 In today's lecture, we are going to see the energy concepts related to excitation, mostly a harmonic excitation of a single degree of freedom system. We are going to look into how the energy transfer takes into place, so how much of energy is being input in the system and if it's a damp system then how much of that energy is being lost through that damping or the energy dissipation. We are also going to see how to obtain the equivalent viscous damping for a damped system. Now, as you know that viscous damping is not always a realistic mechanism for any structure, what it provides us a linear damping model that is suitable for mathematical solution purposes. So, we prefer to use viscous damping, so we will see that today as well how to actually equate a system that has viscous damping to a realistic system and how to obtain the coefficient of viscous damping. So, till now we have been studying basically how to find out the solution to this equation subject to this type of harmonic excitation. Now, what happens in many scenarios that in order to apply this harmonic load, it can be applied either using a motor which is constantly vibrating at frequency omega t or the frequency omega with amplitude p naught or it might also be applied with let us say an actuator in a laboratory. Now, if you look at it an actuator is nothing, but a device that is used to apply load for let us say we have a column here and we can apply what kind of displacement history or load history we want to apply through this actuator. Now, it might not always be possible in a laboratory setup to apply huge excitation or harmonic excitation using these actuators ok. So, what we are going to learn today how to apply basically harmonic excitation using vibration generator ok. Now, these vibration generators ok it could be a deliberate design to actually produce vibration to a structure ok and the practical example would be where you would actually like to find out the modal properties ok using the vibration generators and other could be cases of unbalanced load. For example, if you have a rotating machinery ok and if it is unbalanced load is there then it would apply some force ok. A very common example that many of you see in your houses is actually if you have let us say washing machine ok. So, many of the washing machine now they have like you know come up with different technology to balance the load, but usually what happens due to clothes rotating inside the mass distribution is not symmetric about the center and that leads to unbalanced load and consequently due to the frequency unbalanced load there is a basically load that is applied to the supports here ok. And that you can observe in any any situation any common scenario where the washing machine is being operated ok or we could actually design a vibration generator in which for example, let us say here ok. I have two rotating masses ok which are vibrating about. So, these are connected through a common sleeve here at the center and these are rotating at an angular frequency of omega. So, that after time t the angle traversed through these masses are omega t and let us say the masses are m by 2 and m by 2 ok and let us say this radius here is r ok. So, as we know if we have a mass rotating about the fixed rotating about a fixed axis it would apply a certain centripetal force ok. So, if this mass is rotated it would apply a centripetal force which is directed along this radius and the magnitude of that centripetal force is actually m by 2 omega square r ok. Basically mass times square of the angular frequency times the radius of rotation ok. Now, if you look at it the system here the horizontal component of these centripetal forces are going to cancel off. However, the vertical fold components are actually going to add up ok. So, if you take the vertical component it would be sin omega t and plus of course, for the second mass same expression would be there ok. So, net vertical force net horizontal force is actually equal to 0, but net vertical force is this which is nothing, but m omega square r sin omega t ok. And basically the rotation is actually through an external source let us say it could be a so like an electricity source or motor that applies these rotational velocities of these two masses. Now, what happens if you take this rotating setup of unbalanced load ok and if you attach to any kind of structure ok, if you attach to any kind of structure let us say here ok what will happen it will depending upon at what frequency it is operating it will apply this much of excitation. So, if you consider something like this p naught sin omega t ok if you attach this vibration generator to any structure it would apply the same amount of force to that structure and then the response of that structure subject to this excitation force can be found out using the conventional methods that we have discussed previously ok. Except there is one small difference here if you see p naught now is m omega square r ok which is a function of excitation frequency. So, initially remember when we had p naught sin omega t p naught was independent of any excitation frequency here it is a function of the excitation frequency. So, as we increase the excitation frequency our amplitude also increases ok and that is also the reason we cannot apply a very small amplitude load here ok or we can say like we cannot apply a load statically because to apply a load statically I have to have the frequency omega which needs to be very very small ok. So, for a static I need to have omega very very small and that would lead to p naught which is very very small. So, the p naught would be very small. So, the response of the structure would be like you know very small or like you know it cannot be realistically measured. So, that is why it is difficult to apply static load using this method of vibration generator ok. Now, given that it depends on the excitation frequency let us find out if I attach to a structure and that a structure is now being idealized ok for example, let us just leave