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ME429 Fall 2009 HW2 Solution

# ME429 Fall 2009 HW2 Solution - Assigned on 12.10.2009...

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Assigned on 12.10.2009 Monday Due date 19.10.2009 Monday, beginning of the lecture ME429 Fall 2009 HW2 Solution 1) In the given system the support vibrates harmonically, which implies that the mass also vibrates in the same frequency. Hence, the response amplitude of the mass can easily be determined using transmissibility formula, as follows: ( ) ( ) ( ) ζ ζ + = - + 2 2 2 2 1 2 1 2 r X Y r r ω ω = n r ω = n k m a) Since the mass response is harmonic as ( ) ω = sin x t X t , acceleration amplitude (which is indeed the maximum acceleration) of the mass can be written as ω = dotnospdotnosp 2 max x X Substituting the transmissibility equation for X, one can obtain: ( ) ( ) ( ) ζ ω ω ζ + = = - + dotnospdotnosp 2 2 2 2 2 max 2 1 2 1 2 r x X Y r r Note that it is possible to observe that, for harmonic excitation and response, relation between the accelerations of the mass and the support is the same as that between their positions. ( ) ( ) ( ) ζ ω ω ζ + = = = - + dotnospdotnosp dotnospdotnosp 2 2 2 2 2 2 1 2 1 2 r x X X y Y Y r r b) Force is transmitted through the spring and the damper. The sum of these forces is = dotnospdotnosp T F mx Note that the maximum transmitted force is obtained when the maximum acceleration is reached. So, ω = = dotnospdotnosp 2 ,max max T F mx m X

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Assigned on 12.10.2009 Monday
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ME429 Fall 2009 HW2 Solution - Assigned on 12.10.2009...

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