Page 471 - Dynamics of Mechanical Systems

0593_C13_fm Page 452 Monday, May 6, 2002 3:21 PM

452 Dynamics of Mechanical Systems

To solve Eq. (13.7.4), let x have the matrix form:

ω
x = Ae it (13.7.7)
where ω is a frequency to be determined and A is the array:
 A 
1
A =   A 2   (13.7.8)
  A  
3
Equivalently, the coordinates of x may be written as:

ω
ω
ω
x = A e it , x = A e it , x = A e it (13.7.9)
1 1 2 2 3 3
By substituting from Eq. (13.7.7) into (13.7.4) we have:
+
2
−ω MA KA = 0 (13.7.10)
or
+ )
2
− ( ω MK A = 0 (13.7.11)
That is,

 − ( mω 2 + ) k 2 −k 0   A 
   1 
 −k − ( mω 2 + ) k 2 −k  A 2  = 0 (13.7.12)

 − ( mω 2  A 
k 2
 0 −k + )  33 

Equation (13.7.12) is equivalent to the scalar equations:
− ( mω 2 + ) k A − kA = 0 (13.7.13)
2
1 2
2
−kA + − ( mω 2 + ) k A − kA = 0 (13.7.14)
1 2 3
2
k
−kA + − ( mω 2 + ) = 0 (13.7.15)
2
As with the eigenvalue problem we encountered in Sections 7.7, 7.8, and 7.9 in studying
inertia, we have a system of simultaneous linear algebra equations for the amplitudes A ,
1
A , and A . The system is homogeneous in that the right sides are zero. Thus, we have a
2
3
nontrivial (or nonzero) solution only if the determinant of the coefficients is zero. Hence,
we have:
 − ( mω 2 + ) k 2 −k 0 
 
det  −k − ( mω 2 + ) k 2 −k  = 0 (13.7.16)
 − ( mω 2 
 0 −k + ) k 2 