it here ok. Let us say this is now being idealized as a spring mass damper system all right ok and with mass m all right and to which I am applying p that is equal to m omega square r sin omega t ok. So, do not confuse between this small m and the capital M ok. The response that need to be found out is for the capital M ok, but the excitation frequency is applied using this small m ok. So, in this case the u t would be found out as u naught sin omega t minus phi ok. Now, if you look at here u naught I can again obtain as r d times the static displacement which I can further write as p naught by k. Now, in this case p naught is m omega square r divided by k times r d ok. So, u naught would be m ok. So, I can write u naught as if I divide this by m and then multiply also by m ok. I can write m r by m times omega by omega n whole square times r d ok. So, for a frequency dependent force ok for a force whose amplitude is actually frequency dependent ok like in the case of vibration generator this is the expression that we get for peak dynamic displacement all right. Now, similarly we can also obtain the expression for basically the acceleration ok. Now, as you know acceleration u naught is actually p naught by m ok times omega by omega n square ok times or let me just write here p naught by m times r a ok where r a is acceleration modification factor remember these all comes from the previous classes that we have discussed. If you remember basically you had u t by p naught by k is equal to r d times sin omega t minus phi ok u dot t divided by p naught under root k m is actually r v cos omega t minus phi and the third expression was for acceleration which was p naught by m minus ok this was r a here. So, this equal to r a sin omega t minus phi ok. So, basically we are using the same expression the amplitude here for u naught would be p naught m times r a which I am writing it here. So, I can further write this as m omega square r remember this would be capital M here because the mass of the single degree of freedom system is actually capital M. So, just keep that in mind ok. So, what I would have here is capital M again times r a ok and if I would like I can further write this as m r omega n square divided by capital M times omega by omega n square r a ok. So, the variation of u naught and the acceleration ok u naught double dot would actually if you try to plot it now it would become a divergent function ok. So, let us say I am trying to plot here u naught divided by m r omega n square by m ok it will start from a value of 0 for a very small value of damping it would look like something like this ok. So, it would look like something like this because I have r a which is actually which used to approach to 1 for large value of omega by omega, but now it is multiplied with omega by omega n square. So, now at large value of r a it actually approaches to infinity ok. Now the question becomes where do we actually use these kind of things. So, we utilize vibration generator as we discussed to provide harmonic excitation to different type of system ok which are usually not possible using actuators in a laboratory setup. So, let us say and this actually has been done let us say our goal is to apply harmonic excitation to a structure like multi-storey building which is outside and or like in a big dam that is there. So, I cannot take an actuator and like an even if I wanted I could not apply that kind of force. So, what I do I apply and like in this case I need to attach this vibration generator to either building or the dam and then keep increasing the frequency and we will see that at certain frequency ok. The response will start to gradually increase ok and that is basically the time period of the or the natural frequency of the structure if the damping is very small ok. So, that is basically a beneficial use of vibration generator. In many cases you have unbalanced machinery for example, let us say you know we talked about washing machine ok. I could have an air conditioning unit which is like you know supported at these two points or it is supported to let us say wall and then again it is rotating ok. So, you know there are different applications of this kind of vibration generator ok. Now, we utilize the vibration generator to find out what is the modal properties. So, we want to find out omega n and damping in the system and for that what do we do actually we perform using this vibration generator we apply the excitation. Now, if you remember for damping ok for sorry at resonance if you remember for a small value of zeta I could write as u naught by u s t is equal to 1 by 2 zeta ok for small value of damping ok. Now, the easiest way to find out damping is to find out ok. So, remember this u naught is basically at frequency omega equal to omega n ok. So, zeta I can directly write as u s t naught divided by 2 u naught omega by omega n. So, let us say ok let us say if I can find out what is the amplitude at frequency omega equal to omega n in the system through this vibration generator ok. So, then and if I can find out u s t naught somehow then I should be able to obtain zeta. The problem with that is we do not a priori know what is the value of omega by omega n plus as previously described it is very difficult to apply static load using vibration generator ok. So, in realistic situation what do we do we actually obtain frequency response curve ok. So, basically we obtain frequency response curve ok in which what do we do ok. We let us say rotate this vibration generator ok. So, that it applies at certain frequency ok to a structure or it could be like you know any type of structure, but let us say we are considering a single degree of freedom system ok. So, let us say this load is being applied to the system here ok. When this load is applied we can find out for this frequency omega n ok. Remember for this structure omega n is fixed ok I am only varying the excitation frequency I can find out what is the peak response either I can do it in terms of displacement or I can do it in