 Welcome back everyone. We are going to continue from our last lecture in which we started obtaining the response of a single degree of freedom system subject to harmonic load. Today, we are going to focus on damped system. And then see how the response of a damped single degree of freedom system subject to harmonic load. How does it differ from an undamped system? Okay. In previous class, we discussed what is the response of a undamped system subject to harmonic excitation. Okay. But we said that in reality or for all practical utility, all the system would have certain amount of damping in them. Okay. So, today what we are going to discuss is damped or maybe I should say viscously damped system, damped SDUF subject to harmonic excitation. Okay. And as we have been doing like you know till now, what I am going to do, I am going to write down my equation of motion for this. So, what do I get here as MU dot plus there is a damping term now CU dot plus stiffness from KU dot that should be equal to the applied force which is harmonic in this case, which can be represented as P naught of P naught time sin of omega t. Okay. So, now our goal is to obtain the response or solve this differential equation and get the expression for U of t and then physically interpret all the results that we get from it for different values of the parameter. Okay. Now, we already know that if we have a linear differential equation, second order differential equation, the total response can be written as sum of particular solution plus complementary solution where particular solution is any unique solution that satisfies this equation and complementary solution is the solution to the homogeneous part of this equation by setting the right hand side equal to 0. Okay. Now, if you look at this carefully, let us first look into the particular solution of this. So, what do we basically want is the expression for U U P of t. Okay. That should satisfy this equation that I have here. Okay. So, let me just write it as sin omega t. Right. Now, when the system was undamped and this term was not there, we simply said that because the left hand side correspond like you know contain only U P and double differential of U P. Okay. Which is equal to some constant times sin omega t. The solution would be in the form of some constant times sin omega t because if you take sin omega term and then double differentiate it, then it would have some value times again the sin omega term. However, in this case, I have single differentiation of U P as well. So, I cannot use that solution. In this case, I would have to assume now that my U P would be some constant times sin omega t plus another constant times cos of omega t. Okay. And our goal is to find out the values of this constant C and D. All right. So, if you take this solution and you substitute in this equation here. Okay. Basically, you would get something like an expression in terms of C and D times sin omega t plus another expression in terms of C and D and of course, zeta m all those terms would be there. Okay. And then it would be cos omega t and then I would get this should be equal to P not of sin omega t. Okay. Now, this would give me. So, as I said it would be some term plus C plus some terms times D similarly here some terms times C plus some terms in D. Okay. So, if you compare the coefficient of sin omega t and cos omega t on both side of the equality, you will get two equations in terms of C and D and then you can solve it to obtain the expression for C and D and I would leave that for you to do. Okay. I am just going to write here the final expression for the value of the constant C and D. Okay. So, I can write my C as P not by K. Okay. 1 minus omega by omega n square and in the denominator I will have this term here plus 2 zeta omega by omega n square. Similarly, the value of t I can write it as P not by K but the numerator is now going to change. I would have this term here and denominator is going to remain the same what I had for C. So, let me just write it here again anyway. Alright. So, once you get the values of C and D, you have your particular solution. Now, you want your homogeneous solution. Okay. So, you want what would be the uc of t that satisfies what equation for homogeneous for complementary solution or homogeneous solution you need to set up the set the right hand side of the differential equation equal to 0. Okay. So, I can write here as C and then, okay. And as I have discussed in previous lectures, the solution for this equation would be of the form uc of t would be zeta omega nt. Okay, which is to the power exponential power minus zeta omega nt and then I would have a cos omega d term here which is the damped frequency and then I would have sin omega dt. Okay. So, the final expression for ut which is sum of both solution, okay, can be written as sum of up of t and then uc of t. So, let me write it this as okay. And then I have the particular solution which is C sin omega t plus d cos omega t knowing that C and D have already been evaluated. Okay. So, we know the values of constant C and D. What we do not know are the constants A and B, you know. And if the initial conditions are given to you, for example, if u of 0 and velocity u dot of 0, if they are given to you, then you can substitute in this expression here to obtain the constant A and B. All right. Now, again, like the previous case of an undamped harmonic motion, consider here, we have two frequencies