 Welcome back everyone. In last class we started free vibration of a single degree of freedom system. In today's class we are going to study how to obtain the response or the free vibration response of a damped system. As you know in reality most of the structure would have some amount of damping and some cases we can ignore it if the effect of damping is not much. But realistically damping need to be considered in the response and we are going to see if we have a damped system then how do we obtain the expression for the displacement and the amplification factors related to displacement ok alright. So let us get started. And we are going to study today's class, last class we saw that what is the equation of a motion and what is how we saw how to solve the equation of motion for a undamped free vibration right. So we said that the equation of motion for undamped free vibration is nothing but this and then we saw that the response of this can be represented in terms of initial conditions as u0 cos omega nt plus u dot 0 sin omega nt alright. And we said that omega n is basically the natural circular frequency of the system right. And we tried to plot the ut and then we tried to understand discuss what is the physical meaning of each of these parameters. Now in this class today what we are going to focus on is damped free vibration. So it is still a free vibration so the external port is still be 0 however for this case there would now be a non-zero value of damping ok. So we are going to consider damped free vibration ok. Now for damped free vibration we are going to consider a specific type of damping that I have previously stated that is viscously damped system ok. So for viscously damped system as I have previously stated the damping force can be represented as a linear function of velocity ok. So that my equation of motion that I need to solve it becomes mu double dot plus cu plus ku that should be equal to 0 right. So now today we are going to solve this equation that we have here and we are going to follow the same procedure. So again it is a second order linear homogeneous differential equation ok and the solution would take the same form as the last class so I can represent my solution to t as an exponential function e to the power lambda t alright. So once I make that substation back to this equation of motion for damped free vibration ok I will get as lambda square m or just write it like this m lambda square plus c lambda plus k should be equal to 0 ok. Now e to the power lambda cannot be equal to 0 ok that does not give me any feasible solution. So I am going to equate this expression here equal to 0 and that is a quadratic equation ok and from your quadratic like you know from your previous knowledge the solution to this equation can be written as lambda equal to minus c plus minus c square minus 4 k m divided by 2 m here ok so denominator is 2 m if you have forgotten basically I am trying to write the solution of the quadratic equation is as x equal to minus v plus minus v square minus 4 ac divided by 2 m ok. So just correspondingly I have got this expression for the roots of this difference of this quadratic equation ok I can further write this equation as minus c by 2 m plus minus and then I will take that 2 m inside the square sign so that it becomes 4 m square ok or I can further write it as this minus k by m ok. So I have got 2 roots for the lambda corresponding to the positive and negative sign ok. Now if you notice here carefully I have a term inside the root here ok. Now this term would be positive or this term could be negative and that would determine actually what or what kind of motion resulting motion the system would have ok. So let us consider the first case in which we say that or we consider that the term under within that square root sign is actually greater than 0. So what I am saying here this term here is actually greater than 0 ok that would mean lambda has or I can write my lambda 1 as this is positive so my lambda 1 and lambda 2 would be real I will get 2 real roots ok. So the first would be this and lambda 2 would be again minus c by 2 m however negative sign here. Now if you look at carefully both these lambda 2 lambda 1 and lambda 2 are negative would be negative values because if you look at it here I have minus c by 2 m and I have under whatever the term that I have under square root is less than c by 2 m ok because remember I have c by 2 m here but then I have another term k by 1 which is actually subtracting from it. So the positive term here is actually less than here so this would again be a negative number here and this would of course be negative because there is a negative sign here ok. So let us say that my 2 roots are actually minus k1 and minus k2 where k1 and k2 are like you know positive numbers ok. So the solution as I said for this type of equation is written as a linear combination of both roots so I can write it as a e to the power lambda 1 t plus b e to the power lambda 2 t alright and for this particular case here I can write it as a e to the power minus k1 t plus b e to the power minus k2 t ok. And if you look at this function both e to the power minus k1 t and e to the power minus k2 t are exponentially decreasing function so if I write it here or if I try to plot this function here I have u of t and then I have t a function would look like something like this ok alright and like you know of course depending upon the value of k1 and k2 if k1 and k2 are very large it would decay very fast if that they are not very large it would decay much slower ok. So the rate of decay actually depends on the value of k1 and k2 ok however one thing to notice here is that now my undamped or my damped system here ok when this condition is there when under the term under the square root is actually greater than 0 I get 2 real positive basically 2 real negative roots ok. I do not see any vibration so do