Exercises — Q factor — quality of oscillator

Kya diya hai: fractional energy loss per period $\frac{\Delta E}{E}=0.01$. Kaun sa tool: energy definition — ye literally wo ratio hai jo hume diya gaya hai. $Q=2\pi \cdot \frac{E}{\Delta E}=\frac{2\pi }{0.01}=200\pi \approx 628.$ Sanity check: chhoti loss ⇒ bada $Q$. Tuning fork sach mein kaafi der tak ring karta hai, isliye hundreds mein value believable hai. ✓

Kaun sa tool: bandwidth form — hume ek peak frequency aur width diya gaya hai. $Q=\frac{f_0}{\Delta f}=\frac{50}{0.5}=100.$ Matlab: peak apni centre frequency se $100$ guna zyada narrow hai — ek sharp, selective resonance. ✓

Kaun sa tool: damping form — ${\omega }_0$ aur $\gamma$ dono directly diye hain. $Q=\frac{{\omega }_0}{\gamma }=\frac{40}{0.5}=80.$ Kyunki $Q=80\gg \frac{1}{2}$ hai, system underdamped hai, isliye formula valid hai. ✓

Step 1 — natural frequency. Pehle kyun: $Q$ ko "kitni tez swing karna chahta hai" wali number chahiye. ${\omega }_0=\sqrt{\frac{k}{m}}=\sqrt{\frac{80}{0.2}}=\sqrt{400}=20\text{rad/s}.$ Step 2 — damping rate. Kyun: normalized ODE define karta hai $\gamma =b/m$. $\gamma =\frac{b}{m}=\frac{0.16}{0.2}=0.8{\text{s}}^{-1}.$ Step 3 — combine karo. $Q=\frac{{\omega }_0}{\gamma }=\frac{20}{0.8}=25.$ Underdamped ($25\gg \frac{1}{2}$), isliye saare light-damping formulas apply hote hain. ✓

Kaun sa tool: ring-down count. Amplitude ka $1/e$-decay hota hai $\approx Q/\pi$ oscillations ke baad. $N_{\text{amp}}\approx \frac{Q}{\pi }=\frac{25}{\pi }\approx 7.96\approx 8\text{oscillations}.$ Ye kahan se aata hai (taaki magic na lage): envelope hai $A_0{e}^{-\gamma t/2}$, jo $1/e$ par pahunchti hai time $\tau =2/\gamma$ par. Is time mein phase advance hoti hai ${\omega }_0\tau =2{\omega }_0/\gamma =2Q$ radians. Har oscillation mein $2\pi$ radians se divide karo: $2Q/2\pi =Q/\pi$. ✓

Kaun sa tool: RLC Circuits analogue. $L$ mass ki jagah play karta hai, $1/C$ stiffness ki, $R$ damping ki. $\sqrt{\frac{L}{C}}=\sqrt{\frac{2\times 10^{-3}}{8\times 10^{-9}}}=\sqrt{2.5\times 10^5}=500\Omega .$ $Q=\frac{1}{R}\sqrt{\frac{L}{C}}=\frac{500}{5}=100.$ Ek cleanly tuned circuit. ✓

Step 1 — ring-down relation ko invert karo. $N_{\text{amp}}\approx Q/\pi$, isliye $Q\approx \pi N_{\text{amp}}=\pi \times 40\approx 126.$ Step 2 — energy time constant. Energy decay hoti hai $E_0{e}^{-\gamma t}$ ke according (from Energy in Oscillations), isliye uski $1/e$ time hai ${\tau }_E=1/\gamma$. $Q={\omega }_0/\gamma$ use karo jahan ${\omega }_0=2\pi f_0$: $\gamma =\frac{{\omega }_0}{Q}=\frac{2\pi (256)}{126}\approx 12.8{\text{s}}^{-1},{\tau }_E=\frac{1}{\gamma }\approx 0.078\text{s}.$ Toh energy approximately $78\text{ms}$ mein $1/e$ tak drop ho jaati hai. ✓

Kaun sa tool: bandwidth form $Q={\omega }_0/\Delta \omega$, isliye $\Delta \omega ={\omega }_0/Q$. Equal ${\omega }_0$ ke saath, width inversely proportional hai $Q$ se: $\frac{\Delta {\omega }_B}{\Delta {\omega }_A}=\frac{Q_A}{Q_B}=\frac{300}{60}=5.$ B ka peak 5 guna wider hai — kaafi kam selective (dekho Resonance & Forced Oscillations). ✓

Energy definition: $Q=2\pi /0.126\approx 49.9\approx 50.$ Light-damping assumption check karo. Exact per-period loss hai $\Delta E/E=1-e^{-\gamma T}$. Light damping isse approximate karta hai $\gamma T$ se. Yahan $\gamma T=-\ln (1-0.126)=0.1347$ hai, jabki approximation $0.126$ hai — dono mein sirf $7\%$ ka fark hai. Comment: $Q\approx 50\gg \frac{1}{2}$ ke saath system firmly underdamped hai, isliye $Q={\omega }_0/\gamma$ kuch percent tak reliable hai. Approximation tabhi break down hogi jab $Q$ near $1$ ho. ✓