now here. I have omega d and then I have omega which is the forcing or excitation frequency. So, the question becomes at what frequency will the resultant response be whether it would be at omega d or omega. Now, what I see here, okay, let me just delete this, what I see here is that I have now an exponential term which is e to the power negative power zeta omega nt and this term is actually decreasing with time. Okay. I mean this whole term is decreasing with time. So, after a certain amount of time, this term would become 0. Okay. And as we said, the term that vibrates at the frequency of the structure which in this case is omega d is called the transient solution. Now, the wording or the name transient was not very clear when you are discussing undamped harmonic motion. However, in this case, now you can see why we call it transient solution. If you look at here, after some time, okay, and the time like, you know, it depends, you know, on the value of what is zeta and omega n, but after some time, we know for sure that this term would become almost negligible and equal to 0. So, I would be only remaining with this term, the second term here. Okay. So, this is called transient solution, okay, or the transient vibration and the second part which is a particular solution, it is called steady state. Okay. Because steady state is reached, okay, it is a state that is steady after a certain amount of time. Okay. All these vibrations actually died out and what you are left with is this steady state solution. Okay. So, like we did for undamped harmonic motion, let us again try to plot this function. Okay. So, now we know that our total vibration consists of two type of vibration. One is the transient response, another one is the steady state response. Transient response is basically damped, free vibration. Okay. And the steady state is like forced part of the forced vibration at frequency, which is equal to the forcing frequency. Okay. So, first, let me just draw the both responses here. Okay. So, if I draw the first part, let us call this u1 and let us call this u2. Okay. So, if you try to draw u1 depending upon the initial condition or whatever, it would look like something like damped free vibration. Okay. So, it will, the amplitude will keep on decreasing and after sufficient amount of time, it will go to zero. Okay. But then I have also the other part which is the steady state and I can see that my steady state is actually, it is not decreasing in amplitude. Okay. It is remaining constant. All right. So, let us see. It starts at this point and then it has certain amplitude, something like that. So, this is steady state and this is transient. Okay. So, u1 here and u2 here. So, the total response is basically the sum of these two responses. Okay. The total response is actually some of these two responses and how I can write, basically draw it, it would look like something like this. Okay. Pardon my drawing. I will try to do as best as I can. So, it would look like something like this. Okay. So, what is basically happening? I am summing up these two responses. So, initially you can see that the free vibration is about the steady state response with the time this, these vibrations about the steady state vibration actually died out. The transient vibration actually died out and then the total response. So, this is the total response. This is also the total response. Okay. After sufficient amount of time, total response actually converges to steady state response. Okay. And how fast does it converge? Well, that depends on the value of this term here, right, because that is what is going to decide that how fast the exponential term is decaying. So, how fast is the steady, how fast is the initial response converges to the steady state response or the total response converges to steady state response that depends on the value of damping and the frequency, natural frequency of the system. Okay. All right. So, this is why, so now I hope this definition, why is it called transient and the steady state different part of solution should be clear to you. Know that the maximum response of a system could still be governed by the, during the transient phase. Okay. So, as you can see, my steady state is smaller than the total response and this additional contribution is because of the transient vibration. Okay. However, after certain regular sufficient amount of time, the total response converges to the steady state response and that is why for the future discussion, we would be focusing on the steady state response. All right. So, let us go ahead with our steady state response. Okay. So, now we would be focusing on this term here which is nothing but c sin omega t plus d cos omega t. Okay. But before going to that, we discussed about the concept of resonance in the last lecture and we said that a resonant frequency is the frequency at which Rd or the deformation response factor or you can just simply say that resonant frequency is the frequency at which the response becomes maximum. Okay. And the process is actually called the resonant frequency. So, for that case, it was when omega was equal to omega d. All right. That was the case when the response was becoming maximum and it was becoming, so for undamped case, okay, it was getting unbounded. Right. Okay. Let us see what happens for a damped system. Let us see if we still get a similar kind of behavior. Okay. So, what we want to do for omega is equal to omega