 Hello everyone. In today's lecture, we are going to learn about frequency response factor. And this is specific to single degree of freedom system subject to harmonic load, but the concept can be extended to any other loading as well. Knowing that any loading to Fourier transformation can be transferred to a sum of systems with different frequencies. So, one of the way to measure the dynamic effect of a load is to look at the amplification with respect to static condition and that is what the frequency response factors are to measure that dynamic effect. So, let us get started. Till now what we have studied is that how to find the solution to the harmonic excitation of an undamped and damped system. So, let us say we have the equation of motion ok. We have the equation of motion for a damped system which I can write as this alright should be Pt here and this is P naught sin omega t. Now, we saw that basically there are two part of the solution. The response for damped system and undamped system ok for undamped system. Let us write down the solution for a damped system. We found out that this is the solution and then there was this part of the solution and we said that well there are two type of vibration. The first one is basically the transient vibration ok which dies out over sufficient amount of time because of this damping term that we have here and second is the steady state solution and then we turned our focus on the steady state solution ok. So, a steady state solution basically the system is oscillating at the excitation frequency omega ok and this these parameters C or these constants C and D these were basically functions of omega by omega n which is the frequency ratio ok excitation frequency divided by the natural frequency of the system alright and basically we defined or we expressed our solution u of t and this is specifically for a specifically for steady state solution u of t is equal to u naught which is the dynamic amplitude here omega t minus phi where phi is the phase angle ok. So, the dynamic amplitude is written as static amplitude times r d times sin omega t minus phi ok and r d is nothing, but it is called displacement response or deformation response factor ok and it is given as this expression here which we can write as this. Now, till now basically we have only considered the displacement response factor ok, but as you could imagine different kind of systems right would have different response quantities of interest. For example, right now we are only considering displacement here, but it might so happen that somebody would like to know what is the velocity modification factor and other would be the acceleration modification factor ok. So, what we are going to study now basically dynamic response factors ok. So, these factors combined are called dynamic response factors and then we will obtain each of these one by one. So, as I said u naught I could write as u s t naught times r d times sin omega t minus phi ok and you know that the static amplitude is nothing, but the applied the amplitude of the applied harmonics force divided by the stiffness of the system. So, I can further write this as p naught by k is equal to r d times sin omega t minus phi ok. Now, what we are going to do now let me write this naught as u naught sorry this should be u of t ok. So, what we want to do now express I have the displacement history here ok expressed as a dimensionless quantity. So, u t has the same dimension as p naught by k. So, I am basically normalizing it with respect to p naught by k. So, I am I want to obtain velocity history ok. So, u dot of t and the acceleration history which would be double differentiation of u divided by some factor and that function we need to obtain ok. So, let us see how do we do that. Now, we know that basically velocity is nothing, but the differentiation of the displacement. So, what I am going to do here I am going to differentiate this equation on the left and right hand side by one or differentiate it once and then see what do we get ok. So, let us differentiate it with respect to time. So, that I get here omega and this becomes cos omega t minus phi. Now, what I can further do divide this by omega n ok and on this side I also divide this by omega n ok and I know that my omega n is nothing, but under root k by m. So, if I substitute that what I will get here is basically this p naught divided by under root k m ok and that should be equal to i d times omega by omega n cos of omega t minus phi. Now, this expression here I can further write as a new expression called r of v ok this would be omega t minus phi. This r v is called velocity ok, it is called velocity response factor velocity modification factor ok, Scottish velocity modification factor. And r v basically as you see here r v is given as nothing, but frequency ratio times the displacement modification factor ok. So, we got our first modification factor after r quantity that is the velocity modification factor ok. Now, let us again differentiate this equation and see what do we get. Now, just to mention if you see here this is my velocity and if you look at in the denominator this p naught divided by under root k m this is the whole expression has units of units of velocity that is why I tried to write it down like that ok. So, I am normalizing my velocity with respect to this quantity here ok alright. Let us again differentiate this equation and then see what do we get ok. So, I have this expression here I would have negative already omega square by omega n times sin of omega t minus phi ok. And if I further divided by omega n and let us write down divide here by omega n as well what do I basically get as acceleration ok divided by p naught by m and this is equal to r d times omega by omega n square times sin omega t minus phi ok. So, this quantity here and let us neglect the negative term because sin is like you know amplitude varies between positive 1 and negative 1. So, it does not matter anyway. So, I am going to write here my r a as equal to r d times the frequency ratio whole square ok. And as you know this is also equal to if you write in terms of velocity this would be r v times just frequency ratio ok. So, now you can see that this has units of acceleration. So, this is a normalized acceleration expression for normalized acceleration that