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ASE 365 - Lecture 6 - SDOF harmonic response x k m F(t = F...

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SDOF harmonic response EOM: A = F/k represents excitation amplitude in displacement units (static response). k c m x F(t) = F cos ω t (or F sin ω t)
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becomes EOM so , and Then t X t x t X t x p p ϖ ϖ ϖ ϖ cos ) ( sin ) ( 2 - = - = t kA t X k t X c t X m ϖ ϖ ϖ ϖ ϖ ϖ cos ) cos ( ) sin ( ) cos ( 2 = + - + - case. undamped the in do they as cancel simply not do terms dependent - time the case, damped the For ??? ???
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? about How t X t X t x s c p ϖ ϖ sin cos ) ( + = ) sin cos ( ) ( ) cos sin ( ) ( 2 t X t X t x t X t X t x s c p s c p ϖ ϖ ϖ ϖ ϖ ϖ + - = + - = , Then or , becomes EOM t kA t X t X k t X t X c t X t X m s c s c s c ϖ ϖ ϖ ϖ ϖ ϖ ϖ ϖ ϖ cos ) sin cos ( ) cos sin ( ) sin cos ( 2 = + + + - + + - t kA t kX X c X m t kX X c X m s c s c s c ϖ ϖ ϖ ϖ ϖ ϖ ϖ cos sin ) ( cos ) ( 2 2 = + - - + + + -
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. and for , equations 2 the solve , From s c s c s c s c s c s c s c X X kX X c X m kA kX X c X m t kA t kX X c X m t kX X c X m 0 cos sin ) ( cos ) ( 2 2 2 2 = + - - = + + - = + - - + + + - ϖ ϖ ϖ ϖ ϖ ϖ ϖ ϖ ϖ ϖ ϖ balance" harmonic of method "
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response. the and excitation the between lag phase the is where , try important, is response the of amplitude the since Or, φ φ ϖ ) cos( ) ( - = t X t x p ) sin sin cos (cos ) cos( ) ( ) sin cos cos (sin ) sin( ) ( ) sin sin cos (cos ) cos( ) ( 2 2 φ ϖ φ ϖ ϖ φ ϖ ϖ φ ϖ φ ϖ ϖ φ ϖ ϖ φ ϖ φ ϖ φ ϖ t t X t X t x t t X t X t x t t X t X t x p p p + - = - - = - - = - - = + = - = , , Then t kA t t kX t t X c t t X m ϖ φ ϖ φ ϖ φ ϖ φ ϖ ϖ φ ϖ φ ϖ ϖ cos )
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