 Hello everyone. So, till now in multi degree of freedom system, we have seen how to set up the equation of motion. Now, the next step is to learn methods to solve those equations of motion to get the response of a multi degree of freedom system. When we considered single degree of freedom system, we saw that there is only one way in which a single degree of freedom system can respond. However, a multi degree of freedom system, which is made up of several degrees of freedom can respond in multiple ways and the total response actually is combination of different type of responses that the structure can uniquely respond in and we are going to look into that. So, we are going to also learn about mode shapes, which basically reflects or which basically describes one of the ways in which a structure can reflect and we are also going to learn about frequencies. Now, a multi degree of freedom system would have multiple frequencies. So, we will see that how to actually find out those frequencies. Okay. So, let us get started. We have already studied how to set up the equation of motion of a multi degree of freedom system and we saw that now the forces are represented in terms of vectors and their displacement are represented in terms of vectors and there are mass matrices and stiffness matrices. So, as against to a single degree of freedom system, okay, we set up the equation of motion for a multi degree of freedom system in terms of these forces, okay and we saw that how we can get the mass matrix, the damping matrix and the stiffness matrix as this. Let us see once we have obtained our equation of motion as this, okay. How do we solve this equation of motion? Okay. Can we employ the same analytical solution that we have derived for single degree of freedom system to solve the multi degree of freedom system? Okay. And if we can, then how do we do that? Okay. So, first thing that we are going to talk about is basically free vibration. Okay. And let us first consider undamped free vibration. So, right. Damping matrix is 0 and there is no external force. So, I am considering that 0. So, my equation of motion becomes mass times acceleration vector and then k times displacement vector equal to 0. Okay. All right. So, we are going to discuss about the solution of this equation. But to get into that, first we need to discuss some basic characteristic of vibration for a multi degree of freedom system. Okay. Now, we know that for a single degree of freedom system, let me just draw it here. Okay. I had only one degree of freedom u and there was only one possible way in which a single degree of freedom system could move. So, by definition, I needed only one degree of freedom to represent the default position with respect to its initial equilibrium position. However, let us consider a two degree of freedom system, something like this. Okay. So, such that I am going to consider different two degrees of freedom such that the total mass remains the same. Okay. However, the distribution is different little bit. So, let me draw. In the first, let me draw the one degree of representation. This is the one degree of representation of a system. Now, let us consider. So, there is only one way it can deflect. Right. We saw that we only consider deflection like this. So, it is a deformable system. Consider deflection like this. Okay. So, we know how it would deflect. However, let us consider system like this in which I have two masses, okay, which are joined first from the support to the first mass. I have considered the same stiffness, very stiffness k and stiffness case. However, mass m1 and m2 and m1 is greater than much greater than m2 here. In the second case though, let us consider a different representation. So, again in this case, I have m1 m2 and in this case, m1 is much smaller than m2 here. Okay. And in the third case, I am going to consider m1 and m2 which are comparable to each other. Okay. As you can imagine, physically if you look at it, there are multiple ways or in this case, if it is a two degree of freedom system, okay. There are two ways in which this structure or this system can deform. One would be simply like this. However, if this mass is very heavy, then it is likely that the movement of this mass is going to dictate the overall response. Okay. Now compare this to the second system that I have drawn here. In this case, if the mass m2 is much larger than mass m1, then the overall response is going to be dictated by the movement of mass m2. Okay. But in general, reality lies somewhere in between where mass m1 and m2 are comparable and basically the deformation or relative deformation of both m1 and m2 actually going to dictate the overall response. Okay. And for two degree of freedom system, it can either respond like this or it can also respond like this. There are only two ways in which this structure can deform in two dimension. Okay. So, for multi-degree of freedom system, okay, the total mass and the stiffness are not the only parameter. The distribution of mass and stiffness also plays a role. And we saw that for the distribution of mass here. I can do the same thing where I keep the masses same, but I consider different stiffnesses. Okay. And in this case also, you can imagine that depending upon the stiffness between the masses and the support, okay, the overall response is going to change. Okay. Now, in previous chapters, we saw that we used to assume a deflected shape of a continuous system and