ch02_section_3

ch02_section_3 - Chapter 2, Section 3 24 July 2001 Exercise...

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Exercise 2.44 Nodim time: τ = ω nat t q τ () 3 τ 2 3 cos τ . τ 0 0.04 , 2 π . .. 02468 40 20 0 q τ τ 024 2 0 q τ τ Find instant for maximum: q’ τ 2 τ . 3 sin τ . t2 t max root q’ t () t , t max 1.496 =
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ζ 0.4 ω d 1 ζ 2 q τ () 3 8 ζ 2 . 2 ζ . τ . τ 2 exp ζ τ . 3 8 ζ 2 . cos ω d τ . . 5 ζ . 8 ζ 3 . ω d sin ω d τ . . . q’ τ 2 ζ . 2 τ . ( ) exp ζ τ . ζ 38 ζ 2 . cos ω d τ . . 5 ζ . 8 ζ 3 . ω d sin ω d τ . . . ω d ζ 2 . sin ω d τ . . 5 ζ . 8 ζ 3 . ω d cos ω d τ . . . + ... . Max time based on envelope function: τ max 4 ζ N 2001 n1 N .. τ n N1 τ max . q val q τ q’ val q’ τ 051 0 100 50 0 q val n τ n max q val 0.019539 = Small positive maximum ==> τ is small 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 0.2 0 0.2 q’ val n τ n q’ val T 1 72 73 74 75 76 77 7.418•10 -3 4.561•10 -3 1.616•10 -3 1.417•10 -3 -4.54•10 -3 7.752•10 -3 = τ 75 0.37 = q val 75 0.014424 = Max positive q = 0.019537 @ τ = 0.555
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ω nat 1 ζ 0.05 λ 0.1 ω d 1 ζ 2 qt () 1 λ 2 2 ζ . ω nat . λ . ω nat 2 exp λ t . ( ) exp ζ ω nat . t . cos ω d t . λζ ω nat . ω d sin ω d t . . . . T 4 ζω nat . N 201 n1N .. t n n1 N1 T . 0 1 02 03 04 05 06 07 08 0 1 0 1 2 n t n
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M5 ω nat 50 ut () 1 M ω nat 2 .
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ch02_section_3 - Chapter 2, Section 3 24 July 2001 Exercise...

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