Ok. You have the system
where and are both symmetric. For a system to be conservative, what must happen?
I'm going to attempt to write up your hint, here. You're saying that
.
Is that correct?
It can be shown that for n-many masses the system of masses and springs can be written in time form M x + K x = 0 ( equation 1)
where x's are vectors and the first x has two dots over it.
[K]_ij = [K]_ji => K^T = K
where K is a matrix of spring constants and M is a matrix of masses
show that equation 1 is a conservative system.
Hint: d/dt(x^T * Mx) = 2x^TMx
where the x^T 's have a dot above them and the lone x has two dots
x with a dot above it = dx/dt where x is a vector
can someone please go through this step by step? I missed this lecture in class and the teacher said he went over this and I can't figure out how to do it.
It makes more sense that it should be:
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If so, we have
[LaTeX ERROR: Convert failed]
The multiplication Mx is: an n by n matrix by a vector, say n by 1. This gives an n by 1 vector. Then x^T multiplied by Mx will give a number, since x^T is 1 by n. So, by the same reason [LaTeX ERROR: Convert failed] must be just a number. Hence it is equal to its own transpose.
[LaTeX ERROR: Convert failed]
Combining this with what we had above, we get the result
[LaTeX ERROR: Convert failed]
Correct. It's a relatively standard notation in physics: dots mean time derivatives, and primes mean spatial derivatives. The more primes or dots, the more derivatives.
So the question right now is this: what do you suppose has to happen in order for the system to be conservative?