utilize that which we call as a shape function and utilize that to reduce is to single degree of freedom system. Okay. But that was an approximate method and we assume that the deflected shape can be represented by static deflected shape or some other function. However, through the analysis of multi-degree of freedom system, okay, we wish to find out what would be the exact deflected shapes of a multi-degree of freedom system. Okay. And to get that, okay, we are going to do or solve the equation of motion that we have just written here. Okay. Now, so, as you can imagine, the total response of a multi-degree of freedom system, okay, depends on the distribution of mass and the distribution of stiffness. Okay. And it can deflect in multiple ways, depending upon the number of degrees of freedom. Okay. So, the total response, okay, so let us say I have the total response as vector U can be written as linear combination of different deflected shapes. Okay. So, let us say this is phi 1. Okay. So, I am using phi 1 to represent a deflected shape. This is similar to the shape vector that we had used in previous chapters. So, let us say I have a deflected shape which is something like this, okay, linearly increasing with height. So, if the height is h here and height is h, then this deflected shape can be written as, okay, half and one where this is the first degree of freedom and this is the second degree of freedom. Similarly, if I have a deflected shape, okay, which looks like something like this, I can write my phi as let me say minus 1 and 1. All right. So, these are some of the possible ways in which a system can deflect. Now, this is a two degree of freedom system. You could have multiple degree of freedom system. Okay. And there could be different shapes. So, these are the possible ways in which a structure can deflect but I still do not know that which deflected shape is going to be the close or going to be going to determine the overall response. So, what do we do? We consider all deflected shape and we say that the total response is actually going to be some of all such deflected shape, okay, a linear combination, okay. And these factors actually determine the relative contribution of each of these mode shapes to the overall response, okay. So, this is how I imagine my multiple response of a multiple degree of freedom system. And we are going to see how to find out these shapes exactly and how to find out these factors, okay. So, let us start with an example of a two degree of freedom system, okay. Let us say this is m1, m2 and this is k1, k2, okay. And let us say it has been given some vibration, initial vibration, okay. So, it might be given some initial displacement, okay. Let us say it looks like something like this, all right. So, this is the initial displacement at these two degrees of freedom. Now, if we give initial displacement to these two masses and then plot the resulting motion from these two masses, we are going to see that, okay. In general, the motion of these masses are not harmonic, okay. Similarly for this, so this is u1 plotted against time and this is u2 plotted against time, okay. So, if I give some initial displacement to a multi-degree of freedom system and let it vibrate, the response of each degree of freedom in general is not harmonic. Now, compare that to single degree of freedom system in which when we provided initial condition u0 and u.0, we said that the resulting motion would be harmonic and it was given as u0 cos omega nt plus u.0 divided by omega n sin omega nt, okay. But that is not the case here, okay. So, in general, the resulting motion is not harmonic plus as these two masses, m1 and m2, okay, vibrates, we will see that the structure or this two degree of freedom system here, the shape of this two degree of freedom system is going to change with time, okay. So, after some time, it may look like something like this, okay or it may also look like something like this, okay. So, the shape is actually going to change, okay. But, okay, there is a possible combination of initial displacement given to these degrees of freedom, okay, such that if two degree of freedom system given initial displacement in such a fashion, okay, there exists a combination of u10 and u20 such that the resulting motion is actually going to be harmonic. So, let me retry it with initial displacement, this is going to be harmonic, okay. And u1 is also going to be harmonic, okay. So, there exist few combinations, okay, such that if the motion is initiated by providing those initial displacement to each degree of freedom, then the resulting motion of each degree of freedom is actually harmonic, okay. Those combination of degrees of freedom or those combination of initial displacements are actually called characteristic shape or also called mode shape, okay, of a multi-degree of freedom system. Now, for two degree of freedom system, we would have two such characteristic shapes or mode shapes such that if it is given the initial displacement, okay, with those proportions, then it would have resulting motion at the degrees of freedom which would be harmonic in nature plus it would maintain its shape, okay. So, for example, this two degree of freedom system, okay, if I give initial displacement like this, okay, it is going to vibrate such that it maintains its shape. So, if you can look at here, the relative, okay, at any time t, the relative shape of the structure is not changing, okay. Of course, that