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Problem set — damping, resonance, and the quality factor

Problem set: damping, resonance, and the quality factor

Exam-style problems. Work them with a pen before touching a computer; the last two are meant to be finished numerically. Solutions and marking rubrics live with the instructor material and are deliberately not on this site.

Throughout, γ≡b/m\gamma \equiv b/m, so amplitudes decay as e−γt/2e^{-\gamma t/2} and Q=ω0/γQ = \wnat/\gamma. If you are checking against a textbook that defines γ=b/2m\gamma = b/2m, say so in your working and convert once, at the start.

Problem 1 — classifying a decay

A cart of mass 0.50 kg0.50\ \mathrm{kg} on a spring of stiffness 8.0 N m−18.0\ \mathrm{N\,m^{-1}} is released from x0=0.10 mx_0 = 0.10\ \mathrm{m} at rest. The damping coefficient is b=0.20 kg s−1b = 0.20\ \mathrm{kg\,s^{-1}}.

(a) Compute ω0\wnat, γ\gamma, ωd\omega_d and QQ. Which regime is this, and by what margin?

(b) By what factor has the amplitude fallen after ten full oscillations? After how many oscillations has the energy fallen to 1%1\% of its initial value?

(c) What value of bb would make this system critically damped? Show that at that value the solution (C+Dt)e−ω0t(C + Dt)e^{-\wnat t} is what the underdamped form becomes as ωd→0\omega_d \to 0, rather than a separate case bolted on.

(d) An overdamped oscillator crosses x=0x = 0 at most once. Prove it from the general overdamped solution c+er+t+c−er−tc_+e^{r_+t} + c_-e^{r_-t} with r±r_\pm both real and negative.

Problem 2 — the peak that is not where you expect

(a) Starting from mx¨+bx˙+kx=F0cos⁡ωtm\ddot{x} + b\dot{x} + kx = F_0\cos\omega t, substitute x=Re⁡ ⁣[Xe−iωt]x = \Real\!\left[X e^{-\ii\omega t}\right] and derive X=(F0/m)/(ω02−ω2−iγω)X = (F_0/m)/(\wnat^2 - \omega^2 - \ii\gamma\omega) in three lines. State exactly where the assumption “the transient has died” entered.

(b) Maximise ∣X∣|X| and obtain ωpeak=ω01−1/(2Q2)\omega_{\text{peak}} = \wnat\sqrt{1 - 1/(2Q^2)}. Show that a real maximum exists only for Q>1/2Q > 1/\sqrt{2}, and say in words what the response curve looks like below that threshold.

(c) Show that ∣X(ω0)∣=Q F0/k|X(\wnat)| = Q\,F_0/k exactly — no approximation — and that the phase lag there is exactly π/2\pi/2 whatever the damping.

(d) A structural engineer reports “the resonant frequency of this footbridge is 1.85 Hz1.85\ \mathrm{Hz}” from a swept-force test, and separately measures Q=25Q = 25 from a ringdown. By how much does the bridge’s ω0\wnat differ from the reported figure? Was the engineer wrong to conflate them?

Problem 3 — energy bookkeeping at resonance

(a) Show that the energy of a lightly damped free oscillator decays as e−γte^{-\gamma t}, and hence that Q=2πE/∣ΔE∣per cycleQ = 2\pi E / |\Delta E|_{\text{per cycle}} reduces to ω0/γ\wnat/\gamma. Where in that derivation did you need Q≫1Q \gg 1?

(b) For the driven steady state, compute the cycle average of the power delivered by the drive, ⟨Fx˙⟩\langle F\dot{x}\rangle, and show it equals ⟨bx˙2⟩\langle b\dot{x}^2\rangle. Confirm the result is 12γmω2∣X∣2\tfrac12\gamma m\omega^2|X|^2.

(c) Show that this power peaks at exactly ω0\wnat for any damping, and explain the physical difference between that statement and the result of Problem 2(b). Which of the two would a microphone measuring loudness report?

(d) Near resonance, reduce ⟨P⟩\langle P\rangle to a Lorentzian and identify its full width at half maximum. A quartz crystal at 32 768 Hz32\,768\ \mathrm{Hz} has Q=105Q = 10^5: what is its half-power bandwidth in hertz, and how long does it ring after a kick?

Problem 4 — the RLC translation

A series circuit has L=10 mHL = 10\ \mathrm{mH}, C=100 nFC = 100\ \mathrm{nF} and R=20 ΩR = 20\ \Omega, driven by V(t)=V0cos⁡ωtV(t) = V_0\cos\omega t.

(a) Write the equation for the charge qq on the capacitor and identify the mechanical analogue of each term. What plays the role of mass, of damping, of stiffness?

(b) Compute the resonant frequency in hertz, QQ, and the half-power bandwidth. Is this circuit good enough to separate two AM stations 10 kHz10\ \mathrm{kHz} apart at 1 MHz1\ \mathrm{MHz}?

(c) You need to double QQ without changing the resonant frequency. Give two different ways, and state a practical cost of each.

(d) The current, not the charge, is what a receiver measures. At which frequency is the current amplitude largest, and why is that not the same question as (b)?

Problem 5 — resonance on trial, numerically

Using wavelab.oscillators (in the browser laboratory or locally):

(a) Compute amplitude-response curves with steady_state_response at Q=0.6Q = 0.6, 1.0, 3.0 and 20. Locate each maximum numerically and plot ωpeak/ω0\omega_{\text{peak}}/\wnat against 1/Q21/Q^2 alongside the predicted line. At which of your four values does the peak disappear, and does the numerical maximum warn you or fail quietly?

(b) Integrate from rest at exact resonance with simulate at Q=10Q = 10. Extract the amplitude envelope and compare it with Q(F0/k)(1−e−γt/2)Q(F_0/k)\left(1 - e^{-\gamma t/2}\right). How many periods pass before the amplitude is within 5%5\% of its final value, and how does that compare with Q/πQ/\pi?

(c) Repeat (b) at Q=40Q = 40. Confirm that the steady-state amplitude is four times larger and that reaching it takes four times as long. Write two sentences on what this trade-off means for a radio receiver, a musical instrument, and a shock absorber.

Problem 6 — challenge: designing a shock absorber

A car body of 400 kg400\ \mathrm{kg} per wheel sits on a spring chosen so that the suspension’s undamped natural frequency is 1.2 Hz1.2\ \mathrm{Hz}. Hitting a kerb displaces it by 8 cm8\ \mathrm{cm}; the ride is judged acceptable once the displacement stays within 1 mm1\ \mathrm{mm}.

(a) Compute the settling time to 1 mm1\ \mathrm{mm} as a function of bb, for bb from well under to well over critical, and find the value that minimises it.

(b) You will find the optimum lies slightly below critical damping, not at it. Explain why — what does a small amount of overshoot buy?

(c) Real suspensions are tuned nearer Q≈1Q \approx 1 than to the settling-time optimum. Name one thing the model above ignores that could account for the difference, and say which specification bullet on the module page licenses that ignorance.