Exercises — Q factor — quality of oscillator

What we have: fractional energy loss per period $\frac{\Delta E}{E}=0.01$. Which tool: the energy definition — it is literally the ratio we are given. $Q=2\pi \cdot \frac{E}{\Delta E}=\frac{2\pi }{0.01}=200\pi \approx 628.$ Sanity: small loss ⇒ big $Q$. A tuning fork does ring for a long time, so a value in the hundreds is believable. ✓

Which tool: bandwidth form — we're handed a peak frequency and a width. $Q=\frac{f_0}{\Delta f}=\frac{50}{0.5}=100.$ Read it: the peak is $100$ times narrower than its centre frequency — a sharp, selective resonance. ✓

Which tool: damping form — both ${\omega }_0$ and $\gamma$ are given directly. $Q=\frac{{\omega }_0}{\gamma }=\frac{40}{0.5}=80.$ Since $Q=80\gg \frac{1}{2}$ the system is underdamped, so the formula is valid. ✓

Step 1 — natural frequency. Why first: $Q$ needs the "how fast it wants to swing" number. ${\omega }_0=\sqrt{\frac{k}{m}}=\sqrt{\frac{80}{0.2}}=\sqrt{400}=20\text{rad/s}.$ Step 2 — damping rate. Why: the normalized ODE defines $\gamma =b/m$. $\gamma =\frac{b}{m}=\frac{0.16}{0.2}=0.8{\text{s}}^{-1}.$ Step 3 — combine. $Q=\frac{{\omega }_0}{\gamma }=\frac{20}{0.8}=25.$ Underdamped ($25\gg \frac{1}{2}$), so all light-damping formulas apply. ✓

Which tool: ring-down count. Amplitude $1/e$-decay happens after $\approx Q/\pi$ oscillations. $N_{\text{amp}}\approx \frac{Q}{\pi }=\frac{25}{\pi }\approx 7.96\approx 8\text{oscillations}.$ Where this comes from (so it's not magic): the envelope is $A_0{e}^{-\gamma t/2}$, which hits $1/e$ at time $\tau =2/\gamma$. In that time the phase advances ${\omega }_0\tau =2{\omega }_0/\gamma =2Q$ radians. Divide by $2\pi$ radians per oscillation: $2Q/2\pi =Q/\pi$. ✓

Which tool: the RLC Circuits analogue. $L$ plays the role of mass, $1/C$ of stiffness, $R$ of damping. $\sqrt{\frac{L}{C}}=\sqrt{\frac{2\times 10^{-3}}{8\times 10^{-9}}}=\sqrt{2.5\times 10^5}=500\Omega .$ $Q=\frac{1}{R}\sqrt{\frac{L}{C}}=\frac{500}{5}=100.$ A cleanly tuned circuit. ✓

Step 1 — invert the ring-down relation. $N_{\text{amp}}\approx Q/\pi$, so $Q\approx \pi N_{\text{amp}}=\pi \times 40\approx 126.$ Step 2 — energy time constant. Energy decays as $E_0{e}^{-\gamma t}$, so its $1/e$ time is ${\tau }_E=1/\gamma$. Use $Q={\omega }_0/\gamma$ with ${\omega }_0=2\pi f_0$: $\gamma =\frac{{\omega }_0}{Q}=\frac{2\pi (256)}{126}\approx 12.8{\text{s}}^{-1},{\tau }_E=\frac{1}{\gamma }\approx 0.078\text{s}.$ So energy drops to $1/e$ in about $78\text{ms}$. ✓

Which tool: bandwidth form $Q={\omega }_0/\Delta \omega$, so $\Delta \omega ={\omega }_0/Q$. With equal ${\omega }_0$, width is inversely proportional to $Q$: $\frac{\Delta {\omega }_B}{\Delta {\omega }_A}=\frac{Q_A}{Q_B}=\frac{300}{60}=5.$ B's peak is 5 times wider — much less selective (see Resonance & Forced Oscillations). ✓

Energy definition: $Q=2\pi /0.126\approx 49.9\approx 50.$ Check the light-damping assumption. The exact per-period loss is $\Delta E/E=1-e^{-\gamma T}$. Light damping approximates this as $\gamma T$. Here $\gamma T=-\ln (1-0.126)=0.1347$, versus the approximation $0.126$ — they differ by only about $7\%$. Comment: with $Q\approx 50\gg \frac{1}{2}$ the system is firmly underdamped, so $Q={\omega }_0/\gamma$ is reliable to a few percent. The approximation would only break down for $Q$ near $1$. ✓

(a) Bandwidth. $\Delta f=f_0/Q=32768/10^5=0.32768\text{Hz}$ — extraordinarily narrow, which is why quartz keeps such precise time. (b) Amplitude $1/e$ time. ${\tau }_{\text{amp}}=2/\gamma$ and $\gamma ={\omega }_0/Q=2\pi f_0/Q$: ${\tau }_{\text{amp}}=\frac{2}{\gamma }=\frac{2Q}{2\pi f_0}=\frac{Q}{\pi f_0}=\frac{10^5}{\pi (32768)}\approx 0.972\text{s}.$ (c) Oscillations in that time. $N=f_0{\tau }_{\text{amp}}=Q/\pi =10^5/\pi \approx 3.18\times 10^4$ — over thirty thousand swings, matching $N_{\text{amp}}=Q/\pi$. ✓