overall amplitude is changing, but the relative shape is not changing, okay. And if I have a n degree of freedom system, then I would have n such characteristic shape. For example, this is the first type, first characteristic shape or first mode shape, okay. Second mode shape could be something like this, okay. I can have something like this and something like this and then, so each of these, the shape is actually not changing. So, this is second mode shape, okay. So, let us just recap, okay. We said that in general, if you initiate the motion, the free vibration motion of a multi degree of freedom system by providing initial displacement to different degrees of freedom, the resulting motion at each degree of freedom would not be harmonic and the structure is going to change its shape at each time instant, okay. However, there exist few characteristics shape, okay, with which if the initial displacements are assigned proportional to those characteristics shape, then the resulting motion is going to be harmonic plus the structure is going to maintain its shape, okay. Those characteristics shapes are called mode shapes, okay and a n degree of freedom system would have n such mode shape or characteristic shapes, okay. All right. Now, we define for a single degree of freedom system, the natural time period as 2 pi under root m by k. The question comes, how do we define or analytically we said that it is time taken to complete one cycle of motion, okay, for vibration. Now, as I said in general, the resulting motion here is not harmonic, right. So, then how do we define this time period for a multi degree of freedom system, okay. So, for a multi degree of freedom system, the time period is not defined for the overall response, but it is defined for each mode shape, okay. So, the time period we define for each mode shape. So, let us say I have t1 here for the first mode shape and then t2 here is the second mode shape, okay and t1 or let us say a natural time period of vibration, natural time period of vibration is basically the time taken to complete one cycle of vibration in a particular mode, okay. So, t1, t2, t3 like that for a multi degree of freedom system, okay. Now, as we have previously studied, okay, these relationships still hold that tn would be 2 pi by omega n or frequency would be 1 by tn, okay, all right. So, we have defined or we have discussed that how a structure can deflect in multiple ways, okay. In general, if we initiate the motion by providing initial displacement and velocity or whatever, okay. The resulting motion at each degree of freedom is not harmonic, but there exist some shapes according to which if the structure is assigned initial deflection and allowed to undergo free vibration, then in those cases the resulting motion would be harmonic and the structure would it maintain its shape while vibrating in those characteristic shapes, those are called mode shapes, okay. So, as we have discussed the total response of a multi degree of freedom system can be represented as some linear combination of each mode shape, okay. So, let us say I have n degree of freedom system. So, it would be some linear combination of first mode shape, then second mode shape, okay, like that, all right. Now, the free vibration response of a multi degree of freedom system, the total response ut, okay, in the nth mode, let us say this is the nth mode can be written as the mode shape. So, I am just considering one of these mode shapes, okay. Let us consider the contribution of the nth mode shape to the total response by writing it as un of t and that would be equal to phi n the mode shape times the time variation q and t, okay. And that we have previously also discussed that if the deflected shape of a structure is represented through some shape function, okay, then the total response can be represented as a product of that shape function times some time variation function that represents the evolution in time, okay. Now, as we have said this phi n is actually constant in time does not vary with time, okay. So, the time variation is actually presented by q and t, okay. And this q and t, okay, as we have said that if it is vibrating in one of its mode shape, the resulting motion would be harmonic, okay. So, can I say that q and t can be written as a n cos omega n t plus b n sin omega n t. This is the general equation of motion for the vibration or general equation of motion for the harmonic vibration, okay. a n cos omega n t plus b n sin omega n t and this is for the nth mode, alright. So, the response contribution in the nth mode can be represented as shape factor for the nth mode or mode shape times q and t which we have written as a n cos omega n t plus b n sin omega n t, okay. And our equation of motion for free vibration is this, okay. So, what we are going to do here, we are going to take this, okay and substitute in this equation of motion. Now, if I double differentiate this term here, okay, let us see what do we get, okay. So, q n whole dot t would be basically equal to phi n times, if this is differentiated twice, it would be the same function times minus omega n square, okay. So, I am going to write down minus omega n square times here and this becomes q n of t. So, I am going to substitute that here. I get m n m as this, alright, times q which I am going to write this as and phi n and then q and t plus k times u which would again be phi n times q n of t, okay. And this can be further written