Differential equation systems eigenvector plus#
The second normal mode has a frequency of little K plus 2 big K over M, square root and it's when the two masses are moving opposite each other. So the first normal mode has a frequency of the square root of little K over M, is when the masters are both moving together to the right and to the left. The motion doesn't look simple but in fact it is simple. So, if you look at this first clip of the video, you'll see this random motion that we saw initially.
Let's go back to the MATLAB simulation that I did and see how the seemingly random motion can now be separated into these two simple motions. So these are the two normal modes of motion. So you have a factor of two here in that spring constant. The compression of big K is double, right? As being double because it's coming from the left end also coming from the right. Why is it two times big K? Because when the mass one is moving to the right, mass two is moving to the left. The little K is coming from the springs on the sides. Here, the spring constant or the effective spring constant is little K plus 2 times big K. So the motion is coming in, out, in, out looking like this. The second normal mode has eigenvector 1 minus 1. So, the frequency associated with this oscillation doesn't depend on big K, the middle spring is just square root of little K over M which is simply the frequency of an oscillation of one mass connected to a spring, of spring constant little K. The middle spring is not changing length, right? So, this oscillation is as if the middle spring is not even there, it's not changing length. X1 equals X2 means that the masses are moving together to the right and to the left and the oscillation is looking like this, right? X1 equals X2. First eigenvector means that the solution X equals VE to the RT, satisfies X1 equals X2. Now we can use the principle of superposition, but the main point here is that I want to understand what is the physical significance of having found these eigenvalues and these eigenvectors? So let's look at the first eigenvector 1, 1. Here, the frequency is omega 1, is square root of little K over M and here the frequency is omega 2 which is square root of little K plus 2 big K over M. So, well the eigenvalue I wrote down as minus K, but I discussed how this eigenvalue leads to frequency. I should also write down the eigenvalues here. So we get the second eigenvector, V2 is going to be 1 minus 1. Here then V2, 2 has to be opposite sign of V1, 2. So then we get the matrix here plus big K, big K, big K and big K times our second eigenvector which is row 1 column 2, row 2 column 2, equals 0. So here we are subtracting this lambda 2 from the diagonal which is the same thing as adding a little K and adding two big K. The second eigenvector is coming from lambda 2 equals minus little K minus 2 big K and again we're doing the problem A minus lambda 2I times V2, equals 0. So our first eigenvector V1 is, we can just write that as 1,1. You can read off the eigenvector here, is just that V1,1 equals V2,1. The second equation is just negative of the first equation. So that would be row 1 column 1, row 2 column 1 and that's supposed to be 0. I can write that in the way I've always been writing it as the first column of a two by two matrix. Adding little K, we have minus big K, big K, big K and minus big K times our eigenvector V1. So we need to subtract negative K from the diagonal which is the same thing as adding little K and we end up then with the matrix. So remember we're doing this two by two matrix, I write as A, so we have A minus lambda 1I times the eigenvector V1 is suppose to be 0. So we have lambda 1 is this minus little k, is the eigenvalue and we're trying to find the eigenvectors. So let's look at the two eigenvalues one by one. We did our usual ansatz of X equals VE to the RT, converted the problem to an eigenvalue problem AV equals lambda V, where lambda equals MR squared and then we've already found the two eigenvalues associated with this matrix. We've got the governing equations in terms of matrix equation. So we're solving this couple of oscillator problem.
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