 So, I think there are couple of questions that we try to answer what we will do, we will try to look at a small tutorial problem that we have. So, if we can just try to solve that in 5 minutes time. So, please take your time to you know just solve the first problem of the tutorial. So, let us just try to work on that and I need that you know answer from you. So, we are in now problem number 1 of the tutorial again since we have discussed the spring mass system we want to find out what is the time period natural frequency of vibration and the velocity and acceleration for the problem given. So, can you get the equivalent stiffness of the entire system what is the equivalent spring constant yes some remote centers are actually giving the correct answer. Once we know the spring constant then we should be able to get the time period and the frequency. So, f and capital T f and tau yes the spring constant is going to be equivalent of the total system is 4.11 10 to the power 3 that will be the Newton per meter or 4.11 kilo Newton per meter is the k and from that we can get the frequency. Frequency is 2.77 hertz and time period is 0.364 I think there is a perfect answer from what is this center 1140 1140 that answer is absolutely correct. So, we all hope that this is now clear. Now, I am going to display the solution clearly as such it is a very simple problem two springs are in series and one is in parallel. So, if we look at the solution click carefully. So, ultimately these two spring as we know they are in series. So, therefore, what happens if you give it a delta then this will have some delta 1 this will have some delta 2. So, based on that logic we can say that equivalent spring constant of spring 1 and 2 that is k 1 k 2 divided by k 1 plus k 2 that is for spring 1 and 2. Now, once the spring you know 1 and 2 has been mapped with a equivalent constant k prime. Now, you have k prime and k 3 now they are in parallel connection we see here. So, therefore, such to be in a displacement will be same now here. So, therefore, what is happening that we know that k equivalent must be equals to k prime plus k 3. So, ultimately we look at here that k equals to k prime plus k 3. So, we are going to get this. So, equivalent k can be determined that is equals to 4.11 kilo Newton per meter and from that we can get the time period tau n and f n that is the frequency. So, we have now kind of gone through a very simple problem then we are going to discuss other type of systems. So, simple spring mass system we have studied that was the first part. Now, we are going to generalize this concept and all I am saying that now look we can actually represent it a system by an static equivalent component and static equivalent system and we can still do sum of force you know equals to 0 and sum of moment equals to 0. Now, so far you know when we are looking at it the mass was just a particle. Now, suppose we are now going to study for you know bars. So, suppose you have bars like this which is again idealized from you know some kind of frame structures let us say. So, we have already studied this. Now, what is this? This is rotational inertia. So, rotational inertia of a bar about its own mass center that we have already studied. So, mass moment of inertia about its mass center we know it is going to be m L square over 12 where m is the total mass remember capital M is the total mass. So, suppose now let us say I have pin somewhere in the body and body is rotating counter clockwise again very small oscillation. So, that theta is coming into play. So, how do I show the rotational inertia force? Simply going to be like this in a clockwise fashion which is actually opposing the motion of the bar. So, therefore, we can say that this should be capital M L square by 12 theta double dot. So, we will need this you know when the body is rotating about its mass center. So, then we can actually have the inertia force which is the rotational inertia force in this way. Similarly, let us look at this now. Suppose the body is trying to rotate about a 0.0 a which is not its mass center. So, what happens? Now, remember D'Alembert's principle always state that we take the effect of the inertia forces specially the translational inertia force that should be at the mass center. So, we must have the translator inertia force as well as rotational inertia force. So, remember as per the D'Alembert's principle we have to now localize this translational inertia force and the rotational inertia force about its mass center. Therefore, how to do this problem? Remember what it means that if I have a bar like this and it is rotating by an amount theta. So, it is oscillating about its equilibrium configuration then I can decompose this problem into two parts. One is pure translation and another is the rotation about its own mass center. So, therefore the essential meaning is that I have this rigid body that rigid body has a lumped mass that lumped mass is at the center. So, what that lumped mass will essentially do? It will perform two type of motion. One is the pure translation and another one is the rotation about its own axis. So, therefore how do we determine the pure translation? Remember the degree of freedom is theta only one degree of freedom is there and the all know other degree of freedom the y let us say if I use y as a translational degrees of freedom then that is always related to the theta. So, this is called the generalized degree of freedom where I know the shape as it rotates. So, I can either interpret it in terms of translational degrees of freedom or rotational degrees of freedom. So, the shape is always going to be known to me shape of the deformation in this case shape of the rotation. So, now therefore inertia forces effect