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Unformatted text preview: he process for the second eigenvalue, λ2 = 6 , yields,
−1 x 1 0 4 − λ2 = −2 5 − λ2 x 2 0 4 − 6 −1 x1 −2 −1 x1 0 −2 5 − 6 x = −2 −1 x = 0 2 2 Again letting x1 = 1 , and solving for x2 , yields, −2 −1 1 0 −2 −1 x = 0 2 1 [ −2 −1] x = −2 − x2 = 0 2 Which yields, x2 = −2 . Therefore, the second eigenvalue/eigenvector pair is: 1 λ2 = 6 , { x}2 = −2 Lecture Notes 17 06/16/06 12:25 PM Mechanical Vibrations I Final Notes
Looking at the form of the eigenvalue problem one more time, it can be shown that it can actually be used to solve certain first order differential equations directly, those of the form y ' = Ay . Assuming that the vector { x} is of the form { x(t )} = { X } eλt , then the eigenvalue problem can be written as. [ A]{ x} = λ { x} [ A]{ x(t )} = λ { x(t )} [ A] { X } e λ t = { X } λ e λ t [ A] { X } = λ { X }
Which is really no different than the example problem solved above. There are a few issues, which will be covered later, applying this method to the second order differential equations that arise in structural dynamics, specifically the need to use statespace expansion to convert the set second order differential equations to a larger system of first order differential equations. Finally, when using MATLAB to solve eigenvalue problems, while the eigenvalues should be the same, the eigenvector will in general be scaled differently. MATLAB uses unity vector length for its scaling. Lecture Notes 18 06/16/06 12:25 PM Mechanical Vibrations I #4  Homogeneous Solution of the Equation of Motion
For the single degree of freedom (SDOF) system shown, the equation of motion is:
f(t) x(t) m c k mx(t ) + cx(t ) + k x(t ) = f (t ) This differential equation can be solved by any of several methods, including: Laplace Transforms, Fourier Transforms, or by an assumed solution. In this case, assuming a solution appropriate for a secondorder, constant coefficient differential equation. The general multidegree of freedom problem can be formulated in matrix form as: [ M ]{ x(t )} + [C ]{ x(t )} + [ K ]{ x(t )} = { f (t )}
Using a solution appropriate for a linear, constant matrix coefficient, differential equation, (namely complex exponentials), we have: { x(t )} = { X } e st & { f (t )} = { F } e st NOTE: that X, F and s are complex scalars. By focusing on the homogeneous solution (i.e. f = 0 ), we first evaluate the derivatives of the assumed solution, { x(t )} = { X } e st :
d dt d2 dt 2 { x(t )} = { x(t )} = { X } se st { x(t )} = { x(t )} = { X } s 2e st By substituting the assumed velocity and acceleration into the original matrix expression and collecting terms, we have: (NOTE: e st is non zero for all time and so may be eliminated.) [ M ] s 2 + [C ] s + [ K ] { X } = {0} We should recognize this as a form of eigenvalue problem (or a null space solution involving s.) Also, recognize that the solution { X } = {0} is the trivial solution and does not contribute any useful information. There...
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 Fall '11
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