lecture14

Lecture14

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Unformatted text preview: 1.2 + 2x = 5 cos 4t , 2 5 dt dt 1 x (0) = , x (0) = 0 2 Example 6 - Continued Given the solution x (t ) = e −3t 25 86 50 38 cos t − sin t + − cos 4t + sin 4t 51 51 102 51 transient steady −state transient term : xc (t ) s.t. limt →∞ xc (t ) = 0 steady-state term: xp (t ) Example 7 - Transient/Steady-State Solutions The solution of d 2x dx +2 + 2x = 4 cos t + 2 sin t , 2 dt dt x (0) = 0, x (0) = x1 is given by x (t ) = (x1 − 2)e −t sin t + 2 sin t Undamped Forced Motion: β = 0 Solve the IVP d 2x + ω 2 x = F0 sin γ t , dt 2 x (0) = 0, x (0) = 0 Example A mass weighing 16 pounds stre...
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This document was uploaded on 04/06/2014.

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