if I want to take I will have in this case what is translation now that translation is given by y. So, what is y? y is equals to L by 2 multiplied by theta. So, that is the translational motion right displacement. Therefore, what is the inertia? If the displacement is L by 2 theta then the acceleration translational acceleration will be L by 2 theta double dot right. So, basically you have mass times acceleration that is the translational inertia force and it has to be shown downward. Similarly, what is my rotational inertia about the mass center that is I already know from the previous you know study that it is going to be capital M L square by 12 right that is my mass moment of inertia about it is center multiplied by the alpha alpha is the angular acceleration. So, that is theta double dot ok. So, therefore when I represent this body although it is rotating about this point I will take the effect as per the D'Alembert's principle of the inertia forces at its mass center. So, at its mass center remember if the body is rotating about a I have two effects one is the translational and one is the rotational. So, I have a translational inertia force and I have a rotational inertia force. Now, what happens remember in this problem if I really take the moment about this point. So, now what is the equivalent system of this? The equivalent system can be interpreted by this what is given here right. So, this system and this system will be equivalent in terms of the rotational aspect right. So, what we can simply say that then how can I find out what is J A which is mass moment of inertia about A. So, let us go back to the very first lectures on equivalent system then we take the moment right. So, that will be the moment resultant about A. So, the moment resultant about A of this system will give me what J A theta double dot right. So, that is m L square by 12 theta double dot. So, that is this moment right here plus you have this force vertical force multiplied by the distance is L over 2. So, that is going to total is going to be m L square by 3 theta double dot. As we all know we have already studied what is the mass moment of inertia about A it is actually m L square over 3. So, therefore as we can see clearly if I am adopting D'Alembert's principle and I take the effect of inertia forces at the mass center I need not to worry about what is the mass moment of inertia about elsewhere or in turn what will happen that when I take the moment of these forces and set it to equals to 0 everything will come automatically. So, there is no parallel axis theorem is coming into play to get the mass moment of inertia about A. So, only thing is that I need to understand how the body is making the movement and from that movement I have to find out the inertia forces that are coming into the body at its mass center. Remember there may be many links. So, I can add one more link to here and I can try to study this problem. So, that will be our next level of discussions. So, let us look at this problem it is a idealized model again an actual model you see here. So, some beam it is pivoted here that means there is a hinge here and we have the spring here. Now, we are going to apply some kind of loading right here. So, ultimately it becomes a spring and mass system. Let us say I have also mass of this bar I can have a large mass at this tip and I do have springs. So, ultimately I am again it is representing this one by a spring mass system and I want to find out let us say you know we can do lot of stuff now because we have learnt it. So, natural frequency of vibration can we find out can we found out the period of vibration even you can go back to the you know let us say if we are looking at the free vibrations that means I displace the mass system and release it what kind of velocity acceleration I am expecting. Now, what is most critical to understand this as I said the first of all I have to understand that what is the static equilibrium configuration of this system. Where is my static equilibrium configuration? So, that means if we do include if we include the gravity effect first let us try to find out where is my static equilibrium configuration. So, I solve the problem statically. So, at the first part mass of this bar is ignored. So, there is no mass of this bar rigid bar, but we have only a large mass attached to this rigid bar. So, therefore, now the way to solve this problem will be is that first we must understand what is the displacement diagram. If I say the mass is displaced by delta s t then remember the spring you know how much the springs are stressed or compressed that can also be identified based on the concept of similar triangle. So, here I have only theta s t that is the static rotation that is connected to the static displacement of the tube mass. So, now let us look at what are my spring forces since I know the displacement. So, this spring is going to be compressed the right spring right. If this is under compression then what kind of force is going to come on to the bar as it is shown. So, force has to be directed upward. Similarly, for this spring the force has to be directed upward as well because this is going to be in tension. If it is under tension then I have to have outward right from the body and this is under compression. So, this is inward force on the body, but direction will be same in both cases. So, that is how we tell the students this is the clear free body diagram. Now, from the static take the moment about a 0. So, therefore, this is my equation right that is at equilibrium. Now, we for sure know what is the value of delta s t in terms of the mass and the springs constants. So, next thing would be as I said how about now if I displace the system from the static