we have obtained. So, as you can see we have obtained the expression for the displacement modification factor, velocity modification factor and acceleration modification factor ok. And these quantities can be further written as this expression that you see here if I write it like this ok this should be equal to r v and this should be equal to r d times omega n ok. Now we are already we already know what is the variation of r d with respect to omega by omega n ok. And we know that what happens when omega by omega n is very small when it is very high and when it is close to 1 ok. Now based on that we can also obtain the variation of first r v as a function of omega by omega n and then r a as a function of omega by omega n. So, let us write down these functions ok and see what do we get. So, I have r v is equal to omega by omega n divided by same ratio here that expression that we have been using till now ok. And your r a is nothing but omega by omega n square ok. And then 1 minus frequency ratio square to the power square plus this quantity here. Now as you can see here my displacement ok my displacement modification factor was 1 when omega by omega n is equal to 0. However, if you look at this expression now I have a omega by omega n term in the numerator. So, when omega by omega n would be very small then the velocity would be equal to 0. So, let us see that for the case when omega by omega n is much smaller than 1. Remember that r d which is basically what you see here I mean let me write it anyway. So, there is no confusion as such ok. This is the expression for r d that we have been using till now alright. So, r d as you can see when omega by omega n is very very small r d tends to be 1 ok. However, if you look at r v and r a let us see what do we get. So, r v we get as remember that in denominator this term would become 0 this term would become 0. But now I have numerator this term omega by omega n. So, now this would become 0. Similarly, r a if you have any omega by omega term in the numerator it would again go to 0 alright. So, this is one difference. Second difference is when omega by omega n is much greater than 1. So, if it is a very very large value let us see what happens r d as we know if it is a very large value it goes to 0 that is directly that can be directly observed from here ok. So, it goes to 0. However, r v the velocity modification factor if you look at it now I have terms in the numerator and denominator as well. However, for very large value of omega by omega n if you look at the denominator I have a power of omega by omega 1 to the power 2, but in the numerator it is power 1 ok. So, the denominator here it actually increases at a higher rate than the numerator ok. So, that is why when the value become very large it actually goes to 0 as well ok. Now, compare that to third case for the acceleration modification factor the highest power in the numerator is omega by omega n to the power 2 and then the denominator also it would be 2 because this is the factor ok square to the power square is going to be contributing to the highest power of omega by omega n ok. So, this one actually tends to be 1. So, if you have system for which omega by omega n is very very large then r a actually approaches to 1 ok. So, these are the conditions for slowly varying force ok because by omega by omega n is much smaller than 1 and these are the conclusions for rapidly varying force ok. And you know I mean we are doing just here analytically we could use any of the numerical tools to actually plot the variation of all these and then see how do they look like for different values of damping. So, I have already done that let me just go ahead and copy that here and see how does it look ok. So, going to copy it here alright. So, if you look at it here ok if you look at it what do we see you can see here this is my r d this is r v and this is r of and you can see the variation r d starts with a value of 1 at a small value of omega by omega n and then as the you increase the value it approaches to 0 ok. r v starts with a value of 0 when omega by omega n is equal to 0 and then again for a large value of omega by omega n it also approaches to 0, but at a slower rate ok, but at a slower rate then the displacement ok. Compare that to the acceleration modification factor r a here it starts with 0 and that it converges to a value of 1 for very large value of omega by omega n ok. And these you can think in terms of like you know flexible system rigid system slowly varying force rapidly varying force ok. So, you can think it in those aspects ok alright. So, once we have that figured out ok let us see how do we find out because now we have 3 modification factor r d r v r a ok. And remember we had used r d to find out what would be the maximum response right you could use that. So, r d is basically like you know for different value of frequency ratio what is the displacement correct. Now in this case let us say we want to find out what would be the frequencies ratio at which these modification factors would have their peak values ok. Now remember the definition of resonances or definition of resonant frequency is the excitation frequency excitation frequency at which the response becomes maximum. Now one might ask well what response are you talking about are you talking about displacement or are you talking about velocity or you talking about acceleration because for different like you know for different people different response quantities could be of interest ok. Acceleration might be interest to someone ok if they are studying for let us say if you are trying to study the vibration in a vehicle then acceleration would be of interest. If you are trying to study about the displacement of the shock absorber or absorbing system then displacement could be of interest or velocity could be of interest ok. So, we need to know if we need to optimize something that if we are talking about resonance then it is with respect to what quantity ok. So, what we are going to do here we are going to define three type of resonance one is displacement resonance then velocity resonance ok and then the third one is acceleration resonance. Now one might ask that why would they not be at the same frequency like you might till now you might be under impression that when omega is equal to omega