terms of acceleration ok. So, I do it for one frequency omega n then I again rotate the same machinery at different excitation frequency ok. So, let us say it is a higher frequency ok and then again I do the same thing ok. I measure what is the peak response either in terms of let us say displacement or acceleration usually what happens acceleration can easily be measured using accelerometer in a using you know instrumentations and in many cases finding out the excel displacement is little bit difficult ok. So, we measure actually the acceleration. Now remember for all these cases my acceleration is for the force amplitude right that depends on the excitation frequency ok. So, in order to find out the acceleration that is independent of the frequency of the force amplitude we divided by that frequency. So, we divided by omega n square we divided by omega 2 square. So, that whatever we get is actually independent of the frequency alright and then we do it for several such several such excitation and then we try to plot ok the response ok and see the variation ok. Once the variation is obtained remember this is omega by omega n ok. Once the variation is obtained ok then we can go ahead and use the half bandwidth method ok half bandwidth method where we apply R this is R by root 2 and then this is let us say omega a and omega b ok. Using this method we can find out the damping ok formula you already know it is omega b minus omega a times omega b plus omega a ok and either you can directly find out the frequency here ok omega by omega n at which your response is maximum ok or you can take average of omega a and omega b to find out the omega n ok. So, these techniques can be employed without having to find out the UST naught or the response at the maximum value of omega n. So, using this technique ok experimentally we can find out the modal properties alright ok. Once this is clear let us move on to the next topic which is basically as we discussed we need to measure in laboratory acceleration displacement. So, the question comes how do we actually measure it and what instruments do we use and what are the principle behind those instruments ok. So, the first thing that we are going to discuss is in vibration measurement instrument ok vibration measurement instrument. We are going to discuss acceleration measurement devices ok many times it is commonly called as accelerometer ok. So, we will first talk about acceleration measurement ok. Now, acceleration in the laboratory is measured using as I said something called accelerometer ok. In its simplest form an accelerometer is nothing, but a spring mass damper system ok which is inside a small rigid box ok. This box is something like this ok and if I have to measure acceleration to any point on a structure all we do we just take this small box and then connect it to that rigidly to that surface ok. So, this provides us whatever the acceleration of the connecting point ok and how do we do that let us see. Basically let us say if it is connected to even a top of a structure like this ok. So, let us say it is connected at this point ok. So, in terms of excitation it is nothing, but ok a spring mass damper system ok. So, I can further rearrange this and I can write this as let us say support excitation in terms of u g of t ok. So, this system basically like that it does not matter which point you attach this accelerometer to this point this point it is basically a spring mass damper system being excited through a support excitation ok and this is a representation that we have been discussing till now ok. Now, we know that for this case our u t can be written as u s t naught alright times r d sin omega ok this is not omega n this is omega t minus 5 ok and excitation basically is in the form of u g t is equal to ok. So, this is acceleration here it is in the form of u g t times sin omega t alright. Now, you might argue that the actual response here is support excitation is not harmonic and I completely agree with you, but like we have previously discussed even if we have a random excitation like an earthquake ok it is made up of several such frequency ok right. So, the goal here is that if we can understand or find out the response to one such frequency ok and then combine all those frequency ok then we should be able to find out the total acceleration at that point. Now, if you look at here ok. So, I want to devise a instrument which is giving displacement history like this ok I can further write this as remember u s t naught here is nothing, but p naught by k which is minus m times ok u g double dot or I can write this as u g naught divided by k ok. So, this can be written here as minus m divided by k r d and this is u g naught omega t minus phi ok, sin term here ok. Now, this expression is nothing, but u g t minus phi by omega if you substitute the expression here you will see in this one ok that you will get the same expression ok. So, if you substitute in this expression let us see sin omega t minus phi by omega ok. This you get as sin omega t minus phi ok. So, basically this is the same expression is this ok for a time variation t minus phi t minus phi by omega. So, I can write this expression ok I can write the expression for u t as minus omega n square times r d and this as t minus phi by omega alright. Now, if you look at this expression carefully ok this is what we want to measure isn't it? We want to measure at whatever excitation ok the support is being going through ok. So, we want to measure the support excitation ok. So, the question is how do we do that how do we measure the excitation. So, that support excitation actually is nothing, but the displacement ok multiplied with some factor if you take like you know these factor r d and this ok and with a phase difference of phi by omega ok. Now, we can say that in this case ok that my u t right now depends on the frequency ok. So, the displacement actually is depending upon the frequency ok because r d also depends on the frequency and phi by omega also depends on the excitation