Mathematical Methods for Physics and Engineering : A Comprehensive Guide

(lu) #1

NORMAL MODES


neous equations forαandβ, but they are all equivalent to just two, namely


α+β=0,

5 α+β=

4 Mω^2
k

α;

these have the solutionα=−βandω^2 =k/M. The latter thus gives the frequency


of the mode with eigenvector


x(5)=(0 1 0 − 10 −101)T.

Note that, in this mode, when the spring joining masses 1 and 3 is most stretched,


the one joining masses 2 and 4 is at its most compressed. Similarly, based on


reflection symmetry in they-axis,


x(6)=(1 0 − 10 −1010)T

can be shown to be an eigenvector corresponding to the same frequency. These


two modes are shown in diagrams (e) and (f) of figure 9.5.


This accounts for six of the expected eight modes, and the other two could be

found by considering motions that are symmetric about both diagonals of the


square or are invariant under successive reflections in thex-andy- axes. However,


sinceAis a multiple of the unit matrix, and since we know that (x(j))TAx(i)=0if


i=j, we can find the two remaining eigenvectors more easily by requiring them


to be orthogonal to each of those found so far.


Let us take the next (seventh) eigenvector,x(7), to be given by

x(7)=(abcdefgh)T.

Then orthogonality with each of thex(n)forn=1, 2 ,...,6 yields six equations


satisfied by the unknownsa,b,...,h. As the reader may verify, they can be reduced


to the six simple equations


a+g=0,d+f=0,a+f=d+g,
b+h=0,c+e=0,b+c=e+h.

With six homogeneous equations for eight unknowns, effectively separated into


two groups of four, we may pick one in each group arbitrarily. Takinga=b=1


givesd=e= 1 andc=f=g=h=−1 as a solution. Substitution of


x(7)=(1 1 − 111 − 1 − 1 −1)T.

into the eigenvalue equation checks that it is an eigenvector and shows that the


corresponding eigenfrequency is given byω^2 =k/M.


We now have the eigenvectors for seven of the eight normal modes and the

eighth can be found by making it simultaneously orthogonal to each of the other


seven. It is left to the reader to show (or verify) that the final solution is


x(8)=(1 − 111 − 1 − 1 −11)T
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