- #1
boeing_737
- 12
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Hi all,
I have a question about Multi DOF vibrating systems. For free vibration of undamped MDOF systems, we have the equations of motion as :
[itex]M[/itex] [itex]\ddot{q}[/itex] + [itex]K[/itex] [itex]{q}[/itex] = [itex]{0}[/itex] (1)
Where,
[itex]M[/itex] - n x n mass matrix
[itex]K[/itex] - n x n stiffness matrix
[itex]{q}[/itex] - n x 1 vector of generalized coordinates
Most vibrations book try to obtain the eigenvalue problem by assuming the solution to (1) as
[itex]{q}[/itex] = [itex]{Q}[/itex] [itex]e^{j \omega t}[/itex] (2)
For a 2-DOF system, (2) is
[itex]q_{1}[/itex] = [itex]Q_{1}[/itex] [itex]e^{j \omega t}[/itex] (scalar eqn, 1st component of [itex]{q}[/itex])
[itex]q_{2}[/itex] = [itex]Q_{2}[/itex] [itex]e^{j \omega t}[/itex] (scalar eqn, 2nd component of [itex]{q}[/itex])
My question is why do we assume the same exponent for all components of [itex]{q}[/itex]? Why not assume (for a 2 DOF system) the solution as
[itex]q_{1}[/itex] = [itex]Q_{1}[/itex] [itex]e^{j \omega_{1} t}[/itex]
[itex]q_{1}[/itex] = [itex]Q_{1}[/itex] [itex]e^{j \omega_{2} t}[/itex]
ie, [itex]{q}[/itex] = [itex]W[/itex] [itex]Q[/itex], where [itex]W[/itex] is a diagonal matrix containing the exponent terms?
Thanks,
yogesh
I have a question about Multi DOF vibrating systems. For free vibration of undamped MDOF systems, we have the equations of motion as :
[itex]M[/itex] [itex]\ddot{q}[/itex] + [itex]K[/itex] [itex]{q}[/itex] = [itex]{0}[/itex] (1)
Where,
[itex]M[/itex] - n x n mass matrix
[itex]K[/itex] - n x n stiffness matrix
[itex]{q}[/itex] - n x 1 vector of generalized coordinates
Most vibrations book try to obtain the eigenvalue problem by assuming the solution to (1) as
[itex]{q}[/itex] = [itex]{Q}[/itex] [itex]e^{j \omega t}[/itex] (2)
For a 2-DOF system, (2) is
[itex]q_{1}[/itex] = [itex]Q_{1}[/itex] [itex]e^{j \omega t}[/itex] (scalar eqn, 1st component of [itex]{q}[/itex])
[itex]q_{2}[/itex] = [itex]Q_{2}[/itex] [itex]e^{j \omega t}[/itex] (scalar eqn, 2nd component of [itex]{q}[/itex])
My question is why do we assume the same exponent for all components of [itex]{q}[/itex]? Why not assume (for a 2 DOF system) the solution as
[itex]q_{1}[/itex] = [itex]Q_{1}[/itex] [itex]e^{j \omega_{1} t}[/itex]
[itex]q_{1}[/itex] = [itex]Q_{1}[/itex] [itex]e^{j \omega_{2} t}[/itex]
ie, [itex]{q}[/itex] = [itex]W[/itex] [itex]Q[/itex], where [itex]W[/itex] is a diagonal matrix containing the exponent terms?
Thanks,
yogesh