00:01
Here i'll review how to solve a system of two coupled oscillators using what are called the k, the spring stiffness constant matrix, and the mass matrix.
00:13
So what is true is you can put both of the motions of two masses in as a vector, which we're going to call x.
00:21
So x is a vector, x1 and x2, which would stay decoupled if there weren't springs between them.
00:31
And k represents the stiffness acting between them.
00:37
And here we simply mean a positive number.
00:41
Okay, so you've got to kind of think through.
00:43
Well, mass 1 is going to be pushed by both spring 1 and spring 2.
00:52
So for that position, we have k1 plus k2 acting on mass 1.
01:01
And then between them, we have k2 is going to be in the opposite direction off the diagonal.
01:14
Okay, so equal and opposite forces going on on the two masses.
01:19
And then mass two is going to be pushed by k3 plus k2.
01:25
And typically, you use the diagonal for the positive ks and the off diagonals for the negatives.
01:33
And that can get a little tricky, but it sort of depends whether you want to write hook's law with the negative in front of that k matrix or not.
01:42
Okay, the mass matrix, on the other hand, is usually you don't have masses coupled, so it's just the identity matrix with each of the masses multiplied by the diagonal ones.
02:00
And then usually what you do is the same thing you would in a second, order differential equation.
02:06
That is, you can assume an exponential solution for the x vector in this case.
02:19
And here we're going to write down x as an oscillating solution with an amplitude out in front of it as a vector, and then an oscillating exponential.
02:33
And then we can turn this into an algebra equation.
02:37
So x double dot is equal to minus omega squared, e to the, well, there's an a in front, and then e to the i omega t.
02:52
And if we put that into our equation, then we wind up with the k matrix.
03:04
Okay, let's just leave that as the k matrix for now.
03:09
The minus sign in front of it, dot, a, e to the i omega t is equal to minus m omega squared.
03:26
Whoops, the m is matrix.
03:29
The omega squared is just a scalar times a, e to the i omega t.
03:46
Okay, so what's good about this is we are left with ignoring the e to the i omega t, and we now have to solve an algebra equation, which is basically either, yeah, it doesn't matter which side.
04:04
You bring over to which side.
04:05
So we'll start with the m -omega -squared minus k dot a is equal to zero.
04:19
So we've carried over our two matrices with the scalar in front.
04:23
All right.
04:24
So this is where we are setting up to find the, eigenvectors and the eigenvalues of the matrix.
04:42
Okay, so that's the technique that kind of this relates to.
04:46
But what we know is in order for that left -hand side dotted into an arbitrary vector a, that the determinant, if that's going to be equal to zero, the determinant of m omega -squared minus kx2.
05:07
Has got to equal 0.
05:13
And that will give us a quadratic equation that we can solve for omega squared.
05:26
So omega squared is the variable.
05:28
It'll wind up with a fourth power in it.
05:30
But we'll kind of work that out.
05:33
So here's an example.
05:36
And then we'll go work out the eigenvectors as well.
05:40
What do we have? we have k1.
05:47
K1 is, 5, k2 is 10, and k3 is 5.
06:02
And both masses are equal, so m1, equal m2, equal 5.
06:10
So we could do some things with those fives, but we'll go ahead and work out the k matrix.
06:17
And like i said, i'm defining it where the on -diagonal pieces are positive.
06:23
So we basically have k1 plus k2, so that's 50, and then we have minus 10, minus 10, and k2 plus k3 is 15.
06:37
And the m matrix is simply just five on the diagonals and zero off.
06:47
Okay, so now we can put that together.
06:50
M omega squared minus k is a new matrix with five omega squared.
07:01
Minus 15, minus 10.
07:06
Minus 10, there's nothing on the diagonal...