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Resonant forcing at Ω=ω0 and the two regimes of the amplitude maximum

Example

Consider the forced model m x′′+c x′+k x=f0cos⁡(Ωt) on time domain I=R with the values and conclusions below (The damped driven oscillator model: second-order equation, phase-plane system and characteristic data, Sinusoidal forcing: the steady response, its amplitude and phase, and resonance at Ω=k/m, The amplitude of the sinusoidally forced damped oscillator: location and value of its maximum).

  • (i) Resonance: m=k=f0=1, c=0, Ω=ω0=1. Then xp(t)=12tsin⁡t is a solution of x′′+x=cos⁡t, it satisfies xp(0)=0, xp′(0)=0, and ∣xp∣ is unbounded on [0,∞).
  • (ii) Amplitude peak above zero frequency: m=k=f0=1, c=1. Then u∗=1−12=12>0, the maximum of A(Ω)=1/(1−Ω2)2+Ω2 is attained at Ω∗=1/2, and A(Ω∗)=2/3, while A(0)=A(1)=1.
  • (iii) Amplitude peak at zero frequency: m=k=f0=1, c=2. Then u∗=1−2=−1≤0, and the maximum of A(Ω)=1/(1−Ω2)2+4Ω2 on [0,∞) is attained at Ω=0 with value 1, while A(1)=1/2.

Facts & Assumptions

[L1]

In the resonant case c=0, Ω=ω0=k/m the function xp(t)=f0tsin⁡(ω0t)/(2mω0) is a solution of the forced model on R and is unbounded on [0,∞) when f0≠0 (Sinusoidal forcing: the steady response, its amplitude and phase, and resonance at Ω=k/m clause 2). [given]

[L2]

If c>0 the amplitude of the steady response is A(Ω)=∣f0∣/(k−mΩ2)2+c2Ω2; its maximum is attained at Ω∗=u∗ when u∗=k/m−c2/(2m2)>0, with value 2m∣f0∣/(c4mk−c2), and at Ω=0 when u∗≤0 (The amplitude of the sinusoidally forced damped oscillator: location and value of its maximum; the connection to the steady response is clause 1 of Sinusoidal forcing: the steady response, its amplitude and phase, and resonance at Ω=k/m). [given]

[L3]

(sin⁡t)′=cos⁡t, (cos⁡t)′=−sin⁡t; sums and products are differentiated by the algebra of derivatives and the chain rule. [thm-sine-and-cosine-derivatives, thm-algebra-of-derivatives, thm-chain-rule]

Verification

technique · direct; substitute the resonant candidate, and evaluate the amplitude formula in the two regimes of the preceding proposition
1.1givenL1L3algebra

Line (i): with m=ω0=f0=1 the candidate is xp(t)=12tsin⁡t; by [L3], xp′(t)=12(sin⁡t+tcos⁡t) and xp′′(t)=12(2cos⁡t−tsin⁡t)=cos⁡t−12tsin⁡t, so xp′′+xp=cos⁡t, as required, and xp(0)=0=xp′(0). At tk=π/2+2πk one has sin⁡tk=1 and ∣xp(tk)∣=tk/2→∞, so xp is unbounded on [0,∞), in agreement with [L1].

1.2givenL2algebra

Line (ii): here u∗=1−1/2=1/2>0 and 4mk−c2=3, so [L2] locates the maximum at Ω∗=1/2 with value 2/3; directly, (1−Ω∗2)2+Ω∗2=(1/2)2+1/2=3/4, so A(Ω∗)=2/3, while A(0)=1 and A(1)=1/0+1=1, both below 2/3.

2.1givenL2algebra∎

Line (iii): here u∗=1−2=−1≤0, so [L2] locates the maximum at 0 with value A(0)=1/1=1; directly, A(1)=1/(0)2+4=1/2<1, and the same formula gives A(Ω)≤1 for all Ω≥0 because (1−Ω2)2+4Ω2=1+2Ω2+Ω4≥1.

Depends on

Dependency tree · 2 levels

Sources