CHAPTER 10.02: EIGENVALUES AND EIGENVECTORS: A physical example of application of eigenvalues and eigenvectors     In this segment, we will take a physical example of what is the application of eigenvalues and eigenvectors. There are many physical applications of eigenvalues and eigenvectors, but we will just concentrate on a simple one which you can follow based on your knowledge of physics. So let's suppose somebody gives you says that “Hey, I'm going to give you a spring here, I'm going to put a mass next to it and I'm going to put another spring here and another mass right here”. Now what I want to do is let's suppose the spring stiffness is k here the spring stiffness is k here we're assuming the the two springs are the same stiffness.   Let's suppose we want to say this is mass m1 and this is mass m2; that’s the mass of the first mass and mass of the second mass. What we want to do is we want to be able to measure what x1 and x2  are so we are trying to figure out, hey, how does this mass m1 and m2 move when you are either forcing it to move by maybe let’s suppose you might pull it or you might keep on adding some forcing function to it, but let's suppose we take a simple case where you simply pull this mass m2, m1 also gets pulled with it, and then let's suppose you leave it alone what’s going to happen is that the m1 mass is going to oscillate and m2 is also going to oscillate, or maybe not oscillate, depending on the values of k. In this case, they are going to oscillate because there’s no damping; we’re not assuming any damping. We want to be able to figure out how does x1 and x2 vary as a function of time and then we will see hey how does it relate to our application of eigenvalues and eigenvectors so if you look at the free-body diagram of the mass m1 so you have m1 mass right here let's suppose if I assume x1 to be greater than zero and then I assume x2 to be greater than x1 so I’m not saying that this is what's really happening when the system is moving but let's suppose we are saying that at a particular time x1 is greater than zero and x2 is greater than x1.   So if that’s the case now what’s going to happen is that if x1 is greater than zero this spring is going to stretch so if you look at the k1 spring it is going to stretch so if it is stretching that means that is has to have a tensile force in it. So if that's the tensile force in the spring k1 that means the reaction of that force will be like this on the mass itself so that will be kx1 so because this going to stretch by a distance of x1 and the amount of force should be stored in the spring the force should be experienced by the spring will be kx1 so since this direction is like this that means the equal and opposite force has to act on the mass will be kx1 when we are assuming x2 is greater than x1 than if we take the spring two again if x2 is greater than x1 that means that this spring has stretched because its moving x2 more than x1 so this is stretched out because x1 is moving a little less than what the x2 is moving so it'll be stretched out like this so the amount of force should be existing in the spring would be k(x2-x1) so if that is the case then we know that hey that is on this side right here so that means what we'll have k(x2-x1) going this way.   So it's just the opposite force should be existing so your force is k(x2-x1) going this way on the spring on this side so the reaction from the mass would be that one now these are the only forces acting on mass m1 and what we are going to do is we are going to sum the forces on mass m1 so I’m going to get -kx1 + (kx2 - x1) and what this is going to be equal to so this is minus kx1 because let’s suppose we say negative is to the left side and positive is to the right side so we get minus kx1 from here plus k(x2-x1) from here and then we equal to the inertia force in the mass and that will be M1(d2x1 /dt2) so that’s what I’m going to get from there so if I’m going to simplify this I don’t need to simplify it will be like that so that is what our first equation is going to be for mass m1 .   Let's look at m2. On m2 you can very well see that the only force that should be existing will be this way (kx2 - x1) because that I can note from here that hey since the force in m1 is going to be to the right than the equal and opposite force has to act on m2 in the opposite direction or I can look at it this way that hey the force is going this way in the spring so that's the opposite of that is being experience by the mass by Newton’s 3rd law of motion so with M2 right here so what I get is – k(x2 - x1) is equal to that is the force going in the negative direction there or in the negative x2 direction so the total inertia force will be M2(d2x2 /dt2) so this is first differential equation and this is the second differential equation so what we are getting is that the system is governed by two separate differential equation which are simultaneous in nature because we are finding out the two dependent variables x1 and x2 are in both of these equations right here. So let’s see if we put some numbers in here what happens so let's choose M1=10 so I’m not using any units on that so I’m keeping it unit-less, M2=20 and let’s suppose k=15 so what does that do to our equation? So the first equation which I write here will turn out to be that I get (10 d2x1 /dt2) = -15x1+15(x2-x1) and if I take all of these to be left inside what do I get from there so I have (15x2 - 15x1) so I get (10d2x1/dt2) + 15(2x1-x2) =0 so if I take this to the left hand side this is what I will be able to get that equation one rewritten in terms of the numbers and everything taken to the left hand side similarly the second equation which I’ll write here if I do the same kind of substitution I will get 20 d2x / dt2 -15(x1-x2)=0 and that will be my second equation done simply by substituting M2=20 here, k=15 here and taking that to the left hand side that is what I’m going to get as my two equations now so these are the two equations which are going to govern my system coupled with some initial conditions but let’s go and see how this is related to eigenvalues and eigenvectors.   