equilibrium configuration and let it oscillate about its own equilibrium configuration. So, what happens? So, next problem will be remember this is the static equilibrium configuration. So, from the static equilibrium configuration I am giving it a displacement x t. So, my degrees of freedom is here x t that is the displacement which is related to of course theta right here. So, it is a single degree of freedom, but it is generalized why it is generalized because x t that displacement is now connected to theta or in other words I can always predict its shape shape of deformation right. So, therefore, what is happening now you try to think of first I draw the displacement diagram. That means from the static equilibrium configuration I have clearly represented the displacement if this is x t this should be two third x t this should be one third x t. So, you can clearly understand that what is the stretching of the spring and what is the compression of the spring. So, therefore, now you come to the force. So, free body diagram of the forces what is the you know forces acting first of all look at the inertia force since the body is displaced I have taken x to be positive then the inertia force translational inertia force has to be downward. Remember this rigid bar is massless. So, inertia force is not coming from this bar only thing I am concerned about the tip mass that is coming into play. I have the weight of that tip mass the spring forces now how the spring forces look like look at the k 2. What happens to the k 2 from the static equilibrium configuration I go this much. So, ultimately k 2 if you look at it therefore, we can assume it to be in tension if I assume the displacement this way. So, spring force is downward similarly for this one although it is in compression the spring force has to be downward. So, direction of the spring forces will be always same. So, now what we do I am simply looking at now that is my static equivalent system at any instant of time x any instant of time t for a given displacement x remember again I repeat oscillations is small that is the major consideration we are doing. Therefore, this forces does not change its direction as the body oscillates about the equilibrium configuration. So, what happens under the small displacement assumption we can now get the equilibrium of this and as you can see everything is there ultimately what is the equation of motion m x double dot m is the mass of this that portion that mass attached to the that bar and we have the equivalent stiffness that is k 1 by 9 plus 4 k 2 by 9 x t equals to 0. What is most amazing and interesting is that there is no gravity term can you see this very unique there is no gravity term although we have considered the gravity effect. Remember this gravity effect is going to go away because we have already this part is already equals to 0 and that is coming from the static consideration. So, equation of motion for this particular problem when we are measuring from the static equilibrium configuration is again not entering into the equation of motion. If I do not measure the vibration from the static equilibrium configuration that mg will again come into play and we have to be extremely careful. So, it is always advisable that we measure the vibration from the static equilibrium configuration. So, the next is now once I know the k equivalent and the mass. So, therefore, what we can do we can get the time period of vibration and so on so forth. We can now solve a large class of problem as well for free vibrations. Now, let us assume this bar now has a mass distributed mass uniformly distributed mass is present on this rigid bar. Now, before we proceed into this another thing that is being given is that without you know doing this calculation if we would have ignored the gravity effect from the beginning. Suppose I ignore the gravity effect from the very beginning then problem is very very simple the problem is very very simple. If we do not have gravity then we need not to find out the static equilibrium configuration. We can assume the original configuration to be static equilibrium configuration and we can solve this problem. Therefore, you can see that in the force there is no mg because there is no gravity. So, we have assumed static equilibrium configuration itself as a original configuration. There is no displacement due to the gravity. So, therefore, again what we will end up to ultimately you can see that k 1 by 9 plus 4 k 2 by 9 is not that. So, ultimately what the point here is that if we are including the gravity if we do include the gravity we have to be careful and we have to follow these steps first I solve the static equilibrium configuration and from the static equilibrium configuration I vibrate the system right oscillate the system and I am going to get the equation of motion ok. Now, if I exclude gravity effect I need not to worry about the static equilibrium configuration. So, I can start from the original configuration I can solve the problem. Now, both cases you must get the same answer. Answer has to be same from the both cases that is the main point of discussions here. Now, remember in this problem gravity is not entering into the equation. However, you will be very surprised if I look at a pendulum or inverted pendulum problem gravity will automatically come into the equation of motion. So, now with a distributed mass let us say now my bar has a distributed mass what happens again remember we are ignoring the gravity effect. In other words we know for sure that if I measure the vibration from the static equilibrium configuration gravity will not enter the equation of motion. So, let us assume this is indeed my static equilibrium configuration because in this problem gravity