Step 1 — natural frequency. ${\omega }_0=\sqrt{k/m}=\sqrt{90/0.1}=\sqrt{900}=30\text{rad/s}$, so period $T=2\pi /{\omega }_0$. Step 2 — turn "halves in 10 cycles" into $\gamma$. The envelope is $A_0{e}^{-\gamma t/2}$. After time $t_{10}=10T$ it is halved: $e^{-\gamma t_{10}/2}=\frac{1}{2}\Rightarrow \frac{\gamma t_{10}}{2}=\ln 2\Rightarrow \gamma =\frac{2\ln 2}{10T}.$ With $T=2\pi /30=0.20944\text{s}$: $t_{10}=2.0944\text{s}$, so $\gamma =\frac{2\ln 2}{2.0944}=\frac{1.3863}{2.0944}\approx 0.6619{\text{s}}^{-1}.$ Step 3 — recover $b$. $\gamma =b/m\Rightarrow b=m\gamma =0.1\times 0.6619\approx 0.0662\text{kg/s}.$ Cross-check via $Q$: $Q={\omega }_0/\gamma =30/0.6619\approx 45.3$, and $Q/\pi \approx 14.4$ oscillations for a $1/e$ (=$0.368$) drop; halving (a smaller drop to $0.5$) after $10$ oscillations is consistent. ✓

Proof. Energy decays as $E(t)=E_0{e}^{-\gamma t}$ (from Energy in Oscillations, $E\propto A^2$ and $A\propto e^{-\gamma t/2}$). Energy reaches $1/e$ when $\gamma t_E=1$, i.e. $t_E=1/\gamma$. Number of oscillations in that time (each of period $T=2\pi /{\omega }_0$): $N_E=\frac{t_E}{T}=\frac{1/\gamma }{2\pi /{\omega }_0}=\frac{{\omega }_0}{2\pi \gamma }=\frac{Q}{2\pi }.■$ Apply to car suspension, $Q=0.8$: the formula would give $N_E=0.8/2\pi \approx 0.127$ oscillations. The catch: $Q=0.8$ is not $\gg \frac{1}{2}$, so the system is barely underdamped (near critical). "$0.127$ oscillations" is not a real count — good car suspensions are deliberately near-critical so the car settles in well under one bounce rather than oscillating. The formula's spirit ("dies out fast") is right; its literal count is only meaningful when $Q\gg 1$. ✓

Exact per-period loss. Over one period $T=2\pi /{\omega }_0$, energy goes $E_0\to E_0{e}^{-\gamma T}$, so $\frac{\Delta E}{E}=1-e^{-\gamma T}.$ Light-damping approximation. For small $\gamma T$, $1-e^{-\gamma T}\approx \gamma T$. Then $Q=\frac{2\pi }{\Delta E/E}\approx \frac{2\pi }{\gamma T}=\frac{2\pi }{\gamma }\cdot \frac{{\omega }_0}{2\pi }=\frac{{\omega }_0}{\gamma }.■$ Error at $\Delta E/E=0.5$. True exponent: $\gamma T=-\ln (1-0.5)=\ln 2\approx 0.693$. Approximation used $0.5$. Ratio $0.693/0.5=1.386$ — a $39\%$ discrepancy. So at half-per-cycle loss the two forms disagree badly: this is heavy damping ($Q=2\pi /0.5\approx 12.6$ is still $>1$ but the linearisation of $1-e^{-x}$ is already poor). ✓

Peak. $P$ is largest when the denominator is smallest, essentially at $\omega \approx {\omega }_0$, where $({\omega }^2-{\omega }_0^2{)}^2=0$ and the denominator is ${\gamma }^2{\omega }_0^2$. So $P_{\max }\propto 1/({\gamma }^2{\omega }_0^2)$. Half-power condition. Set the denominator to twice its minimum: $({\omega }^2-{\omega }_0^2{)}^2+{\gamma }^2{\omega }^2=2{\gamma }^2{\omega }_0^2.$ Near resonance write $\omega ={\omega }_0+\delta$ with small $\delta$. Then ${\omega }^2-{\omega }_0^2=(\omega -{\omega }_0)(\omega +{\omega }_0)\approx \delta \cdot 2{\omega }_0$, and ${\gamma }^2{\omega }^2\approx {\gamma }^2{\omega }_0^2$. The condition becomes $(2{\omega }_0\delta {)}^2+{\gamma }^2{\omega }_0^2=2{\gamma }^2{\omega }_0^2\Rightarrow 4{\omega }_0^2{\delta }^2={\gamma }^2{\omega }_0^2\Rightarrow \delta =\pm \frac{\gamma }{2}.$ The two half-power points sit at ${\omega }_0\pm \gamma /2$, so the full width is $\Delta \omega =\gamma$, giving $Q={\omega }_0/\Delta \omega$. $■$ Numbers: $\Delta \omega ={\omega }_0/Q=1000/250=4\text{rad/s}$. ✓