homework11sol.20100508.4be57476291ac1.17356704

homework11sol.20100508.4be57476291ac1.17356704 - ( t ) + c...

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TAM 412, Homework 11, Selected answers Harry Dankowicz Mechanical Science and Engineering University of Illinois at Urbana-Champaign [email protected] 1. Here, ˜ ω + - ˜ ω ˜ ω = r 1 - 2 ζ 2 + 2 ζ R 1 - ζ 2 - r 1 - 2 ζ 2 - 2 ζ R 1 - ζ 2 R 1 - 2 ζ 2 = 2 ζ + O ( ζ 3 ) 2. If the non-dimensional energy ˜ E = 1 2 ( ˜ x 2 + ˜ x 2 ) is periodic in ˜ t with period 2 π/ ˜ ω then i ˜ t +2 π/ ˜ ω ˜ t ˜ E ( τ ) = ˜ E ( ˜ t + 2 π/ ˜ ω ) - ˜ E ( ˜ t ) = 0 On the other hand i ˜ t +2 π/ ˜ ω ˜ t ( - ˜ ωA sin (˜ ωτ - φ ) cos ˜ ωτ - 2 ζ ˜ ω 2 A 2 sin 2 ωτ - φ ) ) = (sin φ - 2 A ˜ ωζ ) , i.e., sin φ = 2˜ ωζA 3. Let y denote the displacement of the block relative to the stationary Foor and let l 0 denote the natural length of the suspension spring. Moreover, suppose that the particle reaches its maximum vertical height relative to the center of the block when ωt mod 2 π = π/ 2. Then, M ¨ y
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Unformatted text preview: ( t ) + c ˙ y ( t ) + ky ( t ) = kl-Mg + meω 2 sin ωt and with y ( t ) = x ( t ) + l-Mg k one obtains the equation for x ( t ). The steady-state amplitude is then given by A = meω 2 r ( k-Mω 2 ) 2 + ( cω ) 2 which implies that 2 ζ A | ω = √ k m = lim ω →∞ A 4. In terms of the damping factor ζ and the ratio r between ω and R k/m , the steady-state amplitude is given by A = Y ± 1 + 4 r 2 ζ 2 1 + 4 r 2 ζ 2 + r 2 ( r 2-2) In particular, the term r 2 ( r 2-2 ) is positive for r > √ 2. 3. An approximate solution of period 2 π is here given by x ( t ) =-1 . 57 cos t-. 00677 cos3 t 1...
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This note was uploaded on 11/17/2011 for the course TAM 412 taught by Professor Weaver during the Spring '08 term at University of Illinois, Urbana Champaign.

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