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Lecture Note Sketches Hermann Riecke - ESAM Home Page

Lecture Note Sketches Hermann Riecke - ESAM Home Page

we get y(t) = δ cosω0t

we get y(t) = δ cosω0t + ˆf ω2 (cosωt − cosω0t) 0 − ω2 ˆf 2 = δ cosω0t + ω0 + ω ω0 − ω sin ω0 − ω t 2 � �� � A(t) sin ω0 + ω t 2 For ω0 − ω ≪ ω0 the amplitude A(t) varies much more slowly than sin 1 2 (ω0 + ω)t: beating. For ω → ω0 one gets Notes: y(t) → δ cos ω0t + ˆ f t sin ωt 2ω • at the resonance ω = ω0 the forcing leads to a linear, unbounded growth of the oscillations ii) linear case with damping: α = 0 General solution Notes: ¨y + ˆ β ˙y + ω 2 0 y = ˆ f cos ωt y(0) = δ, ˙u(0) = 0 y(t) = yh(t) + ���� →0 for t→ ∞ ˆ f (ω2 0 − ω2 ) cosωt + ˆ βω sin ωt (ω2 0 − ω2 ) 2 + ˆ β2ω2 • the homogeneous solution consists of a damped oscillation yh(t) = e −σt (a cosωβt + b sin ωβt) with σ = 1 2 ˆ β ωβ = with a and b determined by the initial conditions. � ω 2 0 − 1 4 ˆ β 2 • the damping leads to a lag of the phase of the oscillation relative to the forcing • for large times the initial condition becomes irrelevant since yh → 0 for t → ∞ • the approach to the steady state is oscillatory in the amplitude of oscillation, reflecting the beating obtained without damping 20 (5)

10 5 −10 0 0 −5 t 50 100 150 200 Figure 4: Solution of the linear case with damping for β = 0.05, ω0 = 1, ω = 1.1, δ = 0.1, f = 1, α = 0. In the steady state reached for t → ∞ the amplitude of the oscillation is given by ˆf R∞ = (ω2 0 − ω2 ) 2 + ˆ β2ω2 � (ω2 0 − ω2 ) 2 + ˆ β2ω2 = Frequency of maximal amplitude Amplitude 12 10 8 6 4 2 −4ωmax ˆf � (ω 2 0 − ω 2 ) 2 + ˆ β 2 ω 2 � 2 ω0 − ω 2 � max + 2βˆ 2 ωmax = 0 ω 2 max = ω2 1 0 − 2 ˆ β 2 Rmax = β 2 =0.01 β 2 =0.1 β 2 =0.2 β 2 =0.5 β 2 =1 β 2 ^2=2 β 2 =16 0 0 1 Frequency ω 2 ˆf � ˆβ ω2 0 − 1 4 ˆ β2 Figure 5: Response curve R∞(ω) for ω0 = 1. 21

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