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ASE 365 - Lecture 21 - Vibration absorbers F K M Or k M 0 M...

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Vibration absorbers K M F m k t i p t i e M K F x Fe Kx x M 0 0 2 0 ϖ ϖ ϖ - = = + t i e F x x k k k k K x x m M 0 0 0 0 2 1 2 1 ϖ = - - + + Or... 1 2 2 2 1 2 0 kx kx x m kx kx x m = + = + - : Rearrange . : equation Lower ! , and If - = = m k kx x x m k 2 0 1 2 1 2 0 0 ϖ ϖ . 0 0 ) det( 2 2 0 2 0 2 0 - = - - - - + = - k m k k k M k K ϖ ϖ ϖ But . or resonance, at is system the that indicates this But M K
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finite. is so frequency, resonant a not is that learned we Also, ". ! , and If " said, We 2 0 2 0 1 2 1 2 0 0 x m k kx x x m k ϖ ϖ ϖ - = = zero! be to is that therefore, conclude, We required 1 x response.) zero a produce we , resonance" " a g introducin (By . 0 0 0 0 0 0 2 1 2 1 2 1 t i t i Fe kx x e F x x k k k k K x x m M ϖ ϖ - = = = - - + + get we , since : EOM 1st From Obj165
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? and with system the of s frequencie natural the are What m k s. frequencie natural the are roots two whose in polynomial quadratic a is ) : Consider 2 2 0 det( ϖ ϖ = - M K . 0 ) ( ) det( . 0 ) det( . 0 ) det( 0 2 2 2 2 2 2 2 0 2 2 2 - < - = - = = = - = ϖ ϖ ϖ ϖ ϖ ϖ ϖ ϖ Mm k m k Kk M K M K M K , large For , When , When . above frequency another at and and zero between frequency one at zero equals Therefore 0 0 2 ) det( ϖ ϖ ϖ M K - . above one and quickly) through passed be must (which below resonance one is There 0 0 ϖ ϖ
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Nonperiodic response (undamped) . ) 0 ( ) 0 ( ) ( 0 0 v x x x F x x = = = + , ICs with , : EOMs t K M N η η F η η η x η x = + = + = = K M U KU U MU U U t U t t U t T T T T by multiply and , let : approach Modal : ) ( ) ( ) ( ) ( 2 , 1 ) ( = = + i t N k m i i i i i , : EOMs Modal η η systems. SDOF undamped for EOMs to form in identical are EOMs modal These ). ( ) ( . 2 , 1 ) ( 2 , 1 ) ( t U t i t i t N k m i i i i i i η x = = = = + using back transform Then , for , : EOMs modal Solve η η η
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t t d t m N t i i i i i i t i i i i ϖ ϖ η ϖ η τ τ ϖ ϖ τ η sin ) 0 ( cos ) 0 ( ) ( sin 1 ) ( ) ( 0 + + - = : case general in response Modal ) 0 ( ) 0 ( ) 0 ( ) 0 ( ), 0 ( ) 0 ( ) 0 ( ) 0 ( 1 1 x η η x x η η x - - = = = = U U U U : from available are ICs Modal etc. transforms Laplace solutions known of ion superposit solutions ary complement and particular : using solved be can (ODEs) EOMs Modal
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Nonperiodic response example 2k 2m m k x2 x 1 k = - - + ) ( 0 2 3 0 0 2 2 2 1 2 1 t F x x k k k k x x m m F2(t) = = = = = - 2 2 2 1 2 2 2 1 ) ( ) ( , 1 1 1 , 2 5 , F F t U t U m k m k T F N so Recall ϖ ϖ ) ( ) ( 3 3 2 2 4 15 2 2 3 2 1 1 t F k m t F k m = + = + η η η η : EOMs Modal
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( 29 : ) ( ) ( ) ( ) ( 2 2 T t u T t t u t T F t F - - - = Suppose F2 T ). ( sin 1 ) ( ), ( t u t t kT F t x t u t T F kx x m n n - = = + ϖ ϖ
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