Forced Oscillations and Resonance — Question 5

Section 15-8 · Physics pages 2380–2381

Question 5 Static response and phase identify all three oscillator parameters An unknown linear oscillator has fixed m, k, b > 0. A constant force 2.0 N displaces its equilibrium by .040 m. With a sinusoidal force of amplitude 2.0 N, the steady displacement lags by 90◦at Ω= 10 rad/s and has amplitude .10 m there. Ω X Steady displacement-response curve; peak and width depend on damping. (a) Infer k, m, b from these three observations. (b) Predict the frequency of maximum displacement. Explain why identifying the 90◦point is different from searching for the largest displacement.
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Question 5 - Solution Static response and phase identify all three oscillator parameters Ω X (a) Derivation and calculation Static equilibrium determines stiffness: k = F/xe = 2.00/.0400 = 50.0 N/m . At a 90◦displacement lag, k −mΩ2 = 0 ⇒m = k/Ω2 = 50/100 = .500 kg . X = F0 bΩ⇒b = 2.00 10(.100) = 2.00 kg/s . (b) Result and interpretation β = b/(2m) = 2.00 s−1, ω0 = 10.0 rad/s. Ωr = √ ω2 0 −2β2 = √ 92 = 9.59166 rad/s . The phase condition locates ω0 while the displacement maximum is shifted downward by damping. Physical check The inference assumes settled response and frequency-independent linear m, k, b.

Question 5

Static response and phase identify all three oscillator parameters

An unknown linear oscillator has fixed m,k,b>0m,k,b>0. A constant force 2.0N2.0\,\mathrm{N} displaces its equilibrium by .040m.040\,\mathrm{m}. With a sinusoidal force of amplitude 2.0N2.0\,\mathrm{N}, the steady displacement lags by 90∘90^\circ at Ω=10rad/s\Omega=10\,\mathrm{rad/s} and has amplitude .10m.10\,\mathrm{m} there.

Diagram shown in the original SVG worksheet.

  1. Infer k,m,bk,m,b from these three observations.

  2. Predict the frequency of maximum displacement. Explain why identifying the 90∘90^\circ point is different from searching for the largest displacement.

Question 5 - Solution

Static response and phase identify all three oscillator parameters

Diagram shown in the original SVG worksheet.

(a) Derivation and calculation

Static equilibrium determines stiffness:k=F/xe=2.00/.0400=50.0N/m.\boxed{k=F/x_e=2.00/.0400=50.0\,\mathrm{N/m}}.At a 90∘90^\circ displacement lag,k−mΩ2=0⇒m=k/Ω2=50/100=.500kg.k-m\Omega^2=0\Rightarrow\boxed{m=k/\Omega^2=50/100=.500\,\mathrm{kg}}.X=F0bΩ⇒b=2.0010(.100)=2.00kg/s.X=\frac{F_0}{b\Omega}\Rightarrow\boxed{b=\frac{2.00}{10(.100)}=2.00\,\mathrm{kg/s}}.

(b) Result and interpretation

β=b/(2m)=2.00s−1,ω0=10.0rad/s.\beta=b/(2m)=2.00\,\mathrm{s^{-1}},\quad\omega_0=10.0\,\mathrm{rad/s}.Ωr=ω02−2β2=92=9.59166rad/s.\boxed{\Omega_r=\sqrt{\omega_0^2-2\beta^2}=\sqrt{92}=9.59166\,\mathrm{rad/s}}.The phase condition locates ω0\omega_0 while the displacement maximum is shifted downward by damping.

Physical check

The inference assumes settled response and frequency-independent linear m,k,bm,k,b.