= = =

II = 2nwi = 2n (13.870585)( 10) = 16.656 Hz (2.32a)2

fi = 2nwz = 2n (8.620434)(10) = 13.130 Hz (2.32b)2

jj = 2nw3 = 2n (2.508981)(10) = 7.084 Hz (2.32c)2

(2.33a) (2.33b)

where Hz = Hertz = 1.0 cycle/sec. The eigenvectors corresponding to Ai to )'3 are determined as follows. For each

eigenvalue Ali = 1,2, 3), find the amplitudes Xz and X3 relative to the amplitude XI by letting Xl = 1.0. Any two of the three equations given by Eq. (2.9) can be used to solve for Xz and X3 with Xi = 1.0. From Eqs. (2.9a) and (2.9c),

(8 - A)X1 - 2Xz - 2X3 = 0 -2XI - 2Xz+ (13 - A)X3 = 0

(2.34c)

(2.34a)

(2.34b)

Solving Eqs. (2.33a) and (2.33b) for X3 and substituting that result in Eq. (2.33a) yields

X = (10 - A) _ A and X z

= (8 - ),) - X 3

(2.33c) 3 15 2

Substituting Al to A3 into Eq. (2.33c) yields: For AI = 13.870586:

XI = [1.000000 0.491779 -3.427072] For AZ = 8.620434:

Xz = [1.000000 -0.526465 0.216247] For )'3 = 2.508981:

X3 = [1.000000 2.145797 0.599712] The modes of oscillation corresponding to these results are illustrated in Figure 2.4.