you see any oscillation here ok. If you plot u of t versus t you see that it is just a monotonically decreasing function that goes to 0 at very large value of t however there is no oscillation about the equilibrium position ok which let us say in this case was u equal to 0 ok if it had like an oscillation it would be something like this ok. So for this condition I do not see any oscillation ok so let us check the other condition ok So in other condition I am assuming right this quantity here is actually less than equal to 0 alright. So if it is less than equal to 0 I can write my lambda as ok minus c by 2 m and because this is less than equal to 0 I will take minus negative or minus 1 outside the square root so that it becomes i ok and inside term the terms inside the square roots are a by m minus c by 2 m square ok. So once I take minus 1 out and I get i now I know that this is positive ok and let us you know represent it with another constant saying this is minus c by 2 m this is plus minus i omega d ok. We will see the physical significance of omega d but let us for the right now for the time being assume that this quantity here is represented by another constant omega d where omega d is a positive quantity ok a real quantity. Now using this now I can write my solution as for the ut ok remember that it corresponds to lambda 1 and lambda 2 corresponding to the positive negative sign. So I can write e to the power lambda 1 2 lambda 1 t and then e to the power lambda 2 t ok if I substitute it here I would get e a e to the power minus c by 2 m plus i omega d t and then plus b e to the power minus c by 2 m minus omega d t ok. I can take e to the power minus c by 2 m t term out ok so that I get this one as ok and inside I am remaining with e to the power i omega d t plus b e to the power minus i omega d t and if you remember from your undamped free vibration ok when we try to derive the solution for that I know that I can write this term e to the power i x as sin x cos x plus i sin x e to the power minus i x equal to cos x minus i sin x and then I can like you know substitute this value arrange this term this would basically it would give you something like this ok. Let us say another constant cos omega d t plus d sin omega t of t alright. Now I know that what is the plot of this function and I know that this is an exponential decreasing function ok. So, if I let us try to plot this function for the second case I know that this function looks like something like this from my from the solution of the free vibration looks like something like this. However, it is multiplied with another term which is an exponential decreasing term alright. So, it is you know I mean it would be something like let us say like this ok. So, when you multiply it the resulted function or the ut would look like something like this ok. A function which is actually decreasing in the amplitude alright, but it is still sinusoidal or still harmonic I should not say sinusoidal, but it is still harmonic ok. So, I can plot it something like this ok. So, the amplitude is decreasing ok and this is my u of t and this is actually t on the horizontal axis alright. So, this is the plot of this function. Now compare this to the function I had here I did not have any kind of oscillation here. However, in this case I get an oscillation because of this term the sign and the costum that I have here ok. So, for this condition when this is satisfied I get an oscillation right and I can represent my function or the response the displacement response as this where the amplitude is actually decreasing with time which is due to the damping of the system alright. So, for a damped system the amplitude is not constant ok. And this here the envelope curve this can be represented as e to the power minus c 2 m by theta ok. So, it is decreasing at this rate alright. So, what I found out that a term that I have here c by 2 m minus k by m ok when it is greater than 0 I get no oscillation. So, the if you give initial displacement or initial condition it would simply come back to its original position after a certain time ok. However, if this is a smaller than 0 then I get oscillatory motion ok. Now what would happen if this is equal to 0? Well if this is equal to 0 I would again get 2 negative roots here because this term would be 0 now. So, the solution would still be similar to what I have here. So, again no oscillation ok because there would be any complex term. So, there would not be any sign in the cost term ok. So, it would not have any kind of oscillation ok. So, I should say now I can what I can do if it is greater than all equal to 0 there would be no oscillation. So, it would need to be strictly less than 0 or to have an oscillatory motion ok. Now, for this case what I am going to do equate this term equal to 0 ok that the c that represent this condition is actually the damping coefficient right at which there is a transition from oscillatory motion to non-oscillatory motion. And this c is actually called c critical ok. So, at this critical value of damping what you will see the transition from an oscillatory to non-oscillatory motion ok and I need to find out that value of damping coefficient ok. So, I am going to equate it to 0 ok. So, that gives me the value of c critical as 2 m times a by m ok and which I can further write it as 2 m omega n ok. Remember that there is a square root term here. So, I can further write it as 2 m omega n. So, the critical damping coefficient can be written as 2 m omega n. Now, in reality actual damping constant can be smaller than this or it can be greater than c critical right. So, I define the ratio zeta which is the actual damping in the system divided by the critical damping coefficient ok. So, the ratio of c by c critical is defined as zeta which is called the damping ratio or at many places you will also see that being referred as you know damping as a value of