as k matrix times phi n minus omega n square times m times phi n times q n t. Now, this equation tells us that either q and t can be equal to 0, okay, which means that there is no motion, okay and which is caused like a trivial solution. So, we are not bothered about that, okay. We are the only thing that is going to give us meaningful solution in this one when we consider k phi n is equal to omega n square times mass matrix times phi n, okay. If you look at this carefully, this is of the form a vector times a vector x is equal to some lambda times another vector b times x, okay. This is called Eigen value problem, okay. And you may have come across this kind of problems in your advanced numerical methods course or mathematics course, okay or matrix algebra course, okay. So, this is basically Eigen value problem, okay. Now, in this equation, my k matrix is known to me, my m matrix is known to me, okay. So, remember if I have a like in a multi degree of freedom system, okay, something like let us say this k to k. I know how to get my k matrix, right, which in this case would be simply 3k minus k minus k and k and m would be 2m 0 0 m. We saw that how to get these matrices. So, in general, if the structural properties are given, my k matrix is known to me and my m matrix is known to me. The unknown parameters are actually omega n square and the mode shape phi n, okay. And through this Eigen value problem, okay, we are going to solve for these unknown parameters, okay. So, basically, let me rewrite that k minus omega n square m times phi n is equal to 0. Now, in this case, this would have a non-trivial solution, okay, if this term here is equal to 0. So, this is like solution of n simultaneous equation of motion, okay. And because this is homogeneous, okay, one solution could be the phi n equal to 0, okay, that is the trivial solution. So, for non-trivial solution, we need to have this equal to 0 or specifically not this actually, but the determinant of the coefficient of the phi n vector is equal to 0. So, you can either write determinant of this or you can just use in this equal to 0, okay. This is how we get the solution to an Eigen value problem. Now, if let us say, these are n homogeneous equation, then this would give me, okay, remember the constant here is omega n square not omega n. So, this give me an algebraic equation, okay, of nth power of this constant here, which would be omega n square, to the nth power. So, it would have n roots, which would be real and positive because my k matrix and n matrix are usually real and positive definite and symmetric, okay. So, in that case, I would have n roots here, okay. So, we can solve those type of equation of motion and we can find out the frequencies, okay, omega 1, omega 2, so on, okay. And typically, we arrange it like omega smallest frequency first, okay. Or if you want to write in terms of time period, we will have time period T1, T2 and we arrange it so the largest time period mode is first, okay, this is just a convention, okay. The first mode is also called the fundamental mode, okay. And once we have found out all these roots, omega 1, omega 2, remember it will give you omega n square 1, omega n square 2, but then you can take the square root and find out omega n, omega 2 and so on, okay. So, once you get that, you can substitute it back to this equation, okay. And you can find out the shape function phi 1, phi 2, okay, more specifically the mode shape here, the shape vector phi 1, phi 2 and so on corresponding to each frequency, okay. And that through this procedure, we can find out the exact mode shapes and frequency of a multi-degree of freedom system, okay. And let us take one example to see how do we do that, okay. So, I am going to take example of just what we have seen above. I have a two degree of freedom system with story stiffnesses as k, 2k and k and this is 2m and m. So, the mass matrix I had assembled as 2m, 0, 0m and the stiffness matrix I had assembled as 3k, minus k, minus k and k. So, let us see, okay, what are the modal frequencies and modal shapes for this two degree of freedom system. So, our equation was k minus omega square m determinant of that is equal to 0. So, let us substitute that and see what do we get. So, I have here 3k and then minus k here, minus k here and k here and this is omega square times 2m, 0, 0m here and the determinant of this whole thing. So, I basically get as 3k minus 2m omega square minus k here again have minus k here and k minus omega square m and this determinant to be equal to 0. So, the equation that I get is 3k minus 2m omega square times k omega square m minus k square and this is equal to 0. So, this would give me a quadratic in omega square, okay and let us see what that equation is. So, it would give me here 2m omega 4 minus 5km omega square plus 2k square equal to 0. So, omega square is minus 5km plus minus 25km square m square minus 16km square m square divided by 4m, okay. So, basically omega square is this is 9km square here which under root could become plus minus 9km, okay. So, what do we have here? Remember this is plus here because this is minus. So, 5km plus minus 93km when we take outside the root this will become 3km divided by 4m. So, first let us take this smaller frequency omega 1 square and it would be k by 2m and omega 2 square is 2k by m, okay. So, omega 1 is k by 2m and omega 2 is 2k by m, okay. So, we have got the two frequency of this two degree of freedom system. Once