n you usually get the resonance like situation. However, in reality that is not always the case especially for system with high damping and I will show you with an example let me just copy a figure here and then show you. So, what I have drawn here on the same plot ok in the same figure I have drawn three plots for the response modification factor of displacement velocity and acceleration ok. So, using the expression for each of these I have drawn. Now if you look at here these quantities and this has been drawn for zeta of 0.2 or 20 percent ok. So, the first one is the displacement modification factor and this is the value omega by omega n equal to 1. So, let us draw this line here you can see the maximum for the displacement does not actually occur at the value of omega equal to omega n. Similarly, if you look at the acceleration here it does not the resonance does not actually occur at omega by omega n equal to 1 ok. Only velocity if you see that occurs at omega by omega n. So, let us see how do we find that out how do we find out the exact resonant frequency at which these or the response becomes maximum the displacement response the velocity response and acceleration response ok. So, the expression for each of these were let me first assume that I am writing omega by omega n equal to r some ratio some parameter ok. So, I can write my r d as 1 divided by ok. So, as you know the maximum of any function can be obtained by differentiating. So, the maximum of any function can be obtained. For example, this functionality can be obtained by differentiating with respect to the parameter here with respect to r and then setting it to 0 ok. If we do that if you differentiate it you would be able to find out the value of r at which the expression r d is attains it maximum value the peak value ok. So, let us do that in the denominator there would be some terms ok terms containing this expression here ok we can put it here and in the numerator again we will have the same term ok, but then we will have a differentiation of the term that is inside and that is exactly what we are going to set equal to 0 ok. So, it would be times minus 2 r plus 2 times 2 zeta of r times 2 zeta and of course, all the term would club here in the denominator and we are going to set this equal to 0 ok. So, what do we get as let us take 4 r outside ok. So, we get as 1 minus r square ok plus 2 zeta square ok sorry this is minus 1 times and if we set this equal to 0 ok the value of r that I get from here would be under root 1 minus 2 zeta square. So, this is equal to the omega by omega n. So, the excitation frequency at which this is actually maximum is nothing, but omega n omega equal to omega n times this expression that I have here ok. So, for displacement resonance the applied frequency is actually not equal to omega equal to omega n, but it is omega equal to omega n times minus 2 zeta square and you know one might think that because it is a damn system the resonant frequency should be omega n times 1 minus zeta square because this is equal to omega d. However, as per this expression it does not happen it is that the resonance actually happens at this frequency ok and once you substitute it back to the expression ok you can find out the maximum value of r d as well ok and you can do that calculation I am just going to write down the final expression for r d which is 1 1 minus zeta square ok. So, these this is the frequency excitation frequency for resonance of the displacement response quantity ok and this is the maximum value of the displacement response in terms of the response modification factor ok. So, similarly we can repeat this procedure for velocity modification factor and as well as acceleration modification factor all right and we will see that for resonance. So, let us now consider velocity ok where r v is nothing, but r divided by 1 sorry here it is you have 2 zeta r to the power square ok and if you do that you will get excitation frequency at the velocity becomes maximum is actually omega equal to omega n and the r v or the maximum value of the velocity response in terms of r v is actually 1 by 2 zeta. Similarly for acceleration ok your r of a is r square divided by. So, again follow the same procedure by differentiating it with respect to r and setting it equal to 0. So, that would give you omega the excitation frequency at which resonance or the acceleration resonance happens is actually equal to omega n divided by 1 minus 2 zeta square ok and the maximum value of the acceleration in terms of r a is actually equal to ok this expression right here which is nothing, but same as the expression we had obtained for r d ok. So, what we have derived in terms of different response quantities what are the conditions for the resonance ok. So, what are the excitation frequency at which resonance happens and at that resonance frequency what is the amplitude the displacement amplitude the velocity amplitude and the acceleration amplitude ok. Now, if you look at carefully for a small value of zeta if zeta is let us say smaller than 20 percent which is the case actually for most of the structural engineering systems. You would see that all of these excitation frequencies are actually almost equal because this term the zeta square term is like you know 0.04 ok. So, in terms of like you know I mean when you take the square root it becomes further smaller. So, you do not see much difference in terms of acceleration response factor or velocity response factor or displacement response fraction and neither their displacement frequency they are also like you know approximately equal ok. So, just keep that in mind alright ok. Let us move on to next topic ok. What we are going to do we are going to utilize certain property of this response modification factor ok to come up with a method to obtain the damping from experiments ok. This method is actually called half power method or half power bandwidth method ok. So, let me write it here half power bandwidth method ok and let us see what happens. Now, we know that let us draw this RAD the displacement modification factor ok. This displacement modification factor this curve if I draw it here for certain value of damping you know that it looks like something like this ok. So, this is RAD here this is let us say omega by omega n equal to 1 or let us