n, not omega t. For omega equal to omega 1, what I want to do, I would like to find out the expression for ut and then plot it and then see as the time evolves what happens to the response. Okay. Does it become unbounded? Okay. Or is it unbounded? So, we will see. Okay. Now, let us say for this value of omega by omega n. Okay. Again, let me consider the total solution including the transient solution. What was the total solution that I had written? It was zeta omega nt. I had a cos omega dt plus b sin of omega dt. This was the transient part and then I had the steady straight part which was c sin omega t plus d cos of omega t. Okay. Now, I know the expression for c and d, right. So, the expression for c is basically nothing but p naught by k. Okay. I have 1 minus omega by omega n square divided by the whole term here. Okay. And then d is also similar to this. However, in the numerator I have minus 2 zeta omega by omega n and then I have it in the denominator here. Now, if I substitute omega by omega n in these two expressions, okay, let us see what do we get. Okay. And so, c I would get as because omega n is a numerator. This term becomes 0. So, c would become 0. d would become here I get as p naught by k. Okay. So, let me write p naught by k as u s t naught which is the peak static displacement. Okay. And then in the numerator I would be left with 2 zeta in the denominator. Remember we had term here or let me just write here as well so that there is no confusion. Okay. 1 minus omega by omega n square plus 2 zeta omega by omega n square. So, when I substitute this, I will get this as 2 zeta whole square. So, this I would get d I would get as u s t naught divided by 2 zeta. Okay. Now, if you substitute these in this expression here, okay, and utilize these initial condition let us say u 0 of 0 and u dot of 0 is equal to 0. So, we are assuming that the system was initially at rest. Okay. We can find out the values of a and b. Okay. So, after substituting, I am not going to do all those calculation. Okay. Just giving me the values of a and b. Okay. So, basically you will get a as u s t naught by 2 zeta and the value of b you would get as u s t naught. Okay. Divided by 2 zeta 1 minus zeta square under root. Okay. So, you can substitute these values to the original expression to obtain the value of the expression u t as let me write it here as u s t of naught times 1 by 2 zeta. Okay. And then I have term here cos of omega d t zeta 1 minus 1 zeta square. Okay. And then sin omega d t. This is the transient part of the solution. And then I have another term which is cos omega n t. Alright. So, I basically get this expression here. Okay. Remember that this was the forced vibration, but because omega was equal to omega n, I am writing it like this. Alright. Now, for a small value of damping, let us say damping is smaller than 10% or for like you know, this typically let us say it is 5%. Okay. I can make some assumption and further simplify this expression here. Okay. So, for a small value of damping, small value of damping here. Okay. I can write this expression. I can simplify this expression further. Okay. So, my omega d would become omega n. This term I can neglect. Okay. Because it becomes very small and multiplied by zeta for a small value of damping. So, what I am going to basically left with my ut as us t naught divided by 2 zeta. And then here I would have this term here into cos omega n t. Okay. Now, we have this system. Let us see if we plot this system, how does it look like? Now, can I say that as you increase the value of t, okay. Basically, this term here would reduce to 0. But overall, my ut would never go to 0. Okay. What it would, what would happen? Remember that this is a harmonic function cos n t. And this is basically represents the time varying amplitude of this function ut. Okay. So, as t increases, this term would vanish. And then I would reach a constant amplitude, which would be represented by us t naught by 2 zeta because this term vanish. Of course, there's a negative sign, but doesn't matter because cos omega n t is basically vibrating between plus 1 and minus 1. Okay. So, if I try to plot, let us say here ut divided by us t naught here. Okay. What would basically happen? Let me draw this envelope curve here. Okay. So, this represents the asymptotic line, which is the constant amplitude. So, in this case, it is 1 by 2 zeta. Remember, 1 by 2 zeta is equal to ut by us t naught. Right. So, the amplitude is actually us t naught by 2 zeta. And let us say this is t by t n. So, what will happen? The vibration will start like this. Okay. Because omega is equal to omega n, the amplitude will start to increase. Okay. However, because the system has some damping, what will happen after some time? Okay. After some time, this term here would vanish. And then, due to applied force, I would have a constant amplitude with constant, basically constant amplitude. So, a vibration with constant amplitude. So, the response does not become unbounded if a system has damping. Instead, the maximum value of ut in this case is actually us t naught by 2 zeta. And as you can see, it is very, very sensitive to the value of damping. Okay. So, if the value of damping is very small or let us say bordering to undamped system, let us say 0.01, this ut would be very high. Right. But if the value of damping is, let us say some reasonable value, let us say 5 percent or 10 percent, then my system would still, the response