 We were looking at two coupled mass systems connected through three springs and we wrote down the equation of motion governing the two masses. We found that we obtained two coupled linear ordinary differential equations and we said that we are going to solve this using the method of normal modes. In this case the method of normal modes looks like this. Once again we are looking for oscillatory solutions and an important thing to note is that that in a normal mode motion the entire system vibrates at the same frequency. Now with this you can see that we have derived the equations and we have written the equations in matrix form. If you compare this equation with the equation for a single degree of freedom you recall that we had found that the frequency in the single degree of freedom case was omega is equal to plus minus square root k by m. So we expect that the frequency will somehow k here in the two degree of freedom case is now a matrix. So we expect that the frequency of this system will somehow be related to some property of this matrix and the mass m. So let us see what property is that. Also notice that this matrix is a symmetric matrix. It is a good idea to think about why it is a symmetric matrix. So now let us substitute our normal mode approximation into the governing equations a and b. As expected it will lead to algebraic equations and we have to solve those algebraic equations. So let us do that. So our algebraic equations are minus m omega square a1 is equal to, so this is minus here. If you write it once again as a matrix equation then we have our column vector becomes a1, a2 is equal to 0, 0. Now we have to find non-trivial values of a1, a2 which means that we are not interested in a1, a2 both being 0. Because these are homogeneous set of equations for a1 and a2, 0, 0 is always a solution. We are not interested in this. So we are interested in non-trivial a1, a2. Now recall from linear algebra that if you have a homogeneous set of equations then for non-trivial solutions the determinant of the matrix needs to be 0. This will give you non-trivial solutions but then those non-trivial solutions will not be unique. This non-uniqueness will have important physical consequences as we will see shortly. So for non-trivial solutions determinant of a matrix has to be 0. That matrix is this matrix. Now you can immediately notice that this matrix is just our original matrix or minus of that and you can see that this quantity minus m omega square is basically the eigenvalue. So it is clear that the eigenvalues of the matrix that we have seen earlier will be related to the frequency of oscillation. Because this is a 2 by 2 matrix we are in general going to get 2 eigenvalues. They may or may not be distinct. Let us find those eigenvalues by setting the determinant equal to 0. So is equal to 0 and this is easy to solve. This basically leads to, so we get 2 eigenvalues or 2 frequencies of oscillation in which the system can oscillate, it is 2 because there are 2 degrees of freedom and the frequencies are distinct from each other. One is k by m, another is 3k by m. Now how do we find out? Now because we have converted this into an eigenvalue problem, now the frequencies are related to the eigenvalues. What information is contained in the corresponding eigenvectors? The eigenvectors corresponding to k by m and 3k by m. Let us find those eigenvectors and then answer this question, eigenvectors. Now let us find the eigenvector for omega square is equal to k by m. If you go back to your eigenvalue problem and substitute this value of omega square, it leads you to the matrix equation k minus k minus k k into a1 a2 is equal to 0 0. As expected this will not produce 2 linearly independent equations, but only 1. You can see that it will give you only one equation which is a1 minus a2 is equal to 0. So both the equations are just this and so we have freedom in choosing one of the 2 unknowns a1 or a2 and this equation will determine the other unknown. So if we choose a1 is equal to a2 is equal to 1, then this equation is automatically satisfied and we get our eigenvector 1 1. This is corresponding omega square is equal to k by m, the eigenvalue k by m or square root k by m. The eigenvector corresponding to k by m is 1 1. You can immediately see that this choice is not unique. You could have chosen 2 2 or you could have chosen minus 2 minus 2. All of these choices will also satisfy this equation. This is the non-uniqueness that I was referring to earlier and it tells you that the choice of eigenvectors is not unique. We will see the physical meaning of this slightly later. Now let us work out the so this is eigenvector a, this is eigenvector b corresponding to 3 k by m. If you once again go back and substitute omega square is equal to 3 k by m in the eigenvalue problem that we had seen earlier, you will get the following matrix equation. Once again this does not lead to 2 independent equations but only 1 which is a 1 plus a 2 is equal to 0. We can choose a 1 to be 1 in that case a 2 becomes minus 1. So we have a 1 is equal to 1, a 2 is equal to minus 1 and so we have an eigenvector 1 minus 1 corresponding to omega square is equal to 3 k by m. Here also you could have chosen 2 minus 2 or minus 2 2 and all of which would also be an eigenvector. You can multiply this 1 1 by any positive or negative number and what you get would still continue to be an eigenvector corresponding to the eigenvalue 3 k by m. The same is true for k by m. Alright, so now we have found the eigenvectors, we have also determined the eigenvalues which are related to the frequency of oscillation. Let us solve the problem, we had written 2 equations a and b. Now let us write the answer to this equation, the general solution to these set of equations in terms of those eigenvectors and eigenvalues that we have found. Now obviously because this is a second order ordinary differential equation with 2 unknowns x1 and x2. So we will have to do something more complicated than what we did last time. So let us generalize that. So the general solution now is written as a linear combination of the eigenvectors. So 1 1 was our eigenvector. So corresponding to the first frequency omega square is equal to k by m, we will again have 2 omegas plus square root k by m and minus square root k by m. So like before we write it as a linear combination of plus and minus. But now we have one more frequency, so we will have to write more quantities. Once again plus square root 3 k by m and minus square root 3 k by m. This is the most general solution to this set of coupled linear ordinary differential equations. What are C1, C2, C3 and C4? These are constants of integration which have to be determined from initial conditions. Like before you can see that this and this are just the complex conjugates of each other. So because the eigenvector is the same here and there and because this quantity has to be real, so C1 and C2 must be complex conjugates of each other, we will find that. Similarly, this and this are complex conjugates of each other. Once again the eigenvectors are the same in both the terms and so C3 and C4 