Lecture_22 - MA 36600 LECTURE NOTES MONDAY MARCH 9 3 Assume for the moment that = 0 We may pick a particular solution U(t = A cos 0 t B sin 0 t for

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MA 36600 LECTURE NOTES: MONDAY, MARCH 9 3 Assume for the moment that ω ° = ω 0 . We may pick a particular solution U ( t )= A cos ω 0 t + B sin ω 0 t + F 0 m ( ω 2 0 ω 2 ) cos ωt for any A and B , so in particular choose A = F 0 m ( ω 2 0 ω 2 ) and B =0 . This gives the function U ( t )= F 0 m ( ω 2 0 ω 2 ) cos ω 0 t + F 0 m ( ω 2 0 ω 2 ) cos ωt = F 0 m ( ω + ω 0 ) cos ωt cos ω 0 t ω ω 0 . Consider what happens as ω approaches ω 0 : U ( t )= l im ω ω 0 ° F 0 m ( ω + ω 0 ) cos ωt cos ω 0 t ω ω 0 ± = F 0 2 0 lim ω ω 0 cos ωt cos ω 0 t ω ω 0 . But this limit should look familiar: lim ω ω 0 cos ωt cos ω 0 t ω ω 0 = ∂ω ² cos ωt ³ ´ ´ ´ ´ ω = ω 0 = t sin ω 0 t. To recap, the general solution to the diFerential equation mu °° + ku = F 0 cos ωt is in the form u ( t )= u c ( t )+ U ( t ), in terms of the transient solution u c ( t )= A cos ω 0 t + B sin ω 0 t = R cos ( ω 0 t δ ) µ ω 0 = k m · and the steady-state solution U ( t )= F 0 m ( ω + ω 0 ) cos ωt cos ω 0 t ω ω 0 if ω ° = ω 0 , F 0 2 0 t sin ω
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This note was uploaded on 11/30/2011 for the course MATH 366 taught by Professor Edraygoins during the Spring '09 term at Purdue University-West Lafayette.

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