is not going to play any role if we measure the vibration from equilibrium configuration. So, we will just bypass that therefore and what is shown here that as if we are ignoring the gravity effect. Now, only thing is that remember we discussed that we now have the inertia force of this bar as per the D R M bar's principle the inertia force has to be taken at its own mass center. So, from the displacement diagram the translational inertia force will be M L 2 theta double dot because M multiplied by translational acceleration which is L by 2 theta double dot and what is the rotational inertia force that is M L square by 2 L theta double dot. So, that is the mass moment of inertia multiplied by the angular acceleration. So, I will show this like that and I take the moment about A equals to 0 for equilibrium. So, ultimately what I am going to get remember these two term when you take the moment of the translational inertia force about this and this one right then only thing remember this capital L will be equals to 3 L. So, M L square over 3 that is capital L square over 3 we know that is the mass moment of inertia about A. So, when you are looking at you know moment of these two forces about that that will become this it has to be. So, moment due to these forces about A should be capital M which is the total mass of this bar capital L square or capital L is equals to 3 L divided by 3. So, in this way we will get the equation of motion in terms of the again tip mass displacement the tip mass displacement is a small x right. So, you can get the equation of motion. So, what is being changed compared to when we do not consider the mass here right when you ignore the mass of the bar the change is only here capital M by 3. So, that is coming into the equivalent mass now. So, equivalent mass becomes small m plus capital M by 3 where small m is the tip mass and capital M is the total mass of this bar. Remember I made an statement one can also use principle of virtual work to obtain the equation of motion. So, everything is connected what is not connected how do I do the virtual work by the way I leave it for the homework. If the system is at equilibrium then I will give it a small rotation delta theta and you do the virtual work done by these forces you will set it equals to 0 you will get the equation of motion it is as simple as that and you have learnt that for a single degree of freedom system virtual work does not give any advantage over the equilibrium equation because what happens ultimately what you are doing you are establishing a moment equation about this point right and in virtual work we are calculating the work done by these forces eventually in when we are calculating the work done which will be very equivalent if you take the moment about a equals to 0. So, from the virtual work we can show that same equation can be established ok. So, by definition of the virtual work let us move on to some other problem it has just you know typical exercises we are going to do now. So, derive the equation of motion of a rectangular block resting on a frictionless surface as shown for a small oscillations in a horizontal plane remember the reason as if you know you let us say you have some kind of disc just for the time being that disc is really rotating ok on a horizontal plane and the surface is frictionless. So, what will happen immediately we do not have gravity right the gravity will not your influence the solution on the plane right because we are talking about a planar motion which is on the horizontal plane. So, let us assume this is connected by spring and mass. So, we have the spring here we have the support this is the rectangular block. So, again how do I derive the equation of motion remember for the rectangular block we already know the mass moment of inertia about its own center what was that capital M a square plus b square divided by 12. So, now with that if we just do the displacement diagram how the displacement diagram will look like as you give it a small theta I have to look at the other displacement components right as you give it a theta remember the center of mass has also changed its position and the center of mass should have a vertical translation as well as a horizontal translation ok. So, therefore, now I can get the spring force spring force is K multiplied by b theta b theta is the small displacement ok. Now, what are my inertia forces it is all here look at it carefully firstly what I said let us assume there is a gamma and this gamma is nothing but mass per unit area if gamma is mass per unit area then gamma times a b is equals to total mass right. So, gamma times area is the total mass. So, let us try to look at this now. So, ultimately you have gamma a b that is always going to be the total mass M capital M. So, therefore, how do I look at the inertia forces now I have two components of translational inertia one has to be horizontal another has to be vertical because there is translational motion in both directions. So, I have shown that in one way it is a by 2 theta double dot right the other way it is b by 2 theta double dot. Similarly what is my rotational inertia capital M a square plus b square by 12 theta double dot ok. So, now all I have to do again sum of moment equals to 0 and you see if I do that definitely I am not going to be concerned about the reactions that are coming into play. So, I just do sum of moment about O equals to 0 and I can solve it. So, ultimately what you see now there is a bit of a problem this small m should be capital M actually there is no difference between small m and capital M. So, ultimately you get see what is this this is actually the mass moment of inertia about point O ok. So, therefore, I get this as my equation of