we have that let us substitute back in the equation the characteristic equation or not the characteristic equation with the eigenvalue equation k times phi n minus omega n square, okay or I can write it simply like this k minus omega n square m, okay, times phi n is equal to 0. So, let us first get the first mode shape. So, I can substitute omega 1 equal to k by 2m or omega 1 square equal to k by 2m. Once I substitute that let us see what do we here, okay. So, basically I would have these expressions inside and if I substitute omega square equal to k by 2m this would be simply 2k minus k minus k and then here k. So, it would be k by 2 this times phi 1, okay. Now, phi 1 can be written as phi 1 1 and phi 2 1 which basically represents the coordinates at the degree of freedom 1 and coordinates a degree of freedom 2 for mode 1. So, remember if you have phi j the i corresponds to degree of freedom and j corresponds to the mode and the same would be true for other response vectors as well like u i j, okay and so on. So, I have phi 1 1 and phi 2 1 equal to 0, okay. So, equation basically I get is 2k phi 1 1 minus k phi 2 1 equal to 0 or minus k times phi 1 1 plus k by 2 times phi 2 1 is equal to 0. Now, if you look at carefully both of these equations are one and the same, okay. The second thing is you are going to notice, okay. You can only get solution for this vector in terms of some multiplicative constant, okay. For example, your phi 2 1 or let us say phi 1 1 is nothing but 1 by 2 phi 2 1, okay. So, here because of homogeneous equation it can have infinite solutions, okay. So, you can only get solution for these mode shapes which are in terms of some multiplication factor and that is okay because when we represent a shape factor or a mode shape, okay. These are always the relative location of each degree of freedom, okay. So, what we do typically in these cases to get the mode shapes, okay. We normalize these mode shapes somehow, okay, such that we assume or we find out the individual elements of these mode shapes. So, one of the ways in which we can normalize and especially if it is a representation of a multi-story building is that we can normalize it with respect to the top story degree of freedom, okay. So, let us say we assume it to be 1 which for phi 1, remember for phi 1 this quantity is phi 2 1 and this is phi 1 1, okay. For phi 2, okay, again the same quantity would be phi 2 2 and phi 1 2 for the second mode, okay. So, let us first deal with the first mode. So, if I assume my phi 2 1 equal to 1 by phi 1 1 become half. So, my first mode shape I get as half and 1, okay. Similarly, I can substitute by omega 2 square equal to 2k by m and we will see I can get my second mode shape as minus 1 and 1 again. In the second mode shape we are normalizing with respect to the second coordinate. So, this would look like something like this. The first remember this one is phi 1 2, this one is phi 2 2. So, first degree of freedom and the second degree of freedom. So, this is minus here. So, let us consider in this direction and this is plus here. So, this would look like something like this, okay. So, in both cases I have normalized with respect to the top story modal coordinate, okay. And there are different ways in which you can normalize it. You can normalize it with respect to some specific degree of freedom, okay or you can also normalize it with respect to some other method. We will see that later, okay. So, using this procedure we can get the mode shape and the frequencies of different modes of a multi-degree of freedom system, okay. All right. Now, remember here in a multi-degree of freedom system we would be doing lot of matrix algebra, okay. And it would be very easy if we learn or if we employ some of the techniques and the properties of this mass matrixes, stiffness matrices, mode shapes, you know, so that it becomes or it makes the job easier for us to solve some of the equation of motion, okay. So, let us learn some of the specific properties and some special matrices that we would be using to actually analyze multi-degree of freedom system using matrix algebra, okay. Now, remember for a n degree of freedom system, I have said that I would have phi 1, phi 2 that represents the mode shapes, okay, of each mode corresponding to that I would have the frequency for each mode shape as well, okay. What I can do? I can combine all these, okay, mode shape vector. Remember, right now each of them are a column vector. I can write a mode shape matrix for which each column is actually one of the mode shape vector like this, okay. So, the overall matrix is basically representing columns that are the mode shape of a n degree of freedom system, okay. And we also write this as phi jn, okay, just to represent, okay. This is called modal matrix, okay, modal matrix. Similarly, I am going to define a diagonal matrix represented through this quantity here, okay, omega square. In this, each diagonal element is actually the frequencies of that individual mode and the off diagonal term are 0, okay. Remember, I am considering frequency square because my eigenvalue is actually frequency square, okay. The lambda in that eigenvalue equation is actually frequency square. So, I am going to define a matrix for which the diagonal terms are the frequencies square of each mode shape. And this is called spectral matrix, okay. Now, let us get back to our eigenvalue equation which was k times phi n is equal to omega n square