frequency. So, if I design this instrument it would give me different values of phi by omega at the value of r d ok, but I would not be able to sum those up ok. So, if somehow ok if this term here and this term here let us say are independent of ok are independent of the excitation frequency then the instrument could provide me ok different values let us say omega 1 ok due to a frequency ok omega. So, I can write this as minus r d time omega n square remember r d depends on the omega 1 as well times u g 1 ok which if it is independent of omega I can again write this as t minus phi and r d is also independent of omega then I can write it i r d and u t again I can write it like that ok. So, I will have several things and the total response I can sum it as u 1 t plus u 2 t and so on ok. So, the instrument or the instrument the accelerometer should be independent of the frequency that it is being excited ok. So, it is like you know goal is to design an instrument such that to make it independent of r d and phi by omega ok because then each harmonic component we can record with the same modifying factor the r d by omega n square ok and with the same time lag ok. So, even in that case if my earthquake is a excitation consists of like you know many harmonics ok we can record the u t ok with the same shape as the support motion ok. So, the shape of the function would be same if it is independent of that ok. So, all it would need is that a constant factor times this r d ok and that constant factor can be found out as instrument instrument factor of the accelerometer ok. So, this is our goal make r d ok and omega phi by omega as independent of the excitation frequency ok and that we can do if you look at if you remember the plot for r d ok and phi by omega ok. If you remember for different value of damping look like this for ok for very large value of damping actually it is started to become like this ok. Similarly, this for 0 value of damping there was a sudden rise 0 let us say 90 and 180 and for large value of damping it was almost like this ok. So, what do we see that if my damping ok is 0.7 ok and omega by omega n is a smaller than 0.5 ok with smaller phi and if this is this curve here let us say middle curve is zeta is equal to 0.7 in this one as well ok. Let us say this is 0 1 ok if I consider that let us say the middle curve is zeta equal to 0.7 ok. For these conditions ok my r d is approximately equal to 1 and phi by omega is actually linear ok or I can say in this case it would be almost equal to you know a constant ok. So, it is a completely straight line ok. So, if that is the case then what will happen this would be another constant and this would be 1. So, all value of displacement ok if you remember the expression here ok let me write this like this ok. If this is phi by omega 1 is equal to actually linear then phi by omega is would be a constant ok and r d is equal to 1 ok. So, then whatever the support excitation is is actually the displacement that is measured ok and now let us say I have this spring mass system ok and there could be a like you know digital recorder or it could be like no simple paper recorder let us say this paper is something like this ok and let us say I have this system here ok and the excitation let us say being applied here ok. So, what happens as you apply the excitation and this let us say this role this is a paper role here ok. So, as it vibrates it would actually record is displacement here ok and if this single degree of freedom system satisfy this condition then the same displacement whatever we are recording here can be converted to acceleration it is actually acceleration ok alright ok. So, in this case we can measure it like that ok and that is the basic principle of an accelerometer alright ok. Now, let us move on to the final topic of this chapter which is basically energy calculations in a in damped harmonic motion. So, we know that for viscous damping ok we had represented our f d or the damping forces c times the velocity u dot ok and we also know that the energy dissipated through any force is basically force times the displacement over and the integration of that basically over a certain time ok. So, if I want to find out what is the energy dissipated due to my viscous damper ok I can write this as f d times du over any certain time t ok and let us say I want to find out e d over a cycle ok during steady state harmonic motion. So, I am only focused on steady state harmonic motion ok in that case over a period it would be 0 to t ok times f d times du alright. Now, this t is basically I can further is 2 pi by excitation frequency 2 pi by omega f d I can write it as cu dot and remember du the small displacement I can also write as u dot times dt. So, my expression becomes c u dot square dt and u dot I already know well u I already know as u naught sin omega t minus phi ok we can find out u dot as u naught omega cos omega t minus phi ok. So, we can substitute this expression ok here as c u naught square omega square cos square omega t minus phi ok dt and integrate it over this ok. If we integrate this you will see that the expression that you get is phi c omega u naught square ok. So, this is the energy dissipation over a single cycle during steady state ok. This expression can also be written as if you substitute the value for the damping coefficient c ok this can be further written as 2 pi zeta omega pi omega n times k u naught square ok. If you do not remember we had defined our damping or the damping ratio as c by c critical which I could write as c by c critical is written I can write this as 2k by omega n ok. So, c I can write as 2 zeta ok k divided by omega n. So, basically I am just I just substituted it here to get this expression ok. What is the utility of this expression? We will see later ok. So, remember these two expression and these were simply arrived by considering f d and integrating it over a certain time ok to find out the total energy dissipation. Now, I want to know that with respect to the remember that it