# Imaginary terms yields f t fr cos t fi sin

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Unformatted text preview: pect to frequency as follows. TODO: Show example F(w), show resulting X(w). (FRF continuous, F&X discrete) Express F(w) = CONST …. Show units of FRF = 1/k, compliance. Work example. Show MDOF FRF for comparison. Lecture Notes -3906/16/06 12:25 PM * Mechanical Vibrations I d d ( H (ω ) ) = dω dω 1 ( k − mω ) + (ω c ) 2 2 2 1 2 2 d d 2 ( k − mω 2 ) + (ω c ) H (ω ) ) = ( dω dω ( ) − 2 2 2 2 d ( H (ω ) ) = − 1 ( k − mω 2 ) + (ω c )2 2 dω ( ) − 3 2 d dω (( k − mω ) + (ωc ) ) 2 d ( H (ω ) ) = − 1 ( k − mω 2 ) + (ω c )2 dω 2 ( ) − 3 2 d 2 ( k − mω 2 ) + 2 (ω c ) dd (ω c ) 2 ( k − mω ) dω ω 2 d ( H (ω ) ) = − 1 ( k − mω 2 ) + (ω c )2 2 dω ( ) ( 2 ( k − mω ) ( −2m) + 2 (ωc ) c ) − 3 2 2 In order to satisfy the maximum (or minimum), the derivative of the magnitude of the d frequency response function, H (ω p ) = 0 , must equal zero. Since only the second dω term in the above relation can be equal to zero (except at ω = ∞ ), the solution can be reduced to solving the following equation. ( ) ( 2 ( k − mω ) ( −2mω ) + 2 (ω c ) c ) = 0 2 p p p ( −2 ( km − m ω ) + c )ω 2 2 p 2 p =0 By dividing by the mass, m 2 , the expression can be simplified by recognizing that k c Ω 2 = , 2ζΩ = and ω p = 0 cannot be the peak response for all possible MCK m m combinations. k c 2 −2 − ω p + = 0 m m 2 −2 ( Ω 2 − ω 2 ) + ( 2ζΩ ) = 0 p 2 2 Ω 2 − ω p − 2ζ 2Ω 2 = 0 ω 2 = (1 − 2ζ 2 ) Ω 2 p Lecture Notes -40- 06/16/06 12:25 PM Mechanical Vibrations I Therefore, the frequency of maximum (peak) response is as follows. ω p = 1 − 2ζ 2 Ω For comparison, recognize that the damped natural frequency is ω r = 1 − ζ 2 Ω r . Resonant Response Starting with the definition of resonance, ω = Ω , (i.e. the response of the system to an excitation at the undamped natural frequency of the system.), the amplitude of the system response can be calculated. Starting with the single degree-of-freedom frequency response function, H (ω ) = 1 −mω + jω c + k 2 from the definition of resonance, ω = Ω , (and k = mΩ 2 ) the resonant response, H (Ω) , equals H (Ω) = 1 1 1 1 1 = = = = jΩc jΩζ cc jΩζ 2 km jΩ 2ζ k j 2ζ k Ω If the damping is “small” (i.e. ζ < 0.1 ), then the peak response can be approximated as H (ω p ) ≈ H (Ω) = 1 j 2ζ k Half Power Method After defining the resonant frequency, it is possible to estimate the damping, ζ , from the shape of the frequency response function in the vicinity of the resonance. Again, assuming that the damping is “small” (i.e. ζ < 0.1 ), the half-power bandwidth points can be calculated and then damping can be estimated from the bandwidth and the resonance location. Starting with the definition of the halfpower bandwidth points, H ωp ( ) H ωp 2 ( ) Lecture Notes -41ωL ω pωH 06/16/06 12:25 PM Mechanical Vibrations I 1 2 H (ω p ) = H (ωL ) = H (ωH ) and recognizing that for a light...
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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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