NORMAL MODES
The potential matrix is thus constructed as
B=k
4
3 − 1 − 2000 − 11
− 13000 − 21 − 1
− 2031 − 1 − 100
0013 − 1 − 10 − 2
00 − 1 − 131 − 20
0 − 2 − 1 − 11300
− 1100 − 203 − 1
1 − 10 − 200 − 13
.To solve the eigenvalue equation|B−λA|= 0 directly would mean solvingan eigth-degree polynomial equation. Fortunately, we can exploit intuition and
the symmetries of the system to obtain the eigenvectors and corresponding
eigenvalues 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 independent
solutions of this form, corresponding to translations in thex-andy-directions.
The eigenvector for the first of these (written in row form to save space) is
x(1)=(10101010)T.Evaluation ofBx(1)gives
Bx(1)=(00000000)T,showing thatx(1)is a solution of (B−ω^2 A)x= 0 corresponding to the eigenvalue
ω^2 = 0, whatever formAxmay take. Similarly,
x(2)=(01010101)Tis a second eigenvector corresponding to the eigenvalueω^2 =0.
The next intuitive solution, again involving no internal vibrations, and, there-fore, expected to correspond toω^2 = 0, is pure rotation of the whole system
about its centre. In this mode each mass moves perpendicularly to the line joining
its position to the centre, and so the relevant eigenvector is
x(3)=1
√
2(111− 1 − 11 − 1 −1)T.It is easily verified thatBx(3)= 0 thus confirming both the eigenvector and the
corresponding eigenvalue. The three non-oscillatory normal modes are illustrated
in 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 a
mode in which all the masses move along radial lines – the so-called ‘breathing