 Some of these, you know, one of the question is that why Newton's law is said to be a law and D'Alembert's principle is a principle. Even what I have learnt, you know, I probably would not be able to give you the fantastic answer, but what I have learnt that law can deviate, but a principle cannot. So what it essentially means that if you look at Newton's law, it depends on what reference frame you are choosing. That means if I say force equals to mass times acceleration and I look at a frame that is moving with the, you know, body, then I do not think Newton's law can be hold there. So that law violates. That means depending on the frame you are analyzing, you have to say resultant force equals to mass times acceleration in that direction, okay. But D'Alembert's principle as we started, it is a principle because it does, it is valid, you know, in any moving frame that you see. That means as long as we say that M A is taken in an opposite manner, the body is at rest. So it does not violate. So it is, you know, in any non-inertial frame, if we choose, then that principle will not be broken down. You have also seen that Newton's law is not valid when the speed of light, let us say, you know, your speed is very high comparable to the light, then there are some violations there also. So principle is something that I learnt that some others, you know, theorem is derived from those principles. There is no violation to the principle. For example, why principle of transmissibility? That was another question. Why do we say transmissibility is a principle? Remember the concept is that as long as equilibrium is concerned, a force can be taken anywhere in its line of action and there is no violation to that. You cannot violate that no matter what you do. So principle cannot be violated, but a law there may be deviations, okay. So that is all I can answer at this moment. But if I do get a better answer from other experts, I will convey that to you, okay. So other issues, I do not think there are many relevant questions that are coming up. If we merge this system in any fluid, which density is D, then what effect of the fluid on this system please explain. I do not know the answer. I can only tell you that, okay, probably you know, when you are dipping it, that means there is a resistance from the fluid. So fluid is entering, you know, in the equation of motion, you only get that in terms of the damping force, viscous force. So fluid friction, fluid friction is going to come into play and that we are going to, you know, look at in the next half, this half as well. So that will be damped vibration. Think of a piston, just think of a piston, okay, and that is let us say on a cylinder that piston, you know, is going into the cylinder and that cylinder is basically filled with fluid. What happens? You will have the, okay, you will have the friction force that is coming as a resistance as the piston is trying to go down. Now from the experiment, we have observed that that force, the resistance force is proportional to the velocity of the piston, okay. Other question, what is the, you know, please specify the reference books for dynamics and vibration. I do not think for engineering mechanics course, our books, you know, set of references are fixed. As we said that very, you know, good book if you want to see, definitely there are two, as we have referred to, B.R. and Johnston and Merriam and Craig. So these are the two books, B.R. and Johnston and Merriam and Craig. In both cases, in fact, we find that dynamics is more than that of statics. So now there are other questions coming up. What is the difference between oscillation and vibration? Vibration is more general, oscillation is very specific, okay. So anything oscillate that means we try to say it is always going from positive to negative. But in vibration it is not necessary that it will always go from a, you know, some positive to some negative, okay. So that is one of the questions. I think how to account inertia forces for the body under motion, I think that is what we are covering, okay. Inertia force is again a resistance to the motion and it has to be taken in an opposite way, okay. So let us, you know, there will be questions that is coming up. I am looking on to this. But let us try to, you know, as such I do not see everything is crystal clear in terms of the problems look like. There was a question in this problem, I think the problem number was, yes, so this problem I think why this force was taken up here, someone said it should be down, just look at it carefully, okay. So Xa is chosen to be greater than Xd that means there is a net tension in the spring. If there is a tension in the spring then how it should act, okay. So we have, we are analyzing this problem. We have assumed the fact, let us say Xa is greater than Xd. It does not matter, just because of that, you know, we are deriving the equation of motion based on that particular notion that, okay, I have Xa and then Xd is coming into play, Xa is greater than Xd. Therefore, I have a tension in the bottom spring. So how does tension will be from the spring is going to come to the body, it should be upward, okay. Similarly, if you look at the spring, this is also under tension, spring under, you know, tension at B as well. So how the body is going to see the force? Newton's third law. So Newton's third law is a coming into play, okay. So it is going to be