would still be bounded. Now, this is the technique that is typically used in different kind of structural engineering systems. So, different structures, let us say, there is always a possibility that due to applied force and the source of force could be seismic, wind or any other type of force that has different frequencies in them. What happens that if we want to avoid damage to the structure, we include some damping in the structure so that even if there is a resonance, my response is still bounded. Okay. Now, how do we provide damping in the system? Well, there are various ways to do it. We could, like people use added damping devices. So, you could use supplementary damping using viscous dampers. You can use friction dampers. There are different type of dampers, but the idea is to provide damping in the system because we might not be able to predict, always predict, like what kind of frequency the loading might apply, because loading then would have different frequencies and like you know, even if you anticipate that frequencies, this would be the frequencies, it is always to better to take countermeasures and include damping in the system because if that there is damping in the system, then the response would always reduce. Okay. So, this is one way to actually safeguard including damping in the structure is one of the ways in which we can safeguard our structure against resonance. Okay. I hope this is clear to you. Okay. Now, what we are going to do here? We have said, we have discussed the basically the resonance phenomena for a damped system. Now, I am going to come back to again my expression for ut. So, the total expression again let me write it and I like to write it again and again so that you can have a look at it. Sometimes these expressions could be overwhelming. If you look at them again and again, it would help you to conceptualize each and every term. All right. So, let me write it here again. I have a cos omega dt plus b sin omega dt plus then I have the second part of the solution, which is c sin omega t plus t cos omega t. All right. And we said that this response, this transient part of the response which is vibrating at frequency omega t is going to die down. Okay. After some time and our main concern here is the steady state response here and that is what we are going to discuss further. So, let us take the steady state response. Okay. So, the steady state response is c sin omega t plus d cos omega t. Where is the expression for c and d? I have already presented. Okay. Now, you know that this kind of expression can always be written in the form of u0 or before writing it as u0. Let me write it like this. c square plus d square. Okay. And I can write this as sin omega t c divided by c square plus d square plus cos omega t t divided by c square plus d square. Okay. So, I can write this expression as c square plus t square times sin of omega t which is the first angle minus phi. Okay. And what is sin a of minus b? Okay. It is sin a cos b minus cos a sin b. Okay. So, if we compare b to phi and then write it here, can I say my tan phi here would be or just for this like you know let me first write down the first the cos phi. My cos phi is c divided by c square plus t square and my sin phi is minus of t divided by c square plus t square. All right. Okay. So, if I write it further, I can write this thing as u0 times sorry not u0. Let me again write it here as ut of this is u0 sin of omega t minus phi where u0 is the dynamic amplitude. Okay. Which is equal to c square plus t square. Okay. And my tan phi here is minus d by c. So, I can write phi as tan inverse minus d by c. Okay. Now, all you need to do is to substitute the values of the expressions d and c. Okay. And then you would be able to find out the value of u0 and so I know that let me just write here again. My c is nothing but p0 by k or I will just write well I could either write p0 by k or I can just write usd of not which is the static my peak static deformation. Okay. So, this was this term here. And then I again had the square of the same term in the denominator. Okay. And my d was p0 by k to zeta omega by omega n. And then I had the same term that I have in the here. Okay. So, when you substitute these expressions, what you would get as u0 as p0 by k times this expression here which is 1 minus omega by omega n whole square to the power square times 2 zeta omega by omega n square and this under root. Okay. Which I can further write as remember p0 by k is nothing but usd of not times and the this factor this whole factor here I write as rd. Okay. And if you substitute tan phi you will get as okay 2 zeta omega by omega n divided by 1 minus omega by omega n square. Okay. So, what we have derived actually that for a steady state my response can be written as u0 times sin omega t minus phi which is nothing but usd not times rd times sin omega t minus phi where the deformation response factor rd okay deformation or displacement response factor okay rd as this expression right here okay. And phi is called the phase angle which is nothing but tan inverse again square. Okay. So, this basically you need to understand. Okay. So, this is the expression for the ut and the expression for this deformation response factor and the phase angle. Now, let us see what do they mean right. Now, as we have previously discussed rd or from here rd is basically the ratio of the dynamic amplitude to the static amplitude and represents the