say let me write it here again ok. This is the frequency ratio on the horizontal axis this is 0 this is 1 this is 2 and so on ok this is 3 let us say ok. Now, this is the value or the maximum value of RAD max ok. Now, what happens in experiments many times we apply what we call a sine sweep ok. So, what a sine sweep is actually sine sweep is it is actually a sinusoidal function with varying frequency. So, the it starts with some frequency ok and then it frequency actually decreases or increases ok. So, sweep means that it sweeps through all the range of frequencies ok. Now, what that does if you have a system if this excitation is actually applied to a spring mass system let us say what will happen the system now it is not a constant frequency remember we had this P naught sine omega t now this omega is actually excitation frequency is actually varying. So, when I apply the sine sweep ok through experiments to a spring mass system or any type of dynamic system then the response or RAD when I calculate it actually varies with the frequency ok. So, you can do the experiment ok and you can obtain curve like this. The other way of would be doing this type of experiments for example, if your machine that is applying the excitation does not have capability to apply sine sweep then basically what you do you obtain this curve displacement modification factor what do you do actually you first apply P naught sine omega 1 t then P naught sine omega 2 t and then so on P naught sine omega n t ok. So, that basically you apply this at all frequencies and then for each frequency you basically try to obtain RAD which is the maximum displacement response and that you can measure from the experiment what is the maximum deformation in the spring or the system ok and just divided by this unit for all of them ok. So, you can do for all of them and then you can again obtain points on this and then you can plot this function RAD versus omega by omega n. Once you have that once you have this plot you know that it would look like something like this at some value of excitation frequency it would achieve its maximum ok. Now, this response function the response modification the response modification factor it has a unique property ok. If you consider an amplitude where RAD is actually RAD max by 2 ok and let us say you draw a horizontal line ok. So, you already know RAD max from experiment you divide it by root 2 it will cut your response curve ok response modification factor curve at two points ok. Let us say one is corresponding to frequency omega a and second is omega b these are the two excitation frequency and of course this is divided by omega n ok. So, let us see what happens. So, what I am exactly now let me express this mathematically what I am saying that RAD is RAD max by root 2. So, let us say this RAD at some it gives me some frequencies ok. Let me write here this is as 1 1 minus R square plus 2 zeta R square ok and this would give you RAD max by root 2 ok. And what is the value of RAD max you have previously obtained it as ok the value of RAD max as 1 divided by 2 zeta 1 minus zeta square ok. So, basically when you solve this you will get a quadratic equation in R square ok it would have some R4 terms then R square terms and then some constant equal to 0. So, that would be basically a quadratic equation in R square all right. So, let me write that out if I expand this I would get basically R4 ok and you can do this calculation yourself and check if you are getting the same expression or not R square ok plus ok. This is the expression that you will get ok. And when you solve this this equation basically what you will get as R as equal to and remember R I have considered as or let me you know or let us say here you will get R square as 1 minus 2 zeta square ok there would be another 2 zeta term here now this expression you can neglect the zeta square term here with respect to 1 and same inside this root. So, that you get R square as 1 plus minus 2 zeta. So, R would be equal to 1 plus minus 2 zeta to the power half and through power expansion we know that if the second term here is very small I can write this as 1 plus minus this power times this term here. So, that would be plus minus zeta ok. So, I have now 2 roots which cross this is nothing, but omega by omega n. So, this gives me 2 root omega a and omega b the higher value let us say or s or the smaller value let us say first is omega n which is 1 minus zeta the higher value is 1 plus zeta ok and you can subtract from the second term to the first term ok to get omega n equal to 2 zeta ok. So, basically once you draw horizontal line and it cuts at 2 frequencies which are basically the roots of those ok this width here is nothing, but 2 zeta here we have just proved that here ok. So, this is here 2 zeta ok and you can read this from your graph ok. So, you know omega b and omega a zeta you can easily calculate and omega n you can also find out right here for a small value of omega n it would be simply the value at which it is maximum or many times what you might also do you can approximate omega n equal to omega a plus omega b by 2 ok we assume that this frequency is symmetrically located ok. So, you can write this as zeta calculate this as omega b minus omega a divided by 2 omega n and if you want you can further write this as ok. So, once you cut the frequency sorry this curve through horizontal line at r d max by 2 it will give you two values of frequencies and utilize those two values of frequency to get the damping in the system ok. So, this is called half power bandwidth method which is utilized to get the basically damping in the system for some harmonic excitation. So, remember that there was another method that we had done in the undamped free vibration in which we had used logarithmic decrement method and we had utilized the logarithmic decrement method to get the damping in the system. Here another method is there which we use in like you know utilizing the harmonic excitation of single degree of freedom system to obtain the damping in the system using this method ok. So, these two methods can be conveniently utilized to obtain the experimentally obtain the damping in the system ok all right. So, I hope these these method this method of obtaining damping is clear to you. So, we are going to conclude our lecture.