downward. So I do not see any problem as such. Is that clear now, okay? So that is more relevant towards what I studied, okay. Now what we will do? We will try to solve few tutorial problems before we go to the damped vibrations and we can do the damped vibration as a separate study and as long as we grasp this concept, I think there is not much issue to deliver these concepts to the students, okay. So all I said or all we have learned is the fact that, you know, we can represent the body in an equivalent static system, okay by using the D'Alembert's principle which states that the inertia forces should be taken in an opposite manner. That means mass times acceleration and I times alpha, they should be taken in an opposite sense on the body and then you can solve that problem at any instant of time t, okay, by using the equilibrium. That means resultant force should be 0 and resultant moment should be 0. That is all we have done, okay. But it is a very systematic and thorough procedure and we have found that students actually, you know, enjoy doing that without getting into lot of nitty-gritty of dynamics as such. So with this I think what we will do? We will try to go to the damped vibration but I would not spend much time on that maybe 15 minutes or so because as such we do not, you know, also in the syllabus of IIT it is not there. As you have seen the main objective is to deliver the concept of drawing the free body diagrams. That is what we have been doing from the beginning. How to draw the free body diagram? How to show the forces? And in this dynamic problem I have simply said, okay, using this principle we are looking at the equivalent static system. So let us now look at the damped free vibrations. The first statement that is being given is all vibration are damped to some degree but by forces due to dry friction, fluid friction or internal friction, okay. It has to be damped because we know that a system cannot oscillate for infinite long time. So amplitude is going to decay and it is going to stop at some point of time. So now here basically we have to represent this, you know, how the damping is coming into play. So again it is a viscous damper or a dashpot. So remember this piston is going into a cylinder which is of, has some fluid which has a viscosity. So if it is water and if it is let us say some honey, the viscosity will be different and all of us know that therefore what is going to happen? That one will, honey will give more resistance than that of water as simple as that. Now experimental evidence shows for small motion the force that the body sees is going to be proportional to the velocity and that proportionality constant is called to be damping coefficient C which is a known, okay. Now you look at the equilibrium of this mass with the dashpot. If we just think of the free body diagram of this entire thing as it is isolated from the spring, again we are measuring the vibration from static equilibrium configuration. The mass is displaced by x or let us say some x0 and released, okay. So it is going to oscillate like this about the equilibrium configuration. So at any instant of time x, sorry at any instant of time t displacement is x. So what is the forces? Now what are the forces acting on that? That is k delta s t plus x that is the spring force. Mass times acceleration that is the inertia force W C x dot, okay. Remember this W is actually equals to k times delta s t. So that is going to go away ultimately. So what we land up getting is m x double dot plus C x dot plus k x, okay. Now I have to solve this equation of motion. So idea is that we substitute for x equals to e to the power lambda t, assume the solution in this form and dividing through the e to the power lambda t. So ultimately this is going back to the mass. So we have, remember x double dot that will be what? Lambda square e to the power lambda t. x dot is simply lambda e to the power lambda t, right and k x. So ultimately this cannot be 0 because then we are not solving any problems. e to the power lambda t cannot be 0. So we can divide this entire thing by that one for non-trivial solution. So for the non-trivial solution we have the characteristic equation that is displaced. So it will have two roots, right. So roots are like this. Now what we have defined here? We have defined a damping in a critical damping coefficient such that this is equals to 0, okay. So that way what will happen? I will have the value of the critical damping coefficient that is 2 m square root of k over m that is 2 m omega n, clear. Now what it will do? Ultimately once I know C c let us say that is the critical damping coefficient. Remember if C c by 2 m, ultimately you have to always look at this if that is greater than omega n or if that is less than omega n that is the whole idea or if that is equals to 0. So ultimately your roots of these equations are changing, okay. So we will write it in a bit different form. So ultimately look how I have written it. I will define rather a damping ratio which is xi. xi is equals to what? C by C c. Now C c is already given 2 m omega n. Therefore now lambda can be expressed in terms of xi that is the damping ratio and the omega n, okay. So negative xi omega n plus minus i omega n square root of 1 minus xi square. So I will write the 2 roots in that form, clear just by defining this damping ratio, okay. Now what is happening therefore I can have 3 different