is a forced harmonic motion. So, even at steady state I am basically inputting energy into the system ok. Basically how I am inputting energy into the system is it through p naught sin omega t ok and I want to see where is that energy going ok. So, let us find out how much is the energy that is being input inputted into the system during a steady state ok. So, I can find that over a cycle ok this must be 2 pi by omega right and I can write this as p t times du ok which I can write this as p naught sin omega t and du I can write as u dot times dt ok which I can further write as u naught omega cos omega t minus phi times dt ok. And you know you can integrate this expression over the limits and then you can find out what is the final expression I am just going to write it as a p naught du naught times sin phi ok. And if you remember the expression for tan phi was 2 theta omega by omega times 1 minus omega n omega n square correct. So, sin phi is basically nothing, but 2 theta omega by omega n divided by this expression square ok for this square and then this square which is nothing denominator is nothing, but what it is already value is not it. So, this is omega by omega n times r d here ok. So, I am going to substitute there ok the expression for r d and see what to be exactly get ok. So, if I substitute it here I will get this as phi p naught u naught times 2 zeta omega by omega n alright. And r d is nothing, but u naught or dynamic amplitude divided by u s t naught which is p naught by k right. So, I can just cancel p naught p naught and this I would get as 2 pi zeta omega by omega n times k u naught square. Now, compare this expression to the expression we had got earlier the energy dissipated in viscous stamping over a cycle is same as the energy that you are inputting into the system ok. So, what does it mean? Well in steady state when steady state is achieved ok during the forced harmonic response of a damped system whatever the energy that is dissipated in the viscous stamping is actually the energy that is being inputted into the system. So, that is why all the energy that you are inputting in that to the system it is being lost into the damping. So, whatever the amplitude you had achieved by the steady state ok that would continue ok. So, there would not be any change in the amplitude of the motion at the steady state and this is also the reason why it is called steady state ok. So, the input energy is actually lost in the viscous stamping alright and amplitude remain constant. Now, you might be curious about that what happens to the spring energy the internal energy which is basically ES ok. So, remember even at steady state your spring is undergoing deformation and then mass is undergoing velocity. So, they would have the potential energy which is basically the spring energy I can write as Fs Tu ok which I can further write as ku times u dot dt and you can substitute again the expression for these alright and then integrate it over 2 pi by omega. Similarly, the kinetic energy you can write it as as ok let us say this is a by e k F i times Tu F i is basically the inertial force ok and if you want to find out over a cycle again you can write to this as mu double dot alright mu double dot times q dot t. Again you can substitute this expression and integrate it over the limits what you are going to find out that these expressions are 0 ok. So, what do we see actually that over a cycle of vibration during the steady state the change in potential energy ok or kinetic energy are equal to 0 ok. So, that is what happens during the steady state to the change in the energy of the spring or energy of the basically the kinetic energy of the mass alright. Now, let us have a look at that we said that our damping force is represented as cu dot and this is of course a linear viscous damping that we have assumed. I want to have a look at how the variation I know variation how the variation would look like if I draw it with respect to velocity right this is F d and this is u dot the variation would be linear right. So, that is why it is called linear viscous damping ok, but I want to see what happens the variation if I consider with respect to displacement because displacement you can easily visualize right it is going from initial displacement to its maximum displacement and then oscillating about its equilibrium position ok. So, I want to find out that relationship F d and the displacement u. So, let us see how do we do that. So, we are going to draw the graphical representation of F d. Now, this is equal to cu dot t which I can write it says c omega u naught this would be cos omega t minus phi ok and this I can write this as u naught and cos omega t minus phi I can further write is as 1 minus sin square omega t minus phi under root ok. And if you remember this is nothing, but u t divided by u naught ok. So, I am going to take this term ok here and this I am going to square it. So, that I get this as and phi rearrange these terms ok. If I rearrange this term let us write this one first this is nothing, but equation of an ellipse if you remember the equation of an ellipse looks like form something like this ok. So, this looks like an equation of an ellipse ok. So, if I try to plot f d versus u ok remember the intercepts are basically whatever is the denominator of those respective variables ok. So, it would look like something like an ellipse with intercept on the x axis this one as coordinate would be u naught 0 and this coordinate here would be 0 c omega u naught ok. And what would be the total energy dissipated here it would be f d du integration which is basically the area under this diagram, but this was the expression for the energy dissipation right. So, the area under this ellipse is basically pi times the first intercept let us say u naught times the second intercept. So, it comes out to be pi c omega u naught square ok which is the same expression that we had earlier derived. So, even through the graphical representation we are getting at the same expression ok. Now in