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lecture_22 - MA 36600 LECTURE NOTES MONDAY MARCH 9...

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MA 36600 LECTURE NOTES: MONDAY, MARCH 9 Resonance. We know that the general solution to the nonhomogeneous equation mu 00 + k u = F ( t ) is in the form u ( t ) = u c ( t ) + U ( t ) for a transient solution u c ( t ) and a steady-state solution U ( t ). We found in the previous lecture that u c ( t ) = A cos ω 0 t + B sin ω 0 t = R cos ( ω 0 t - δ ) where ω 0 = r k m . If we try and force the mass to oscillate at a different frequency ω 6 = ω 0 by applying an outside force, assuming that F ( t ) = F 0 cos ωt then we have the solution U ( t ) = A 0 cos ωt + B 0 sin ωt = A 0 = F 0 m ( ω 2 0 - ω 2 ) and B 0 = 0 . Hence the general solution to the nonhomogeneous equation is u ( t ) = A cos ω 0 t + B sin ω 0 t | {z } transient + F 0 m ( ω 2 0 - ω 2 ) cos ωt | {z } steady-state ( ω 6 = ω 0 ) Figure 1 shows a plot of a transient solution u c ( t ), while Figure 2 shows a plot of a steady-state solution U ( t ). Figure 3 shows the solution u ( t ) = u c ( t ) + U ( t ). The “wiggles” in the graph show the effect of the transient solution on the steady-state. In general, if the amplitude F 0 of the forcing term F ( t ) = F 0 cos ωt is “small” then u ( t ) u c ( t ). But if F 0 is “large” then u ( t ) U ( t ). Figure 1. Graph of Transient Solution u c ( t ) = cos(9 t - 1) -2.5 0 2.5 5 7.5 10 12.5 15 -10 -5 5 10 1
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2 MA 36600 LECTURE NOTES: MONDAY, MARCH 9 Figure 2. Graph of Steady-State Solution U ( t ) = 9 cos t -2.5 0 2.5 5 7.5 10 12.5 15 -10 -5 5 10 Figure 3. Graph of Solution u ( t ) = u c ( t ) + U ( t ) -2.5 0 2.5 5 7.5 10 12.5 15 -10 -5 5 10 We have considered the equation u 00 + ω 2 0 u = F 0 m cos ωt when ω 6 = ω 0 , but what if
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lecture_22 - MA 36600 LECTURE NOTES MONDAY MARCH 9...

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