situations. First one is xi less than 1. If xi is less than 1 that means C is less than C c. So that is C is less than the critical damping. We will call that under damped system, okay. So in case of that remember I have a positive i and negative i. 2 roots are given by exactly what is shown and therefore we can say that we have the general solution in this form. Remember this therefore what will happen that omega d is coming into play that is damped circular frequency of vibration. Therefore you are going to get an periodicity of omega d sin omega d t plus cosine omega d t. So this is basically the math and here you are having a decaying term e to the power minus C by 2 m is what xi omega n, do not forget C by 2 m is simply xi omega n that is my damping ratio, okay. So ultimately I am going to get my general solution in this form. Now I have to solve for C 1 and C 2 which can be done depending on the problem. Let us say we are talking about a free damped vibration then this C 1 and C 2 can be obtained based on the information on the initial conditions, okay. So that means whether at t equals to 0 I have x 0 or t equals to 0 I have v 0. So that is under damped system. What is critical damping when xi is equals to 1 that means C is equals to C c. So xi is equals to 1 therefore what happens you can see clearly if you go back to the math therefore you have C 1 plus C 2 t to the power minus omega n t that is the solution, okay, right. And then third one will be xi is greater than 1. Now if xi is greater than 1 you are going to get real roots, okay. So therefore again we are going to get a solution of this form. So this is called over damped system. So what we have now learned that I have xi less than 1, xi equals to 1 or xi greater than 1. Three situations can happen and for each one of these my solution is bound to change just because of the property of the roots. So that is the you know done in the it is assumed that it is already done in the differential equation class. Now let us take the damped vibration as I was saying so we are just trying to understand the physical you know nature how I am going to get my solution rather than going to the math. So what is displayed is a comparison between a damped case and an undamped case. We have already studied, right. Now in the damped case how do I get my solution given that I have x0 and v0 as my initial condition that will be the complete solution. Remember what will happen if I assume x0 equals to sin phi and this quantity as cosine phi then I can express is again sin omega dt plus phi and with a rho here that rho is going to be the amplitude. So there is again an amplitude of vibration, right. Remember that amplitude of vibration if v0 equals to 0 still has the xi. In this case it is undamped situation if v0 0 it becomes x0. So now let us try to plot these two equations. As I said please replace this figure by saying u equals to x so displacement is x velocity at t equals to 0 is x0 sorry displacement at t equals to 0 is x0 and velocity at t equals to 0 is v0 just replace. So I have x displacement this way and time this way and I am comparing an undamped system that is a faint line and damped system that is a bold line. So what is happening ultimately as expected we can see that the damped system is actually decaying in the amplitude and what is that envelope of the decay that envelope of the decay is completely driven by rho multiplied by e to the power xi omega nt. That is the first thing therefore there is a decay in the motion and there is a envelope that controls the decay clear. Second feature what is the second feature you think what essentially xi you do for an you know undamped system what is the effect on the time period. So I have tau d right and tau n do not you think that time period of oscillation a time period of vibration is going to increase when you have presence of xi of course the motion is going to get slow slowed down right. So it is going to take more time to come back to the original configuration or equilibrium configuration than that of the undamped case undamped case when you have no damping. So there is going to be a small increase in the time period of oscillation and you can show that also. We can clearly see that omega d equals to omega n square root of 1 minus xi square. So therefore the time period td is equals to that means damped time period equals to the undamped time period divide by square root of 1 minus xi square. So for xi less than 1 here we are talking about xi less than 1 everywhere. Your time period of vibration is going to increase a little bit depending on the value of xi. Remember xi still has to be greater than sorry less than 1 okay. So depending on the xi time period will definitely increase. Now there is an interesting thing going on how about if I take two successive peaks of the damped system. Let us say I know the amplitude here and let us say I know the amplitude here. Can I get a relationship directly between these two amplitudes? Answer is yes because this envelope is controlled by rho e to the power minus xi omega n t right. Now the difference between these two peaks what is the time difference? Capital td right that is the time period. So therefore if you take the ratio I will have x 1 divide by x 2 should be equals to e to the power xi omega n multiplied by the time period okay. So now what is happening? If we look at from the logarithmic point of view