will have to be complex conjugates of each other in order to keep x1 and x2 real. This will come out automatically. So we will have to determine C1 and C2 in terms of initial conditions. Before we do that, let us shift gradually to real notation. Once again like before we use e to the power i theta is equal to cos theta plus i sin theta using that and now I am leaving my matrix notation and writing out an expression for x1 alone. You can write that by getting x1 on the left hand side and doing the matrix multiplication with C1 and just writing the first row. So the first row would be x1 of t would be C1 plus C2 cos square root k by m t plus i times C1 minus C2 sin square root k by m t plus C3 plus C4 cos square root 3 k by m t plus i C3 minus C4 sin square root 3 k by m t. Similarly you can write an expression for x2. The first part will be exactly the same. The second part will be different because the second row in the eigenvector is different. So we will have C1 plus C2 cos square root k by m t. So that is our expressions. These are still not completely in real notation because there is an i sitting in these expressions but as it will turn out C1 plus C2 will turn out to be a real number, C3 plus C4 will also turn out to be a real number, i times C1 minus C2 will turn out to be a real number and i times C3 minus C4 will also be a real number making the expression for x1 and x2 completely real. So now let us do the same thing that we did earlier. Let us say that we have initial conditions. Now how many initial conditions do we have? We will have 4 initial conditions because now there are 2 oscillating masses. So we have to specify the initial position and the initial velocity for every mass. This is consistent with what we have done so far in the sense that there are 4 unknown constants C1, C2, C3, C4 and we have to determine them using the 4 initial conditions that can be specified. So let us do that. So if we say that x of 0 is equal to x1, 0 some initial position of the 2 masses and x dot 0 is equal to some v1, some initial velocity of the 2 masses. So obviously x1, 0, x2, 0, v1, 0 and v2, 0 are real numbers and we are now going to express our unknown constants C1, C2, C3, C4 in terms of these real numbers. So we obtain x1, 0 from the previous equation by substituting t equal to 0, we obtain, we are just substituting t equal to 0, we are just substituting t equal to 0 in the expressions written at the bottom. And you can immediately see that the cosine terms will go to unity, the sine terms will vanish and that is what is allowing me to obtain these equations that I am writing. Similarly we can take the derivative of the expressions that I just showed you and substitute t equal to 0 in them. If we do that then we would obtain x1 dot of 0 and this by definition is equal to v1, 0. Similarly x2 dot, dot is the derivative with respect to time, x2 dot of 0 is the first term is the same and this by definition again is v2, 0. So those are linear equations for C1, C2, C3, C4 and if you solve them we can determine C1, C2, C3, C4 in terms of x1, 0, x2, 0, v1, 0, v2, 0. You can immediately see that we do not need their individual values, the difference is just this. We had expected earlier that C1 minus C2, i times C1 minus C2 will be an imaginary quantity and you can readily see that from what I am writing down. Similarly i times C3 minus C4 would be a purely real quantity and that you can see here. So similarly we can obtain expressions for C1 plus C2 and we expect this to be purely real, C3 plus C4 is also purely real and so that completes our exercise. Now substituting this and writing the final answer we obtain, we are now shifting completely to real notation, there are no complex quantities in our final expressions like before. So this is the expression for x1 of t. You can write a similar expression for x2 of t, I am leaving it to you to write that, it is very easy. Now let us come to the physical interpretation of this mathematics, we have seen that there are two eigenvectors for this system, those can be chosen to be 1, 1 which has a frequency omega square is equal to K by M. Another eigenvector which can be chosen to be 1 minus 1 and this corresponds to the frequency omega square is equal to 3 K by M. Now these eigenvectors are what are known as the shapes of oscillation or the shapes of modes of oscillation. Let us try to understand why is it called a shape. Let us look at the first mode, so I will call this the first mode and I will call this the second mode. We have seen that there are two independent normal modes here. How do we get the system to vibrate in normal mode 1? This is an oscillatory solution. So we expect that if we set up the right kind of initial conditions, our system would vibrate only in normal mode 1. Now it is immediately clear that by looking at the eigenvector it is immediately clear what should we done in order to set the system in normal mode 1. We go back to the diagram of our system that we had drawn for clarity, 1, 1 is the eigenvector. So this is telling us that we have to displace mass, we have to displace mass 1 in the positive direction by a unit amount. You also have to displace the second mass again in the positive direction by a unit amount. If we do that then we will set up an oscillatory mode of motion and in this mode the system will have frequency k by m. It is also clear that this is not the only way of doing this. If you did this in the other direction so we recall that 1, 1 was an eigenvector but minus 1, minus 1 was also an eigenvector. So if you gave a unit displacement in either directions then you would have the system oscillating in mode 1. So you take the two masses and either displace them both to the left or both to the right. It is from this thing it is also clear that if you displace both the masses by an equal amount then the length of the spring in between does not change beyond what it is in the base state. In the base state the length of the spring is L as I have indicated in the diagram. So you can immediately see that the middle string does not participate in this mode of motion. It is not getting extended beyond its base state length in this mode. Consequently the two masses behave as if they are moving independent of each other because the connecting spring does not get extended beyond its base state length. Thus it is not surprising that the frequency of motion turns out to be k by m because these are these two masses are behaving as if they are uncoupled to each other. Similarly you can go back and think about what is the displacement 1 minus 1. So this indicates that you take one mass 1, displace it to the right by a unit amount, take mass 2, displace it to the left because the second quantity is minus 1. So you can either do 1 minus 1 or you can do minus 1 1 because if 1 minus 1 is an Eigen mode then minus 1 1 is also an Eigen mode. In other words you have to displace them by equal and opposite amounts. If you do that then you will set up the second mode of oscillation. I leave it to you to think why is the frequency of this mode of oscillation more than the frequency of the first mode.