Complementary portion of the solution is sometimes

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Unformatted text preview: 2j jω at − jω at jω at − jω at e +e e +e f (t ) = A − jB 2 2 Collecting like terms: B B A A f (t ) = − j e jωat + + j e− jωat 2 2 2 2 Therefore, representing the characterisitics of the force at ω a with a complex-valued magnitude yields the general form for the forcing function: { f (t )} = {F } e jω t + {F *} e− jω t a a The response(s) to the previous forcing function(s) will be of the same form: { x (t )} = { X } e { x (t )} = jω { X } e { x (t )} = −ω { X } e p p a jω at jω at − jω a { X * } e− jωat 2 − ω a { X *} e − jωat + { X *} e − jωat p 2 a jω at Substituting the above relationships into the matrix equation of motion and collecting like terms gives the following form for the positive frequency terms ( e jωat ). Note that the portion of the solution involving the negative frequency terms ( e − jωat ) is the complex conjugate of the positive frequency and provides no new information. Lecture Notes -76- 06/16/06 12:25 PM Mechanical Vibrations I 2 −ω a [ M ]{ X } + jω a [C ] + [ K ]{ X } = { F } Note that the above linear, algebraic matrix equation involves N independent equations, as long as the system is not undamped with the forcing frequency ω a equal to one of the natural frequencies. Since the complex-valued forcing vector { F } provides N known pieces of information, this system of equations can be solved for the complex-valued response vector { X } . Note also that, if more than one forcing frequency is present, the above particular solution process must be repeated for each forcing frequency. An alternative to the traditional method of undetermined coefficients is the frequency response function approach. In this approach, the forcing function(s) are described, as above, in the frequency domain. The frequency response function(s) between the forcing degrees-of-freedom (DOFs) and the response degrees-of-freedom (DOFs) are computed from: [ H (ω a )] = −ω a2 [ M ] + jω a [C ] + [ K ] −1 The responses caused by the forcing functions can now be found (in the frequency domain) by: { X (ω a )} = [ H (ω a )]{F (ω a )} The particular response(s) can now be formulated in the time domain by converting the frequency domain information back to the time domain using the Euler identities (sine and cosine). Note also that, if more than one forcing frequency is present, the above particular solution process must be repeated for each forcing frequency. Total Solution The total solution, therefore, can be formulated as: { x(t )} = { xc (t )} + { x p (t )} { x(t )} = ∑ (α r {ψ r } eλ t + α r* {ψ r*} eλ t ) + ({ X } e jω t + { X *} e− jω t ) N r * r a a r =1 Lecture Notes -77- 06/16/06 12:25 PM Mechanical Vibrations I Example Solution: For the following three degree of freedom system, formulate the complete solution for the following initial conditions and forcing function. 10 0 0 [ M ] = 0 14 0 0 0 12 50 −30 0 [C ] = −30 55 −25 0 −25 25 0 5000 −3000 −3000 5500 −2500 [K ]...
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This note was uploaded on 09/29/2013 for the course MECHANICAL ME taught by Professor Regalla during the Fall '11 term at Birla Institute of Technology & Science, Pilani - Hyderabad.

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