828_Dynamics 11ed Manual

828_Dynamics 11ed Manual - Engineering Mechanics Dynamics...

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Unformatted text preview: Engineering Mechanics - Dynamics Chapter 22 Solution: kg W ωn = ωn = 6.55 rad s ω = ωn ω = 6.55 rad s Problem 22-57 The block, having weight W, is immersed in a liquid such that the damping force acting on the block has a magnitude of F = c|v|. If the block is pulled down at a distance d and released from rest, determine the position of the block as a function of time. The spring has a stiffness k. Assume that positive displacement is downward. Given: W = 12 lb c = 0.7 k = 53 d = 0.62 ft lb s ft g = 32.2 ft 2 s lb ft Solution: ωn = kg W ζ= b = ζ ωn ωd = cg Since ζ = 0.08 < 1 the system is underdamped 2Wωn 2 1 − ζ ωn −bt y ( t) = B e We can write the solution as sin ( ωd t + φ ) To solve for the constants B and φ Guesses Given B = 1 ft B sin ( φ ) = d φ = 1 rad B ωd cos ( φ ) − b B sin ( φ ) = 0 Thus −bt y ( t) = B e sin ( ωd t + φ ) 826 ⎛B⎞ ⎜ ⎟ = Find ( B , φ ) ⎝φ ⎠ ...
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This note was uploaded on 08/16/2011 for the course EGN 3321 taught by Professor Christianfeldt during the Spring '08 term at University of Central Florida.

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