Figures

# Figures - c = 0 c = 0.01 c = 0.1 c = 1 Figure 3: ( Problem...

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MATH 2930 Summer 2009 Supplementary Figures For Final Exam Solutions 1

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0 0.5 1 1.5 2 2.5 3 -5 0 5 t X(t) Plot of first 2 terms of the solution (Problem 1.e) Figure 1: ( Problem 1 ) x ( t ) versus t . Just two terms are sufcient to produce the exact solution 2
-2 -1.5 -1 -0.5 0 0.5 1 1.5 2 -2 -1.5 -1 -0.5 0 0.5 1 1.5 2 x v v 2 /2 + (x 3 - 1)/3 + (x 4 - 1)/4 = 0 Figure 2: ( Problem 2 ) v ( t ) versus x ( t ) for the given initial conditions. The ±gure makes a closed loop as expected. 3

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0 0.2 0.4 0.6 0.8 1 0 2 4 6 8 10 k, Stiffness Constant Amplitude of Steady State Solution Steady State Amplitude Versus K for various values of damping
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Unformatted text preview: c = 0 c = 0.01 c = 0.1 c = 1 Figure 3: ( Problem 3 ) Amplitude of the steady state solution as a function of the sti±ness constant k for various values of the damping coe²cient c 4 0.2 0.4 0.6 0.8 1 0.5 1 1.5 2 x u(x) Temperature Profile For Various Values of t t = 0 t = 0.01 t = 0.05 t = 0.1 Figure 4: ( Problem 4 ) Temperature profles at various times. Note the exponential decay o± the amplitude about the average temperature 5...
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## This note was uploaded on 01/21/2011 for the course MATH 2930 taught by Professor Terrell,r during the Spring '07 term at Cornell.

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Figures - c = 0 c = 0.01 c = 0.1 c = 1 Figure 3: ( Problem...

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