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Equations ( 2 7 . 6 ) and ( 2 7 . 8 ) can be substituted into Eq . ( 2 7 . 5

Equations (27.6) and (27.8) can be substituted into Eq.(27.5), which, after collection of terms, can be expressed as
(2km1-2)A1-km1A2=0
-km2A1+(2km2-2)A2=0
Comparison of Eq.(27.9) with Eq.(27.4) indicates that at this point, the solution has been reduced to an eigenvalue problem.
MPLE 27.4 Eigenvalues and Eigenvectors for a Mass-Spring System
Problem Statement. Evaluate the eigenvalues and the eigenvectors of Eq.(27.9) for the case where m1=m2=40kg and k=200Nm.
Solution. Substituting the parameter values into Eq.(27.9) yields
(10-2)A1-5A2=0
-5A1+(10-2)A2=0
The determinant of this system is [recall Eq.(9.3)]
(2)2-202+75=0
which can be solved by the quadratic formula for 2=15 and 5s-2. Therefore, the frequencies for the vibrations of the masses are =3.873s-1 and 2.236s-1, respectively. These values can be used to determine the periods for the vibrations with Eq.(27.7). For the first mode, Tp=1.62s, and for the second, Tp=2.81s.
As stated in Sec. 27.2.1, a unique set of values cannot be obtained for the unknowns. However, their ratios can be specified by substituting the eigenvalues back into the equations. For example, for the first mode )=(15s-2. For the second mode )=(5s-2.
This example provides valuable information regarding the behavior of the system in Fig. 27.5. Aside from its period, we know that if the system is vibrating in the first mode, the amplitude of the second mass will be equal but of opposite sign to the amplitude of the first. As in Fig. 27.6a, the masses vibrate apart and then together indefinitely.
In the second mode, the two masses have equal amplitudes at all times. Thus, as in Fig. 27.6b, they vibrate back and forth in unison. It should be noted that the configuration of the amplitudes provides guidance on how to set their initial values to attain pure
BOUNDARY-VALUE AND EIGENVALUE PROBLEMS
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