EG260 Solutions-example_4

EG260 Solutions-example_4 - EG-260 EG-260 Dynamics I...

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EG-260 @Avril Slone 2007 EG-260 Dynamics I - Example Sheet 4 Answers Forced single degree of freedom systems 1. EOM is t F kx x c x m ω cos 0 = + + & & & The steady state solution is of the form ) cos( X θ = t x p then the magnitude of the steady state response X is given by: () 2 2 2 2 0 2 ξω n n f X + = where Resonance occurs when n = which is NOT the maximum response Then 00058 . 0 2 2 2 0 2 2 2 2 0 = = + = n res n n n n res f X f X When n 8 . 0 = (we are told to ignore damping) Magnitude of response for undamped system is: m F f 0 0 =

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EG-260 @Avril Slone 2007 () 00046 . 0 36 . 0 8 . 0 1 2 0 2 2 0 2 2 0 = = = = n n und n und f f X f X ω Therefore we have 36 . 0 00046 . 0 00046 . 0 36 . 0 2 00058 . 0 00058 . 0 2 2 0 2 0 2 2 0 2 0 × = = × = = n n n n n f f f f ξω This gives 143 . 0 58 . 0 * 2 46 . 0 * 36 . 0 = = ξ
EG-260 @Avril Slone 2007 2. Following the previous example, for zero damping 2 0 0 x ω m k f = so the steady state response is () φ + = t m k f t x sin 2 0 Hence the general solution to the equation of motion is t m k B t m k A t m k f t x cos sin sin 2 0 + + + = We evaluate the constants from the initial conditions. 2 0 sin 0 0 m k f B x = = t B t A t m k f t x n n n n sin cos cos 2 0 + + = & A m k f x n + = = 2 0 cos 0 0 0 & so that

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This note was uploaded on 05/18/2010 for the course ENGINEERIN EG 260 taught by Professor Stone during the Spring '10 term at Swansea UK.

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EG260 Solutions-example_4 - EG-260 EG-260 Dynamics I...

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