Mathematical Methods for Physics and Engineering - Matematica.NET

NORMAL MODESThe potential matrix is thus constructed as⎛⎞3 −1 −2 0 0 0 −1 1−1 3 0 0 0 −2 1 −1−2 0 3 1 −1 −1 0 0B = k 0 0 1 3 −1 −1 0 −2.40 0 −1 −1 3 1 −2 0⎜0 −2 −1 −1 1 3 0 0⎟⎝ −1 1 0 0 −2 0 3 −1 ⎠1 −1 0 −2 0 0 −1 3To solve the eigenvalue equation |B − λA| = 0 directly would mean solvingan eigth-degree polynomial equation. Fortunately, we can exploit intuition andthe symmetries of the system to obtain the eigenvectors and correspondingeigenvalues without such labour.Firstly, we know that bodily translation of the whole system, without anyinternal vibration, must be possible and that there will be two independentsolutions of this form, corresponding to translations in the x- andy- directions.The eigenvector for the first of these (written in row form to save space) isEvaluation of Bx (1) givesx (1) = (1 0 1 0 1 0 1 0) T .Bx (1) = (0 0 0 0 0 0 0 0) T ,showing that x (1) is a solution of (B − ω 2 A)x = 0 corresponding to the eigenvalueω 2 = 0, whatever form Ax may take. Similarly,x (2) = (0 1 0 1 0 1 0 1) Tis a second eigenvector corresponding to the eigenvalue ω 2 =0.The next intuitive solution, again involving no internal vibrations, and, therefore,expected to correspond to ω 2 = 0, is pure rotation of the whole systemabout its centre. In this mode each mass moves perpendicularly to the line joiningits position to the centre, and so the relevant eigenvector isx (3) = 1 √2(1 1 1 − 1 − 1 1 − 1 − 1) T .It is easily verified that Bx (3) = 0 thus confirming both the eigenvector and thecorresponding eigenvalue. The three non-oscillatory normal modes are illustratedin diagrams (a)–(c) of figure 9.5.We now come to solutions that do involve real internal oscillations, and,because of the four-fold symmetry of the system, we expect one of them to be amode in which all the masses move along radial lines – the so-called ‘breathing324