motion in terms of the rotation theta. So, again as I said this is going to be you know very systematic approach that we are following as if we are taking the static equivalent system of a dynamic system and we are analyzing that ok by simply taking the moment equals to 0 moment resultant equals to 0. So, the equation of motion if we put all the numbers is going to be that then you can get the natural frequency of vibration you can get the time period of vibration. Now, we are going to come to the next problem before we take the t break. Let us just try to study this problem how it is posed what are the differential equation of motion about the static equilibrium configuration shown and the natural frequency of motion of body A for small motion of B C neglect the inertia effects from B C. So, we do not have let us say ignore the mass of this only mass that we have is at A. If this is the problem first we need to examine is that a single degree of freedom or multi degree of freedom system. However this does not have any inertia force and in a dynamic problem inertia force only you know comes through the mass that means degrees of freedom selection basis is the mass. Now here you do not have any mass here you have mass. So, that means this part has to be statically connected with this part. So, that this B C body has to be somehow statically connected with this part right here of that of this weight A. So, now if you look at it let us assume for the time being that it is 2 degree of freedom system. So, what we have essentially done I said there is a displacement x A here let us say of this mass and how about this there is a let us say rotation theta. So, that rotation will bring you know displacement here as x D and therefore displacement here as 2 x D. So, for the time being let us start the problem with 2 degree of freedom system and let us try to draw the free body diagram and see what happens what relationship exist is there any relationship that can exist between x A and x D ok. So, we will again try to you know deliver all the time what are the typical free body diagrams nothing else we are doing what are the free body diagrams ok. So, here as we see if this is displaced by x A then the spring force has to be k 1 x A right. Now what this spring will do spring k 2 spring k 2 will have x A minus x D. So, that will be tensile in nature, but still if we assume x A is greater than x D to start with to derive the equation of motion ok. Then we have again spring force that is acting downward on this body what is the inertia force since I have assume the motion upward inertia force has to be m A x A double dot ok. So, now you come to B C this bar what are the forces this spring is going to be stretched. So, this is under tension. So, on the body spring force downward this is under compression, but on the body it has to be in this direction right. So, ultimately what is happening now you have the equation of motion for mass A ok. So, for mass A it is already determined that is the equation of motion. Now what happens to this there is no inertia forces here. So, you can clearly see if I take a moment about point C right. Then what happens naturally we can see that x D and x A these two degrees of freedoms are statically correlated ok. So, you have a relationship between x D and x A therefore, what is the equation of motion of this mass. Now we can go back to that previous equation number 1 and once we substitute basically we are going to get this form. So, ultimately we started with a 2 degree of freedom system, but since the bar did not have a mass there was no inertia force therefore, we are able to correlate the degree of freedom of this bar with the degree of freedom of this mass small mass statically ok. And with that we are able to find out what is the natural frequency of vibration because we know now the equivalent stiffness clear. So, let us move on to another problem. So, here a rod A B is attached to hinge A here and two springs each of constant K what is being asked determine the value of K for which the period of small oscillation is A once again and B infinite. So, the here what is my objective I have to somehow find out the value of K such that time period of oscillation is very small and time period of oscillation is very large. Now what do I mean by time period of oscillation small or large? If the time period of oscillation is small that means body is or spring is more of a rigid right. That means I am expecting high number for spring constant K if the time period is very large that means it is oscillating very slow rate right the oscillation is very slow that means spring is very flexible. So, I am expecting a very low number for the K that is the whole issue here. So, the what is what does it mean by time period of oscillation infinite? What does it mean? That means once the body is actually disturbed it is not going to come back to its equilibrium configuration. So, it is a unstable equilibrium. So, for a given value of K right the body will not come back to its original configuration it is going to be very unstable system is that clear? And we have done that if you remember we have done the potential energy approach t equals to infinity time period equals to infinity omega becomes 0. So, if the time period of oscillation is very large what we are saying it is actually you know static problem it becoming omega equals to 0 as if no frequency and all we are trying to understand that in that static problem what is the value of K such that system is unstable that is all and now we go back to the potential energy approach to solve this problem at least this part this part can also be solved from the potential energy approach that is B that is time period of oscillation infinite