Mathematical Methods for Physics and Engineering - Matematica.NET

NORMAL MODESneous equations for α and β, but they are all equivalent to just two, namelyα + β =0,5α + β = 4Mω2 α;kthese have the solution α = −β and ω 2 = k/M. The latter thus gives the frequencyof the mode with eigenvectorx (5) =(0 1 0 − 1 0 − 1 0 1) 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 onreflection symmetry in the y-axis,x (6) =(1 0 − 1 0 − 1 0 1 0) Tcan be shown to be an eigenvector corresponding to the same frequency. Thesetwo 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 thesquare or are invariant under successive reflections in the x- andy- axes. However,since A is a multiple of the unit matrix, and since we know that (x (j) ) T Ax (i) =0ifi ≠ j, we can find the two remaining eigenvectors more easily by requiring themto be orthogonal to each of those found so far.Let us take the next (seventh) eigenvector, x (7) , to be given byx (7) =(a b c d e f g h) T .Then orthogonality with each of the x (n) for n =1, 2,...,6 yields six equationssatisfied by the unknowns a,b,...,h. As the reader may verify, they can be reducedto the six simple equationsa + 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 intotwo groups of four, we may pick one in each group arbitrarily. Taking a = b =1gives d = e = 1 and c = f = g = h = −1 as a solution. Substitution ofx (7) =(1 1 − 1 1 1 − 1 − 1 − 1) T .into the eigenvalue equation checks that it is an eigenvector and shows that thecorresponding eigenfrequency is given by ω 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 otherseven. It is left to the reader to show (or verify) that the final solution isx (8) =(1 − 1 1 1 − 1 − 1 − 1 1) T326