Forced Oscillations and Resonance — Question 2

Section 15-8 · Physics pages 2374–2375

Question 2 A displacement resonance needs sufficiently weak damping For a driven oscillator with fixed force amplitude F0, the steady displacement amplitude is X = F0/ √ (k −mΩ2)2 + b2Ω2. Define ω2 0 = k/m and β = b/(2m). Ω/ω0 X/(F0/k) 1 Example displacement response for b/(2mω0) = 0.25; the peak lies below ω0. (a) Find the positive drive frequency that maximizes X, and state when it exists. (b) For m = 1 kg, k = 25 N/m and b = 4 kg/s, compare this frequency with ω0 and the free damped frequency. What if b2 ≥2mk?
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Question 2 - Solution A displacement resonance needs sufficiently weak damping Ω/ω0 X/(F0/k) 1 (a) Derivation and calculation Maximizing X is equivalent to minimizing D = (k −mΩ2)2 + b2Ω2. D′ = 2Ω[−2m(k −mΩ2) + b2]. Ωr = √ k m − b2 2m2 = √ ω2 0 −2β2 . b2 < 2mk is required for a positive-frequency peak. (b) Result and interpretation m = 1, k = 25, b = 4 : β = 2, Ωr = √ 17 = 4.12311 rad/s . ωd = √ 21 = 4.58258 rad/s, ω0 = 5.00 rad/s. b2 ≥2mk ⇒D′ > 0 (Ω> 0), Xmax = X(0) = F0/k . Physical check Displacement resonance, free damped oscillation, and undamped natural frequency are distinct measurements.

Question 2

A displacement resonance needs sufficiently weak damping

For a driven oscillator with fixed force amplitude F0F_0, the steady displacement amplitude is X=F0/(k−mΩ2)2+b2Ω2X=F_0/\sqrt{(k-m\Omega^2)^2+b^2\Omega^2}. Define ω02=k/m\omega_0^2=k/m and β=b/(2m)\beta=b/(2m).

Diagram shown in the original SVG worksheet.

  1. Find the positive drive frequency that maximizes XX, and state when it exists.

  2. For m=1kgm=1\,\mathrm{kg}, k=25N/mk=25\,\mathrm{N/m} and b=4kg/sb=4\,\mathrm{kg/s}, compare this frequency with ω0\omega_0 and the free damped frequency. What if b2≥2mkb^2\ge2mk?

Question 2 - Solution

A displacement resonance needs sufficiently weak damping

Diagram shown in the original SVG worksheet.

(a) Derivation and calculation

Maximizing XX is equivalent to minimizingD=(k−mΩ2)2+b2Ω2.D=(k-m\Omega^2)^2+b^2\Omega^2.D′=2Ω[−2m(k−mΩ2)+b2].D^{\prime}=2\Omega[-2m(k-m\Omega^2)+b^2].Ωr=km−b22m2=ω02−2β2.\boxed{\Omega_r=\sqrt{\frac km-\frac{b^2}{2m^2}}=\sqrt{\omega_0^2-2\beta^2}}.b2<2mkis required for a positive-frequency peak.\boxed{b^2<2mk}\quad\text{is required for a positive-frequency peak}.

(b) Result and interpretation

m=1,k=25,b=4:β=2,Ωr=17=4.12311rad/s.m=1,\ k=25,\ b=4:\quad\beta=2,\quad\boxed{\Omega_r=\sqrt{17}=4.12311\,\mathrm{rad/s}}.ωd=21=4.58258rad/s,ω0=5.00rad/s.\omega_d=\sqrt{21}=4.58258\,\mathrm{rad/s},\qquad\omega_0=5.00\,\mathrm{rad/s}.b2≥2mk⇒D′>0(Ω>0),Xmax=X(0)=F0/k.b^2\ge2mk\Rightarrow D^{\prime}>0\ (\Omega>0),\quad\boxed{X_{\max}=X(0)=F_0/k}.

Physical check

Displacement resonance, free damped oscillation, and undamped natural frequency are distinct measurements.