So from vibration theory if you have taken the vibrations class you will be quite familiar with this but if you have not I'm just giving it to you that your displacements will be of this form Aisin(ωt - Ø) let suppose so each of the displacement x1 and x2 so i 's go to 1 and 2 will be of the sinusoidal from where Ai is the amplitude of mass i , and ω is the frequency of vibration the angle of frequency of vibration and then let’s suppose Ø is given as the phase shift. So if we assume that is the case then let’s go and see that if we substitute back into our differential equation what happens we have (d2xi /dt2) in both differential equations so if i take the second derivative of this I will get –Aiω2sin (ωt-Ø) that is what I’m going to get from the second derivative of this xi and now I’m going to substitute this into my differential equation so if I look at the first differential equation I’m going to get -10 A1ω2 sin(ωt-Ø) -15(-2A1 sin (ωt- Ø) + A2 sin (ωt-Ø))=0 that’s what I’m going to get from the first differential equation and the second differential equation if I make these substitutions of xi and d2xi/dt2 in this case being x2 I will get -20A2ω2 sin(ωt-Ø)-15(A1sin (ωt-Ø) + A2sin(ωt-Ø))=0 and that will be equal to zero and now you are finding out the sin (ωt-Ø) is a common term so I am going to take that out  because if sin(ωt-Ø) is zero everywhere, which is not really the case, then our previous solution for my displacement, which is not true, so I can very well get rid of this sin(ωt-Ø) here, sin(ωt-Ø) here, sin(ωt-Ø) here, same thing here, same thing here and same thing here. What does that give me? That basically gives me the following thing, it gives me: -10A1ω2 -15(-2A1+A2)=0, -20A2ω2-15(A1-A2)=0 so this is first equation this is second equation this is what I get by taking out the sin(ωt-Ø) term out.   Ok, if I collect now the A1 term gives me (-10ω2+30A1)-15A2=0 that's the first equation and the second equation will be -15A1+(-20ω2+15)A2=0 so we are basically two (?? 12:06) two unknowns if we consider A1 and A2 to be the two unknowns and we don’t know what ω is but let’s suppose we consider it to be two unknowns then in this case now what we are going to find out is that if I simplify this a little bit further and I suppose I divide it by 15 throughout I might be able to so if I divided it by 15 throughout I get (–ω2- 3)A1 - 1.5A2=0 and this one here I suppose I’m going to divide by 15 I get -0.75A1+-ω2+0.75A2 =0 and the reason why I did this because here I divided by ten and I got this here I divided by twenty and the only reason I’m doing this is because I want to make the coefficient of ω2 in both cases of course with A1 I only have one there so -1( A1) here and -1(A2) here so if I now write it like in the matrix form let’s see what happens if I write it in the matrix form I get A1 here and I get A2 here and of course I'll get 0 and 0 here because that's the right hand side so if I look at the quotient of A1 it is –ω2+3 here, it is -1.5 here, -0.75 here, and –ω2+0.75 here.   I can break it down further I can write it like this I can say hey let me only put known quantities here in this matrix and then I'll have something else here an unknown quantities associated with this so what I mean by saying that is if I put A1 and A2 here and I don’t want to put ω2 term here and ω2 term in this matrix here I’ll put 3 and -1.5 here, -0.75 here and 0.75 here and I’ll put ω2 here. So once that is done I can now see that hey let’s suppose if I assume ω2 =λ let’s suppose I want to call this another minus but the ω2 that’s supposed to be λ what do I get from here and I call this I suppose I’ll call this to be the A-matrix I’ll call this to be the X- matrix and x λ is ω2 I’ll call this the X-matrix so what do I get I get [A][X] –λ[X] is equal to zero vector and that [A][X]=[λ][X] so what is going to happen is that this is the eigenvalue problem because λ how is the eigenvalue defined, eigenvalue is defined as if hey if there is a non-zero vector [X] for which [A] times the non-zero vector [X] is equal to some scalar times the non-zero vector [X] than λ is considered the eigenvalue of the problem so if I’m able to find λ I'd be able to find ω2 which will give me the angle frequency within which the system is running so that's why the eigenvalues show us that hey how we can use for the physics this one is called the eigenvalue of course and what is the corresponding vector [X] is called the eigenvector and you can very well see that the eigenvector is a measure of A1 and A2 which is simply a measure of the amplitudes which the spring mass system is having so that's how the physics of the problem and the eigenvalue eigenvector concept are involved. That's the end of this segment.