start the response start to become unbounded what will happen if the displacement exceeds a certain value for example if it is a brittle structure like concrete it will start to break, right concrete will start to get damaged and it will start to break at certain value so you do not get actually infinite response but what happens for the brittle system as the concrete gets damaged its stiffness changes so your omega n actually changes, ok so although initially you had applied a response sin omega t for which initially omega, ok so applied frequency become initial value of omega n t it started the resonance the system starts to get damaged so this omega n changes, ok so let us say this is omega n, ok after sometime f so now this omega is different from the omega n the natural frequency of the system so basically this system would come out of the resonance because the frequency of the system itself like you know changes due to unbounded deformation once it starts to develop other solution like in other situation could be if it is not brittle but let us say it is very ductile like for example steel bailing or some steel structure then it would yield at certain point, ok it won't so again the stiffness would change, ok and the forcing frequency again would not be equal to the applied frequency, ok and then this basically solution or it would come out of resonance and again we can go back to the original solution, ok so I hope this discussion gave you some idea what happens in an undamped harmonic motion of a single degree of freedom system where basically undamped system is an assumption in reality all the system would have some damping so in next lecture we are going to see what happens when the system has damping we are going to look into mathematical formulation and then the physical interpretation of the result, ok so this lecture we are going to conclude it here, all right thank you