At around 35 minutes there is an issue that you can think about and discuss here. It has to do with the origin of non-zero matrix elements in the upper right and lower left -what that means and why it might be a cool thing.
I am looking forward to seeing your questions and comments here.
From George Courcoubetis, here is a solution to the problem from first class. Starting from newton's third law!
https://drive.google.com/a/
I hope everyone had a restful winter break!
ReplyDeleteThat aside, I think the change that was made to the 4x4 matrix is the same as connecting the two masses on the ends by another spring and forming a 4-mass hoop.
Please correct me if I'm wrong, but it seems to me that each row of the matrix corresponds to the spring constants "experienced" by each mass as a result of the displacement of the row's corresponding mass (ignoring signs). Does that make sense?
For example: For row 1 of the 4x4 matrix, I picture a displacement of mass 1. As a result, mass 1 experiences an effective spring constant of |2k|, mass 2 experiences |-k|, mass 3 experiences 0 (since mass 1's displacement doesn't affect mass 3 directly), and mass 4 experiences |-k| (since it *is* connected to mass 1 in this closed circle configuration).
Hopefully I'm not totally wrong. Perhaps there's some intuitive meaning to the positive/negative signs, but since I couldn't think of any, I assume they simply alternate regardless of the system in order to form this real Hermitian matrix.
I applied my reasoning to some other systems and it seems that the sign does not have to alternate in the matrix. I think the element is positive (+) when it is the one being displaced, 0 when it is unaffected by the displaced element, and negative (-) when it is being affected by the displaced element.
DeleteI'll check more systems to test this idea though...
It seems to me the the center of mass of our system remains the same in all of our normal modes. Is this something that holds true for all systems of vibration that we'll be dealing with? Is there some greater significance that can be gleaned from this?
ReplyDeleteI agree. Furthermore, all normal modes must be independent from one another. Since the trivial mode describes the center of mass (COM) motion, then it follows that for any of the non-trivial modes, the COM cannot change its position.
DeleteThanks George. I edited your link into the blog so it can be "active". I hope to see comments on it. What do people think??
ReplyDeleteWow, I can't think of anything to add. That was very comprehensive!
ReplyDelete