times m times phi n, okay. Or maybe we can write omega n square here. Now, this is the equation, the eigenvalue equation for the nth mode, okay. We can combine all the modes together, okay. So, remember, this is for the nth mode of an n degree of freedom system. I would have n such equations, n such separate equations, okay. We can combine all those equations in a single equation, okay. So, that I can write for all modes matrix k times not the mode shape phi n, but the modal matrix phi n is equal to mass matrix times the modal matrix times the spectral matrix. And this basically represents the same thing. This represents, this would have, if you expand it, n such characteristic equations, okay, for n degree of freedom system, okay. So, this is just a compact way of writing all the characteristic equations, eigenvalue problem for all the modes of a multi degree of freedom system, okay, alright. Now, let us look at a very important property of the mode shapes, which is going to help us immensely in solving the equation of motion for a multi degree of freedom system, okay. That property is called orthogonality of modes, okay, orthogonality of modes, okay. Now, the orthogonality of the modes states that, okay, the different modes, okay, different modes corresponding to different natural frequency, okay. So, different modes, let us say corresponding to different natural frequencies, okay, can be shown to satisfy the orthogonality condition, which is basically, if I take a mode shape n and take a transpose of that and multiply it with the mass matrix times the mode shape of another mode, okay, this would be equal to 0. And the same condition can be shown for the stiffness matrix as well, that if I take a stiffness, if I take the mode shape, transpose of a mode shape n, multiply with the stiffness matrix times another mode shape, again this would be equal to 0, okay. And mathematically we can prove this, okay. Let us see how we can do that. Remember that I can write for my nth mode, okay, for the nth mode, I can write my eigenvalue equation as k times phi n, okay, equal to omega n square times m times phi n, okay. This is for the nth mode. Similarly, for the rth mode, which is a mode that is different from the nth mode, I can write down a similar equation, k times phi r is equal to omega r square m phi r, okay, not a matrix here, which is a vector, okay. Now, what we can do here, okay, we can multiply any of this equation with the transpose of other mode. So, let me say, let me multiply this with phi r transpose on both side of the equation, okay. So, that I have omega n square phi r transpose. And then, okay, here also I am going to multiply this phi n and omega r square phi n. So, our nth mode with the phi r to the power transport and the rth mode phi n to the transpose on the both side of equation, okay. Then, I can take basically the transpose of one of the equations. So, let us say I take the transpose of this equation on the both side. Remember, if a matrix is symmetric like m or k matrix, if you take the transpose of this, it would be same, okay. And if you take transpose of two matrix, it would be transpose of the second matrix times the transpose of the first matrix. So, I am going to employ these rules here, okay. So, if I take transpose of the equation one, I would get as phi n transpose times k times phi r is equal to omega n square, okay, times phi n transpose times n times phi r, okay. So, let us say this is equation two and this is equation three. So, now, if I take equation two minus equation three, okay, these two quantities are same, right. And this can be written as omega r square minus omega n square times phi r equal to times m times phi r, okay. Sorry, this is n here and this equal to zero. Now, we have said that these two are different modes. So, basically, if omega r is not equal to omega n, then this term would be here, that would be equal to zero. So, then phi n is equal to zero, okay. And this is basically the orthogonality condition, okay. Now, we have written this for mass matrix, okay. We can derive the same expression for the stiffness matrix as well, just by writing this expression as phi n t and the mass times phi r would be k matrix times phi r divided by omega r square, that would be equal to zero. So, this would effectively become phi n k times phi r is equal to zero as long as this condition is satisfied, okay. So, this orthogonality condition is a very powerful basic conclusion that we utilize to solve our equation of motion. We will just see after discussing this, okay. Now, if these conditions are satisfied, let us define, okay, two square matrices like this, a square matrix k, which is the stiffness matrix, okay. k is defined as the shape factor or the modal matrix transpose times the stiffness matrix. So, this is capital k here, okay, times this. Now, look at this quantity here and similarly we are going to define a capital mass matrix as well, which is basically the product of phi t times mass matrix times phi, okay. Now, if you look at carefully here, what do you have? The modal matrix which I have previously written as each column vector is basically phi 1, each one of the modes phi 2 and then you have phi n here, okay. And then you have the stiffness matrix which would have term k 11, okay, k 12 and so on, k 21, okay, so on. So, let us just general elements here and then phi, okay, here, remember I have transpose here, okay, transpose here. So, this one is