reality what happens ok when we have a structure or any element ok. So, even if we have like you know something like this we said that ok when we apply force on this frame ok it can be represented as a mass spring damper system ok. So, we said that the damping and the stiffness ok this structure would have some damping and stiffness ok and the total mass was concentrated here. So, in general any structural element would have some stiffness and some damping ok small or high depends on the element itself, but it would have some damping. Now, if you know basically the FS versus u for a linear system it is basically simply a straight line ok. So, in reality what happens that when you measure ok especially in an experiment in experiment you do not measure like you know. So, if you take a component let us say column and you start applying a force ok and you start measuring like you know the force behavior it would be basically the sum of this is spring force ok. So, this is here ok some of this spring force and the damper force ok. So, basically some of these two forces and which would look like something like this ellipse rotated by this straight line which would look like something like this ok. So, it would look like something like this ok and how the curve would look like you know how smooth they look like it depends on the actual behavior of the structure ok. So, this is the total force in that structure element versus u ok. If it is a non-linear element ok what you might see actually is that ok. It might also look like something like you know this sorry let me draw that again ok. This is called viscous behavior and this is called hysteretic behavior ok whether it would look like this on this depends on the element properties, but for our cases our assumption is that it looks like something like this alright and this you can observe when you will if you have to perform experiment at some point you will see that your force deformation looks for a regular structural member would look like something like this ok. So, the question comes in many of the situation it would not be actually something like viscous it would be like a hysteretic damper. Now, for this system I know how to solve for the viscous linear viscous damping I know how to obtain the solution we have analytical solution available ok and it is much easier and computational efficient ok if I can somehow analyze or simplify this system to this without losing too much of accuracy ok. So, for that for those situation we do something what is try to find out the equivalent viscous damping so that I can represent a system of a very general ok force deformation loop something like this ok to a viscous damping like this. So, for this what do I need to do I need to find out a stiffness at certain value of displacement let us say this is the displacement amplitude ok alright. So, we need to find out and then what do we do whatever the energy that is dissipated in this random loop equated to a linear viscous damping loop that we had obtained here ok using our own formulation ok. So, let us say we get k from so this is what we got from experiments and we want to idealize this to this. So, once we got this from experiment we can find out what is the k using the these points ok here and here extreme points and we can find out how much is the energy dissipated in one loop let us call this as ed ok. So, the way to go about this is let us say in this case again have the same thing and I am representing this as ok e ok the damper the energy dissipation in the damper. Now, in this situation ok basically what I am doing whatever the energy dissipation I get from the experiment I am equating it to the energy dissipated in a viscous loop which is nothing but remember there is no energy dissipation in the this linear elastic linear behavior of the spring. So, whether it is rotated by an angle does not matter the total energy dissipation is going to be same and I am going to use the second expression which was 2 pi zeta omega by omega k u naught square. Now, if you look at it carefully ok if you look at it carefully this is nothing but twice of the strain energy right twice of the strain energy ok. So, this is here is ESO ok twice of the peak strain energy of the system that we get from the experiment. So, this can be further written as 2 pi zeta omega by omega n ok 2 times ESO. So, zeta I can ok write it as 1 by 4 pi zeta times 1 by omega by omega n times ed by ESO where ed is the experimentally determined value of energy dissipated in a single loop here ok. And ESO is basically the strain energy that is calculated using the effective stiffness and the amplitude ok. If I can find this out I can represent this system using the system that we have derived all the equation for which is the linear viscous damper and how did we do that we did that by equating the energy dissipation in this linear viscous damping system to the experimentally obtained energy dissipation ok. And typically what happens when we perform the experiments we usually do it at the natural frequency of the structure ok. So, in this case what we do typically it is obtained at omega equal to omega n and this case it could be written as ed divided by ESO ok sorry this is not I need to I need to just correct here this is this is actually not 2 pi zeta ok. The value here is actually 1 by 4 pi 1 by zeta by zeta naught times this expression here. So, no no zeta here alright and this you know I mean this is very famously used in like you know many of the equivalent linear analysis equivalent linear analysis we mean that when we actually have in like in reality we have a non-linear system but we want to equate it to a linear system. So, that it is like an amenable to analytical solution and it is more computationally efficient we utilize this expression to find out the equivalent damping ok equivalent viscous damping alright. So, this finishes up the last topic ok of this chapter which is the forced harmonic motion ok. Next class we are going to start a new chapter alright. Thank you.