what is my value of K and very interesting. Now in actual real life you know do not see this kind of problem, but you see in civil engineering application we have a continuous body. So, it is that of a water tank let us say it is a water tank it is a large tower it is a continuous system, but it can be modeled by saying that I have a mass on the top and body is oscillating about its equilibrium configuration. So, again you will see that in case of a you know civil engineering application you will have the spring how the spring is coming spring is determined by the this tower right what is the spring constant of this tower that can be calculated, but remember this will be not a rigid body now this is now a deformable body. Now we are not going to touch upon those type of problems. So, ultimately you can see if the system really oscillates depending on what kind of stiffness I have it can be highly unstable system and we are really trying to determine that given a mass what is that K for which it is highly unstable very very interesting problem as such and we are linking that with the potential energy approach as well. So, as long as let us say my equation of motion is concerned it is very very simple I disturb the system by theta this is my displacement diagram. So, for a small oscillation I have h theta right and this is you know what is this spring force was you know the displacement in the spring is d theta one spring will be under compression another spring will be under tension. So, if you look at the free body diagram of the forces we have K d theta. So, d is this distance do not forget and h is the total height d is the distance from the hinge to the spring and h is the total height that is the height of the mass from the hinge. So, I have inertia force now what is the inertia force h theta is the displacement therefore, acceleration is h multiplied by theta double dot that is my translation you know in acceleration. So, I have m h theta double dot here I have k d theta k d theta. So, I take the moment about a equals to 0 what is my equation of motion is going to be looking like this remember gravity is indeed in the equation of motion you cannot ignore the gravity in this case it has to be there it is a inverted pendulum gravity will always create a disturbing torque the disturbing torque of the gravity will be always in the first order of theta you cannot neglect that I cannot avoid the disturbing torque that is coming from the gravity. So, therefore, what is happening I get the circular natural frequency expression therefore, in the next part what is my tau that is my time period of vibration. So, that is for once again. So, I can calculate a value of k and time period infinity I can again calculate a value of k remember time period infinity means omega equals to 0 that means frequency equals to 0. So, through that I can again get a value of k there are two case and you can clearly see the difference if time period is infinity oscillation is really large it is moving at a very slow you know oscillation system is very very flexible. So, you are having a very flexible spring whereas, here you have a stiff spring you can see that difference. So, as you increase the you know stiffness you are going to go to the more and more rigid system. So, it is vibrating very fast rate time period is reducing. So, frequency is increasing. So, now can we solve this by potential energy approach as I said if time period of oscillation is infinite then I become I am really trying to look it and look into a problem where frequency is 0 that means it is a static equivalent and I am trying to find out at which value of k the system is unstable. So, now here is the main issue. So, how do I again I am going back to the minimum potential energy principle what is the potential energy of this system remember dynamic is already gone because omega equals to 0 I am trying to understand that situation. So, ultimately potential energy of the mass what is that mg h cosine theta this will be also h. So, h cosine theta that is the distance right. So, we have the potential energy of the body from this datum. So, datum will be my hinge. So, that is mg h cosine theta what is the potential energy of the spring is stretched by d theta. So, half k d theta square 2 times half k d theta square. So, for equilibrium what has to happen d v d theta must be 0 does that give theta equals to 0 as an equilibrium configuration yes theta equals to 0 for this indeed implies that theta equals to 0 is an equilibrium configuration question is whether that is stable or unstable at which value of k it will be stable and which value of k it will be unstable. So, we take the higher order of derivative of the potential energy d 2 v d theta 2. So, we just say for stability d 2 v d theta 2 equals to greater than 0 at theta equals to 0 degree d 2 v d theta 2 must be greater than 0 for stability to prevail. So, therefore, we get the d 2 v d theta 2 we said greater than 0 which implies that k must be greater than this number mg h by 2 d square. That means to make it stable k must be greater than 763 which is the same answer as before. So, when k is equals to 763 that means time period is actually infinite it is becoming unstable. Any value that is less than equals to 763 will make the system unstable and we do not discuss it further is that clear. Now, you can see how engineering mechanics is connected a simple principle of potential energy is connected to a dynamic problem in a very nice way. So, what we are going to do now we are going to take a short tree break. We are going to come back and discuss some issues as it appears and then we are again going to solve a very simple you know tutorial problems based on this as logic that we have given.