effect of dynamic behavior of the system on the response of the single degree of freedom system. Okay. And phase angle which is also called a phase lag it represents with respect to the applied force right with respect to the applied force which in this case is pt. What is the lag that is there in ut remember my pt was nothing but sin of omega t okay. However ut was nothing but some constant times sin of omega t minus phi. So, if you try to plot these functions it would simply mean that sin omega t starts at 0 depending upon the value of phi it might start early or it might start late. And this is the difference by which the response would follow the applied force okay. So, it is something like that if you apply the force to the right okay whether your response is to the left and vice versa okay. So, with respect to applied force in which direction is the displacement phi basically gives you information on that okay. So, what I have done here I have considered systems with different values of frequencies ratio okay. I consider let us say 0.5 okay and then I have considered system with omega by omega n is 1 and then 2. And then I have tried to plot the response the dynamic response ut and also the static response ust okay. Not the peak value remember these are the time variation okay. Where ust is nothing but whatever your pt is divided by k without the influence of any mass okay and ut is of course the dynamic displacement so obtained using solving the differential equation okay. So, I have tried to obtain for these value and let us see for these values of frequencies frequency ratio how does the response the static response remember static response because it is simply divided is the force divided by the stiffness it also represents the force response okay. So, they are in completely in phase the static response and the applied force okay. So, let me I have done some calculation and got this okay I will just paste it here for you to see. So, what do you see here basically this is the case where omega by omega n was 0.05 this was the case where it was actually 1 and this was the case where it was actually 2. So, what do you see I mean the blue line is the static one. So, my static deformation or which is also equal to 4 starts at t equal to 0 at 0. However, my dynamic displacement is when it is 0 it is at some negative value okay and it comes to 0 or let us say this represents the equilibrium position here okay after this much of distance okay. So, and this is represented by the phase angle. So, this here is nothing but phi by 2 pi okay how do I get that well I just substitute the value of u0 sin omega t minus phi at which it becomes equal to 0 and that gives me the value of time as phi divided by 2 pi okay I hope that is clear to you okay. Now of course in this case it would not be simply phi divided by 2 pi you would also have the sorry you would also have the term t here the capital T okay but let us assume if this is represented as t by t then it would have been phi divided by 2 pi okay similarly here you can see that here it is lagging just a little bit in the same direction. However, if we consider for omega by omega n we see that when the applied force is 0 okay the dynamic response is actually the negative minimum and when it reaches to its maximum value which is here the dynamic response is actually 0. So, here these responses are lagging by basically 90 degrees okay whatever the response. So, here I can say my remember my phi was nothing but tan inverse 2 zeta omega by omega n 1 minus omega by omega n square. So, when omega by omega 1 is equal to 1 this is actually 90 degrees okay. So, here the phase difference corresponds to a phase angle of 90 degrees okay. Now if you do focus at the third one when omega by omega n starts to increase and becomes a very large value what you would actually see that if you apply the force to the right okay the response would be in the left direction okay and vice versa okay and that is evident from here okay when this is reaches the applied displacement or the applied force reaches to its maximum value my dynamic displacement is in the negative maximum direction okay. So, I hope you understand what does the phase basically means phase is the quantity that is used to show or express the trailing that is there basically between the applied force and the response of the system alright okay. Once that is clear let us put our focus back on the two expressions which were which were for R d okay remember for R d we obtained as R d equal to 1 minus omega by omega n square and then 2 zeta omega by omega n square alright and then 5 was tan inverse okay this expression here okay and what we would like to do remember the most important parameter here is what the frequency ratio. So, like we did for an undamped system what we would like to do here is plot this R d value of R d and the phase angle for different values of frequencies ratio okay and the question might arise why do we do that why what is the important or like you know what are the applications well if I know based on the frequency ratio what is the value of R d and R d represents what it represents the amplification or reduction in the response due to the dynamic behavior of the structure or due to the frequency ratio here right. So, if I know that then I could design my system to have a response in within certain level okay within certain amplification within a certain level or reduction within a certain