530_343lecture08

# 530_343lecture08 - ME 530.343 Design and Analysis of...

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ME 530.343: Design and Analysis of Dynamic Systems Spring 2009 Lecture 8 – Solutions to 2nd Order ODEs Wednesday, February 18, 2009 1 Today’s Objectives Bungee Jumping Example Techniques for solving homogeneous, 2 nd -order ODEs Reading: Palm 3.1, 8.1–8.2 2 Bungee Jumping y = 0 y L L n eq Some data: mass of bungee jumper m = 90 kg (linear) viscous damping const. of air b = 60 Ns/m (linear) spring const. of bungee cord k = 75 N/m local grav. const. g = 9 . 81 m/s 2 natural length of bungee cord L n = 50 m height of bungee jumper from equilibrium y ( t ) initial height of bungee jumper y (0) = 0 m initial downward speed of bungee jumper ˙ y (0) = - 30 m/s time t ζ = b 2 mk = 0 . 365 ω n = q k m = 0 . 913 rad/sec ω d = ω n p 1 - ζ 2 = 0 . 850 rad/sec y ( t ) = ( C 1 sin( ω d t ) + C 2 cos( ω d t )) e - ζω n t y ( t ) = ( C 1 sin(0 . 850 t ) + C 2 cos(0 . 850 t )) e - 0 . 333 t Constants C 1 and C 2 determined from initial values y (0) = 0, ˙ y (0) = - 30 y (0) = 0 1

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˙ y (0) = (0 . 850 C 1 cos(0 . 850 t ) - 0 . 850 C 2 sin(0 . 850 t )) e - 0 . 333 t +( C 1 sin(0 . 850 t )+ C 2 cos(0 . 850 t ))( - 0 . 333) e - 0 . 333 t ˙ y (0) = 0 . 850 C 1 = - 30 C 1 = 35 . 3 y ( t ) = - 35 . 3 sin(0 . 816 t ) e - 0 . 333 t Fear #1: Will I die because the bungee cord breaks? F max = 5000 N maximum stretch force = ky max + mg maximum stretch force = 1541 + 883 = 2533 N Fear #2: Will I die because I hit the ground (at 100 m)?
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530_343lecture08 - ME 530.343 Design and Analysis of...

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