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M334Lec19

# M334Lec19 - Math 334 Lecture#19 3.8 Periodically Forced...

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Math 334 Lecture #19 § 3.8: Periodically Forced Vibrations Undamped Periodically Forced Motion. This occurs when γ = 0: mu + ku = F 0 cos ωt, where F 0 is the forcing amplitude and ω is the forcing frequency. [Recall that the natural frequency of undamped free motion is ω 0 = k/m .] Double Harmonic Motion . This occurs when ω 0 = ω , in which case a particular solution has the form, U p = A cos ωt. Substitution of U p into the ODE gives - 2 A cos ωt + kA cos ωt = F 0 cos ωt A = F 0 - 2 + k = F 0 - 2 + 2 0 = F 0 m ( ω 2 0 - ω 2 ) [ ω 0 = k/m k = 2 0 ] . The general solution is u = c 1 cos ω 0 t + c 2 sin ω 0 t + F 0 m ( ω 2 0 - ω 2 ) cos ωt which is the superposition of two simple harmonic motions of different frequencies and amplitudes, what may be called double harmonic motion . If u (0) = 0 and u (0) = 0, then the solution of the IVP is u = - 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 2 0 - ω 2 ) cos ωt - cos ω 0 t = 2 F 0 m ( ω 2 0 - ω 2 ) sin ( ω 0 - ω ) t 2 sin ( ω 0 + ω ) t 2 [by a trig identity] . The term in the square brackets is a periodically varying amplitude, and the last sine function is a sinusoidal vibration. When the forcing frequency is close to the natural frequency, the sum ω 0 + ω is much larger than the difference ω 0 - ω , so the resulting motion is a rapid sinusoidal vibration with a slowly varying periodic amplitude, what is called a beat or amplitude modulation (AM). Example. Solve the IVP 9 . 082 u + 890 . 0 u = 0 . 5 cos 8 . 799 t, u (0) = 0 , u (0) = 0 .

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Since the natural frequency of the free motion is ω 0 = k/m = 9 . 899, the forcing frequency is ω = 8 . 799, and forcing amplitude is F 0 = 0 . 5, the solution of the IVP is (approximately) u = - 0 . 02677 cos(9 . 899 t ) + 0 . 02677 cos(8 . 799 t ) = 0 . 0590 sin(0 . 550 t ) sin(9 . 349 t ) . Here is the graph of this solution. 0.02 0.01 0.0 ! 0.01 ! 0.02 ! 0.03 ! 0.04 t ! 0.05 0.03 ! 0.06 15.0 12.5 0.05 10.0 7.5 5.0 2.5 0.0 0.06 0.04 Resonant Motion . This occurs when ω = ω 0 . Because F 0 cos ωt is a solution of the associated homogeneous ODE, the form of a par- ticular solution is U p = At cos ω 0 t + Bt sin ω 0 t. The general solution is u = c 1 cos ω 0 t + c 2 sin ω 0 t + F 0 2 0 t sin ω 0 t.
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