actually phi 1, t, not t here, just write down the phi 1, phi 2 and so on, phi n here, okay. Now, if I multiply these elements here, what you will see that for any i of jth term what I will get as in this case. So, phi i of transpose times the matrix k, this overall matrix, okay, times phi 1, phi j let us say here, okay. So, this is any i of jth term. Now, if it is a diagonal term then I can write it as phi ii and any off diagonal term I can write it as phi t k and phi j, okay, where i is not equal to j. Now, if you look at carefully using orthogonality property, this i and j wherever they would be different, those terms would be equal to 0, okay. The non-zero terms would only exist as long as the phi i and j would be both equal to let us say in this case ii. So, only the diagonal terms are going to exist here, okay. And those diagonal terms can be written as k n. So, if you consider the nth diagonal term, okay, the k n can be written as phi n transpose times k times phi n. So, this matrix is actually a diagonal matrix and the each element of these diagonal matrix can be represented as k n equal to mode shape transpose time k times the same mode shape. Similarly, for the mass matrix as well I would get a diagonal matrix, okay. And the nth element of the diagonal matrix can be written as phi n transpose times the mass matrix times phi n. And both these are obtained utilizing the condition of orthogonality of the mode shapes, okay. So, k n is now known to me. So, the nth element of these diagonal matrices are known to me and they are again related by the same expression k n is equal to omega n square n, which you could easily obtain if you just substitute k n phi n is equal to omega n square mass and mass times phi n, okay. And then write the combine the terms and write it as m m. Okay, so the diagonal terms of these matrices k and m are related using this equation right here, okay. So, now the question comes, okay, why we are getting into all these diagonal matrices, why do we need condition of orthogonality and everything, okay. So, to explain that, okay, let us consider first a 2 degree of freedom system, okay. Remember for a single degree of freedom system the equation of motion was this Cu dot plus k equal to 0, okay. So, this was a second order linear differential equation and I could directly solve it. I had only one variable u, okay and I needed to solve it u as a function of t, okay. But now if I have a 2 degree of freedom system like this, let us say m 11, m. So, basically now first let me write down the equation in terms of matrices. So, m mass matrix times acceleration, let us say stiffness matrix times axle displacement vector is equal to 0. In general, I would have elements in the mass matrix as m 11, m 12, m 21, m 22, okay. And this is u 1, u 2, similarly here k 11, k 12, k 21, k 22, okay and this as u 1 and u 2 and this equal to 0. So, when I write down it will give me 2 simultaneous equation, right, which is basically m 11 times u 1 double dot plus m 12 times 2 double dot then k 11 times u 1 and k 12 times u 2 is equal to 0. Similarly, m 21 times this acceleration and then m 22 times the acceleration here k 21 u 1 plus k 22 u 2. The question is how do we solve these two differential equations, okay. These are two simultaneous differential equation and these are coupled, okay. The first equation has term in u 1, u 2 both and the same for the second. So, I cannot directly solve this differential equation, okay. Had they been uncoupled differential equation then it would have been lot easier for me to directly solve this in some terms of some parameter but that is not the case here, okay. But just imagine, okay, if I have diagonal matrices which only have diagonal element and then u 1, u 2. So, let us say this is m 1 here and this is m 2, okay and this is k 1 here and this is k 2. So, no diagonal, no off diagonal terms. Now, if the mass matrix and the stiffness matrix are diagonal matrix then what do I basically get m 1, u 1 dot plus k 1, u 1 equal to 0, m 2, u 2 dot plus k 2, u 2 equal to 0 and now the equations are uncoupled, right. So, if I have n degree of freedom system I get n coupled differential equations which are coupled through the mass matrix and stiffness matrix and if somehow I can diagonalize the mass matrix and the stiffness matrix then I can uncouple those equations of motions and then I can solve n differential equation individually like single degree of freedom system, okay. For which we already have standard results available, okay and that is what we are trying to achieve by utilizing the condition of orthogonality of the motion and by like you know constructing matrices in which the elements, the mass and the stiffness elements are diagonal, okay. So, although here it is not exactly u 1, u 1 because remember u 1 is your phi times q 1, okay. So, these becomes like you know differential equation in terms of q 1 but the idea is if you have mass and the stiffness matrix diagonal then I can get n uncoupled equation and I can solve the multi degree of freedom systems as n single degree of freedom systems and n differential equations, okay and for which the analytical results are already available to me, okay, all right. So, with this I am going to end this chapter here, okay. In next chapter we are going to see how to actually utilize the mode shapes and the frequencies to actually get the overall response of a multi degree of freedom system, all right. Thank you very much.