you take the log of this then ultimately I have xi omega n td. So td can be expressed in terms of tn and tn can be expressed in terms of 2 pi by omega n. So ultimately you see I have 2 pi xi square root of 1 minus xi square. So we say it is a logarithmic decrement between the two amplitudes okay. So if we know the successive amplitude then you can measuring the successive amplitudes we can in principle calculate what is the xi that means the damping ratio and therefore what is the damping coefficient c clear. So physically if we look at now we can actually solve problems through this concept but before that I will show one more thing that okay what happens xi equals to 1 and xi equals to 2. So let us say critically damped that means xi equals to 1 and critically over damped that means let us say xi is greater than 1 so we assume xi equals to 2 what it does? Now all of these remember we can simply take an excel file or let us say we take help of MATLAB okay we take help of MATLAB and we can plot this type of graphs what is t versus displacement right time versus displacement graph. Now this graph is given in terms of velocity equals to 0 the initial condition is at time t equals to 0 I have velocity 0 that means slope of this curve should be 0 velocity 0 means at time t equals to 0 the slope of this curve should be equals to 0 no initial velocity only thing I have is an initial displacement that initial displacement is x0 so it is a normalized curve and we can do that ultimately what is the outcome remember this is a vibratory motion so it is a oscillation right and what happens to this? This is non-vibratory motion both for critically damped and over damp as if they are not going to you know as you let us say you think of this door that door has a dashpot right here and you open the door so you are giving it some theta right and you are releasing this door what it does it is designed in such a way the damping is in such a way that xi is either equals to 1 and greater than 1 that means door is not going to go back to its equilibrium okay so it is not going to go back to the other side and then again come back so in absence of this what will happen if you release it is simply going to hit you and again come back okay that is the whole idea so to prevent that okay I am not going to get hit by opening this door so therefore what is going to happen you simply open the door and leave it it is going to come back and stop at its equilibrium configuration so very nice example to look at so how do you solve these equations I am leaving it to math because that is basically mathematics not going into the details but what is most important to coming back to its equilibrium configuration right it is taking more and more time because time period of vibration is naturally elongated is that clear now so I think with that I will just give a very quick example of that of a loaded railroad car let us say that is weighing 3000 or 30000 pounds it has a mass let us say is rolling at a constant velocity and it couples with a spring and dashboard bumper system okay so it is moving at a constant velocity now this let us say has a negligible mass so what is happening basically this momentum is going to come to here so ultimately it is going to get an initial velocity you have already learned the conservation of momentum right so ultimately this is going to be transferred because this mass is going to be attached so this system mass dampers you know a spring damper system is going to get an additional mass with a velocity v0 at time t equals to 0 so your problem starts now so what happens let us assume that there is a recording device that is measuring the displacement and the displacement is measured like this okay determine the damping constant and the spring constant look how nice this problem is I have a recorded data and from that I am trying to determine what is the spring constant and what is the damping constant required okay so again since I know this successive peaks so therefore I can simply use the concept of logarithmic decrement right so in the next slide first of all we know let us say what is the time period of vibration to complete one cycle that is TD that is 0.41 second that is given I can calculate what is omega D okay now look at the logarithmic decrement x1 by x2 should be e to the power xi omega n TD so therefore e to the power c by 2 m TD right so ultimately I know this I know this ratio x1 by x2 right so therefore I can calculate what is the c because m is also known so c can be calculated and therefore we can go to omega D expression that has k and m both k m and c right and from that I can calculate the k okay so ultimate idea is we do not get into lot of mathematics as such now what is happening we can think of forced vibration we solve the differential equation and so on so forth but ultimately you know in this particular modeling session we have delivered the concept of single degree of freedom system we have looked at what is generalized single degree of freedom system we have introduced the concept of D'Alembert's principle to get the equations of motion for this system and therefore we have also looked at can I solve this type of system for free undamped and free damped vibrations and we have looked at some of the you know applications physical applications that are actually possible okay so I think with that what I will do I will conclude this session.