the critical ok. So, as you can see in terms of c critical or zeta what did we deduced what did we deduce here? We said that if c is less than c critical or zeta is or let us first say yeah let us just say that if this is less than equal to 1 I would have oscillation ok and this is called under damped system ok. So, I will have an under damped system here. When c is equal to c critical or the damping ratio is equal to 1 ok I would not have any oscillation. So, no oscillation is called critically damped system alright is called critically damped system and when c is greater than c critical or zeta is greater than 1 I would again have no oscillation and this is called over damped system ok. And you know if I try to plot the variation of ut for these three type of systems ok. So, now I am plotting ut but now I am dividing or like you know plotting it as a ratio of the initial displacement that is given to it ok and this is t as a ratio of the time period of the system ok. So, what I would see for each of these let me just draw it ok. So, as we discussed this is my under damped system which provide me which provides me the oscillatory motion. So, under damped zeta is smaller than 1 this is an over damped system ok. So, this is an over damped system. So, zeta is greater than equal to 1 and this is again a critically damped system ok. So, this is a critically damped system with zeta is equal to 1 alright. So, this is the only three situation in terms of the resulting motion of a damped free vibration that are possible ok. So, it could. So, if you have a damped free vibration ok it would either be if it is an over damped system or if it is a critically damped system there would not be any oscillation if you provide it initial displacement it would just come back to rest ok from an initial position from the initial displacement. However, if it is an under damped system ok then it would oscillate and then slowly or gradually it would the amplitude would decrease and then come down to 0 and at the rate the rate at which the amplitude actually decreases depends on the value of zeta or the damping ratio ok. So, I hope that is clear to you. Now in reality most of the structure ok. So, the systems that are relevant to structural engineers like or the structures that are relevant to structural engineers like bridges buildings you know most of the systems are actually under damped system ok. So, let us say bridges, buildings ok other type of structure like chimneys, dams etcetera most of these structure have value of zeta which is actually less than 0.1 or 10 percent ok and those system can be categorized as under damped system and that would be the focus of study ok on this chapter that we are we would be primarily dealing with under damped system ok. But you could I mean you know in many cases you can see the example an over damped system. So, where an over damped system would be useful if you think about it any system that you do not want to vibrate, but you still want it to be flexible ok. So, you want it to be flexible ok. However, you do not want it to vibrate. So, it would have stiffness, but it should not vibrate after you give its initial displacement. So, one of the example would be a retracting door mechanism. So, I do not know whether you have observed in many places in the classrooms or halls you know you have actually piston kind of thing that is connected at the top of the door ok which when you push the door it allows basically the door not to vibrate a lot because there are a lot of people coming like you know following you. So, to make it safer what do they have I mean that is one of the purpose to make it safer for other people coming from behind you or the other purpose is also to bring it back slowly to its original position. So, that piston actually provides so much of damping that the system actually work as an over damped system because we do not want it to vibrate back and forth you know once you push it you want it to slowly come down to your original position ok. So, an example of over damped system would be retracting door mechanism and you know in similar scenarios wherever you have something that you do not want to vibrate, but simply you want it to come back to its after giving its initial displacement of velocity you want or you would ideally like it to be an over damped system ok. So, like you know I mean there are there are application of over damped system especially it would be more relevant to mechanical engineers or aerospace engineers ok, but for structurally in the for most cases like building bridges we are mostly focused on under damped system and that is what we are going to study ok. So, we are going to study under damped system ok. Now, we had obtained here the solution or look at here the solution for u of t here ok, but you can see there are two unknown constant ok. So, those constant are c and d. So, again the question becomes how do I find the these constant and that again I would utilize the initial conditions u 0 and u dot of 0. So, let me just write u t as u to the power minus c 2 m by t ok c cos omega dt plus t sin omega dt. So, I need to now solve this equation for the initial condition, but before that remember I said my omega d I am representing it as this quantity c by 2 m square in this. So, let us go back to the omega d value that we had assigned ok. So, our omega d is actually not c by 2, but is a little k by m. So, let me just correct that here ok it is k by m minus c by 2 m whole square alright and I know my k by m is nothing, but omega n square correct. So, I can write this as omega n square what else is there if you remember I wrote that my c by c critical I defined at zeta right and my c critical was 2 m omega n. So, c by 2 m I can write it as zeta omega n. So, this I can again write it as zeta square omega n square ok. So, I can take the value of omega n outside the square root and I am left with