NORMAL MODES
neous equations forαandβ, but they are all equivalent to just two, namely
α+β=0,5 α+β=4 Mω^2
kα;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)Tcan 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 befound 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 byx(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 theeighth 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