level okay and similarly for the phase okay. So, let us see how this one looks like okay now if you remember from the previous lecture for an undamped system okay and what maybe let me just draw it in the side. So, we had something like this alright. So, this is let us say R d which is the displacement response factor or deformation response factor and the horizontal axis was nothing but the frequencies ratio frequency ratio okay and just below this we will also draw our phase angle okay. So, we know that if omega by omega n is very small for example omega by omega n is less less than equal to 1 or let us say close to 0 my R d value actually approaches to 1 right you can see from this expression if omega by omega n is equal to 0 then R d becomes 1 okay. Now when omega by omega n becomes very large like in this case here what I am seeing here my R d becomes 0 okay. So, I can say that starts to increase somewhere here and then for large value it starts to it will converge at some point here and for undamped system we saw that the value was actually unbounded at omega by omega n equal to 1. However if you look at in this case when omega by omega n equal to 1 my R d is now it is not infinity anymore it is not unbounded anymore it is actually 1 by 2 zeta you can substitute here. So, I would get different expression for R d depending upon the value of zeta okay for a value of for an undamped system where zeta is 0 this actually becomes unbounded like something like this okay and for different value of damping I would obtain different curves here okay all of them converging to value of 0 at very large value of omega by omega n. So, let us say this is a very small value of damping 0.01 let us say this is 0.02 this is perhaps 0.05 maybe 0.07 and let us say this is zeta equal to 1 okay. So, as you can see as you increase the damping in the system the R d value actually decreases at all frequencies okay and what happens to the phase angle? So, let us see here okay phase angle the expression remember let me write that again here was 2 zeta omega by omega n okay. So, when zeta equal to very small value or let us say it is 0 we saw that for undamped case it was let me first annotate the accesses. So, this is 0 90 180 this is phase angle this is frequency ratio 0 1 2 and 3. So, for undamped case we saw that it was value for value omega by omega n less than 1 it was 0 and then for greater than 1 it was 180 okay. Now, if we have some damping it would look like something like this okay and as you increase damping the this one the slope would become more gradual okay. So, this is the increasing value of damping okay okay something like this. So, this is increasing value of damping all right okay. So, the way we utilize all these results the design of like you know different type of systems for example if somebody says to you okay that within the limitation of the properties of a system you want your response or dynamic amplification okay not more than let us say if 0 this is let us say 1 2 3 4 okay whatever here okay they want that well the system or the support can sustain only the deformation up to this okay. So, you would ideally like to design a system for which you can appropriately provide damping so that the response is within this level okay or you can design your system based on the frequency okay so that it always falls in this zone. Similarly, if somebody says to you well is it possible that I can reduce the response of the system subject to less than 1. So, the dynamics is actually a dynamic response is actually smaller than the static well you could say that you are going to design a system so you are going to provide either omega n which makes it this value of rd less than equal to 1 here okay. So, these charts are actually very useful in design understand first understanding the behavior of a dynamic system and then design utilizing it to design different type of systems okay. Now, we saw that our value of rd or the dynamic response actually decreases at all values of frequencies if zeta is increased so rd decreases as zeta is increased okay. So, my displacement is displacement of the dynamic system so let me call it dynamic displacement this decreases with damping for all values of zeta okay. So, let us see what happens and this decrease is actually quite sensitive to the value of omega by omega n. For example, if you look at here if you increase the damping from let us say 0.01 here okay 0.01 here to 0.05 you do not see much reduction right only this much. However, for the same thing if you increase the damping at this frequency ratio here let us say this is 0.8 you can see the reduction is actually this much which is significant okay. So, let us now see depending upon the value of the frequencies ratio frequency ratio how would the rd vary and how would the dynamic response can be characterized okay. So, let us consider three cases okay in the first case I would consider a vary or I would consider a case where omega by omega n is less than 1 that means that this forcing frequency is very small frequency small means what it is a slowly varying load okay. So, I can say that it is a slowly varying excitation or force. Now, what happens in that case okay if you look at the value of rd and if you substitute omega by omega n equal to 0 as we previously saw this rd becomes equal to okay rd which is like you know u0 eust by u0 this becomes equal to actually 1 right. So, your dynamic displacements actually