this quantity here alright and this is called the damped frequency of the system and this is different from the natural frequency of the system ok. Note that I am omitting the term circular frequency and that you know at many places we will do that ok. We are I am not going to every time pronounce the whole damp you know the natural basically the natural circular frequency ok. So, if I say frequency and therefore, you just need to understand it like that ok. So, omega d is related to omega n, the damped frequency is related to natural frequency using this damping ratio ok. Now, for the under damped system if zeta is less than 0.2 remember zeta square would be 0.04 ok omega d is approximately equal to omega n alright keep that in mind it is approximately and I can also write remember that using this relationship my td or the damped time period as pn divided by 1 minus zeta square alright. So, if you look at these expressions for any nonzero value of damping what I will have the frequency is actually decreased but the time period is actually increased. So, due to damping my frequency is decreased but time period is increased you know and you know by whatever small amount it may be because as I said if zeta is smaller than 0.2 it is approximately equal the damped frequencies actually approximately equal to the undamped natural frequency ok. But theoretically it would actually lengthen the time period or increase the time period but decrease the frequency ok. So, let us now plot the damped response versus undamped response ok and that we can do after I have solved this equation the equation that I have here for the unknown constant ok. So, let us do that I have these two unknown constants ok and I am again going to write now know that C by 2 m now can be written as zeta omega n. So, I am going to write ut as e to the power minus zeta omega n t and inside the brackets I will have cos omega dt plus d sin omega dt ok. Remember that the damped system is actually vibrating with omega dt this frequency because sin and cos terms have omega d frequency. So, it is actually vibrating with omega d not the omega n. However, the term here outside the exponential term it is omega n you know. So, many times students while writing the equation get confused ok that is why I am doing this derivation. So, that you do not have to remember anything you can pretty much derive all the equation by yourself if you tend to forget it ok. Now, let us utilize this equation for that first I need to do I need to differentiate this equation once. So, I will I am going to use differentiation by parts ok. So, I am first going to differentiate the exponential term this term would still be the same and then I am going to differentiate the parts that I have inside the brackets which would give me minus c omega t this would be sin omega dt plus I would have d omega d times cos omega dt all right. So, let us substitute the value the initial condition. So, u0 is equal to 1 times c times 1 plus 0. So, this equation gives me the value of c as u of 0 ok. Now, let us take the second equation and substitute the value of t is equal to 0 I get that as minus zeta omega n all right and then I will have inside the brackets c plus 0 plus 1 times again I have 0 plus d omega d ok. So, from this equation and substituting the value of c as u of 0 I will get the value of d as u dot 0 plus zeta omega n u of 0 divided by omega p. So, now, I have obtained both constant. So, finally, the expression for the displacement response of a for the damped free vibration can be written as this exponential term here ok times u0 cos omega dt plus u dot 0 zeta omega n u of 0 sin omega of dt. So, this is the expression for displacement response of a damped free vibration ok and I mean you know you can look at look at from here that if zeta is equal to 0. So, if you substitute the value of zeta equal to 0 this system actually transforms to free undamped free vibration right because this term would become equal to 1 and then the second term would be equal to 0 and omega d would be basically omega n because zeta is 0 ok. So, that I can just verify from here, but you know for a undamped free vibration there is no damping so amplitude is actually constant it is not changing. However, I have an exponential decreasing term which would be non-zero if the value of damping is non-zero ok. So, that amplitude of damped free vibration actually decreasing with the time alright. So, if you plot it undamped free vibration or damped free vibration it would look like something like this ok. Let me just try to plot it here. So, let us say this is undamped free vibration so constant amplitude alright and for the damped free vibration it would be something like this. So, if you see the amplitude is actually decreasing ok. So, it is actually decreasing like this ok. So, this is undamped this one is damped ok. One thing that you can observe from this plot that the time period Tn which for undamped let us say describe it like this and for a damped I will describe it using ok. Let me just describe it here these two peaks for the damped system here ok. Depending upon value of the zeta you know they might be close, but they might be different ok that all depends on the value of zeta ok. Now, comparing the response we can see that if a system has very high value of zeta or very high damping the amplitude would decrease much faster and it would come to the original position ok much faster. So, the time taken to come to the zero position would be much smaller ok. And if you keep on increasing there would reach a point where it would become critically damped and it would not vibrate at all ok alright. So, once you understand this ok you can obtain the response of the system alright. You can obtain the response of the system using the expression that we have derived ok. So, we are going to conclude this class at this point of time alright. Thank you.