become equal to the static displacement okay. So, for a very slowly varying force your dynamic displacement is actually equal to static displacement and which is further equal to p0 by k okay and this is actually controlled by the stiffness of the system here. So, what I am basically saying let us say I had a spring mass damper here okay and if I apply a force and I am like you know applying the load which is very slowly applied here like okay. So, what will happen the effect of mass would not be that much okay and the dynamic displacement would effectively be equal to static displacement okay and it is determined by the stiffness of the system here and damping does not play a role here. So, it is the response is actually not affected by the damping here alright. Now, let us consider the second extreme case or before going to that let us consider phi here what happens to the phase angle. Now, can you imagine if you apply the load very very slowly here okay your system would move in tandem to the applied load okay. So, those would be in phase if you apply the load very slowly. So, what happens phi even from the expression when you set omega by omega n equal to 0 phi actually becomes 0. So, those are in phase okay and that you can also like you know see from here that if you apply load very slowly your displacement would move in or the mass would move in tandem with the applied force okay. So, those are in these quantities are in phase okay. In the second case I am considering other extreme situation in which my omega by omega n is greater bit like you know it is significantly large. So, this would be the case when the applied force is varying at a very rapid rate. So, we call it a rapidly varying force okay. So, for a rapidly varying force okay if you substitute this value what you will see your U naught or your Rd basically tends to 0 okay and your dynamic displacement is actually 0 alright and again I would like to explain you through the same example here I again let us assume this is spring mass the dynamic system is being represented by this is spring mass tamper system here. So, what I am basically saying that if the load is applied like you know at a very very high frequency like you know very very high frequency the system would not even respond okay. So, in that case your dynamic displacement is almost 0 alright and let us see what happens to the phase of the system okay for a very high value of this one it becomes the value becomes 180 degree alright and basically whatever the force that you apply your mass would actually whatever small displacement it undergoes okay let us say it is a very small displacement as we said Rd goes to 0 it would effectively be opposite to that applied force okay. So, ut and pt are completely out of phase okay. So, if your pt is applied in the right direction your system would move to the next direction of course we have said that that displacement is very very small okay but in terms of phase they would be completely out of phase alright. Now let us consider another case okay when omega by omega n is actually approximately equal to 1 okay. So, basically the forcing frequency is close to the natural frequency of the system which basically we are trying to like you know discuss about the resonance like situation here. So, in that in this case what will happen if you look at the value of Rd it becomes 1 by 2 zeta okay which is equal to u0 by ust0. So, my dynamic peak dynamic displacement u0 is ust0 by 2 zeta. So, at frequencies that are very close to the resonant frequencies okay your dynamic displacement okay it increases okay significantly than the static displacement because remember this typical value of zeta would be around let us say 5 percent or 0.05, 0.1 you know in that sort of order okay. So, the dynamic displacement is very sensitive to the damping if my frequency applied frequency is close to the natural frequency okay. So, the response is basically controlled by the damping in the system alright and let us see what happens to the phase angle. So, phase angle would become in this case alright 90 degree if you substitute omega by omega n okay denominator would be 0. So, it would be tan inverse of infinity which would be which would effectively give you 90 degrees. So, basically the applied force Pt and ut they are out of phase by 90 degrees okay. So, that would mean that when Pt achieves its maximum value ut would reaches to 0 and then when Pt goes to negative again ut reaches to 0 okay. So, it would be something like what we have discussed here like this situation here okay when it starts as 0 it is minus 90 when it reaches to minus sorry positive amplitude then it goes to 0 okay and when it goes to 0 then basically you have the maximum value of the response okay. So, those are shifted by 90 degrees okay. So, these are the typical scenarios of course loading or like you know it would not be always in these extremes it would always fall it between or like you know closer to one of these scenario but based on these three you would be able to interpret or predict somewhat the behavior of the dynamic system okay alright. So, basically we will study that how to obtain the response of a damped single degree of freedom system subject to harmonic excitation okay and then we talked about the physical meaning of different parameters and how it effects the response of the system alright. So, we are going to conclude our lecture here today alright thank you.