(a) Bandwidth. $\Delta f=f_0/Q=32768/10^5=0.32768\text{Hz}$ — extraordinarily narrow, isliye quartz itna precise time keep karta hai. (b) Amplitude $1/e$ time. ${\tau }_{\text{amp}}=2/\gamma$ aur $\gamma ={\omega }_0/Q=2\pi f_0/Q$: ${\tau }_{\text{amp}}=\frac{2}{\gamma }=\frac{2Q}{2\pi f_0}=\frac{Q}{\pi f_0}=\frac{10^5}{\pi (32768)}\approx 0.972\text{s}.$ (c) Us time mein oscillations. $N=f_0{\tau }_{\text{amp}}=Q/\pi =10^5/\pi \approx 3.18\times 10^4$ — teees hazaar se zyada swings, $N_{\text{amp}}=Q/\pi$ se match karta hai. ✓

Step 1 — natural frequency. ${\omega }_0=\sqrt{k/m}=\sqrt{90/0.1}=\sqrt{900}=30\text{rad/s}$, isliye period $T=2\pi /{\omega }_0$. Step 2 — "10 cycles mein half" ko $\gamma$ mein badlo. Envelope hai $A_0{e}^{-\gamma t/2}$. Time $t_{10}=10T$ ke baad ye half ho jaati hai: $e^{-\gamma t_{10}/2}=\frac{1}{2}\Rightarrow \frac{\gamma t_{10}}{2}=\ln 2\Rightarrow \gamma =\frac{2\ln 2}{10T}.$ $T=2\pi /30=0.20944\text{s}$ ke saath: $t_{10}=2.0944\text{s}$, isliye $\gamma =\frac{2\ln 2}{2.0944}=\frac{1.3863}{2.0944}\approx 0.6619{\text{s}}^{-1}.$ Step 3 — $b$ recover karo. $\gamma =b/m\Rightarrow b=m\gamma =0.1\times 0.6619\approx 0.0662\text{kg/s}.$ $Q$ se cross-check: $Q={\omega }_0/\gamma =30/0.6619\approx 45.3$, aur $Q/\pi \approx 14.4$ oscillations $1/e$ (=$0.368$) drop ke liye; $10$ oscillations ke baad halving (ek chhoti drop to $0.5$) consistent hai. ✓

Proof. Energy decay hoti hai $E(t)=E_0{e}^{-\gamma t}$ (from Energy in Oscillations, $E\propto A^2$ aur $A\propto e^{-\gamma t/2}$). Energy $1/e$ tak pahunchti hai jab $\gamma t_E=1$, matlab $t_E=1/\gamma$. Us time mein oscillations ki sankhya (har oscillation ka period $T=2\pi /{\omega }_0$): $N_E=\frac{t_E}{T}=\frac{1/\gamma }{2\pi /{\omega }_0}=\frac{{\omega }_0}{2\pi \gamma }=\frac{Q}{2\pi }.■$ Car suspension par apply karo, $Q=0.8$: formula deta $N_E=0.8/2\pi \approx 0.127$ oscillations. Catch: $Q=0.8$ ka matlab $\gg \frac{1}{2}$ nahi hai, isliye system barely underdamped hai (near critical). "$0.127$ oscillations" koi real count nahi hai — achhe car suspensions deliberately near-critical hote hain taaki car ek bounce se kaafi kam time mein settle ho jaaye na ki oscillate kare. Formula ki spirit ("jaldi khatam ho jaata hai") sahi hai; uska literal count tabhi meaningful hai jab $Q\gg 1$ ho. ✓

Exact per-period loss. Ek period $T=2\pi /{\omega }_0$ mein, energy jaati hai $E_0\to E_0{e}^{-\gamma T}$, isliye $\frac{\Delta E}{E}=1-e^{-\gamma T}.$ Light-damping approximation. Chhote $\gamma T$ ke liye, $1-e^{-\gamma T}\approx \gamma T$. Tab $Q=\frac{2\pi }{\Delta E/E}\approx \frac{2\pi }{\gamma T}=\frac{2\pi }{\gamma }\cdot \frac{{\omega }_0}{2\pi }=\frac{{\omega }_0}{\gamma }.■$ $\Delta E/E=0.5$ par error. True exponent: $\gamma T=-\ln (1-0.5)=\ln 2\approx 0.693$. Approximation ne $0.5$ use kiya. Ratio $0.693/0.5=1.386$ — $39\%$ discrepancy. Isliye half-per-cycle loss par dono forms kaafi disagree karte hain: ye heavy damping hai ($Q=2\pi /0.5\approx 12.6$ abhi bhi $>1$ hai lekin $1-e^{-x}$ ka linearisation already poor hai). ✓

Peak. $P$ tab sabse bada hota hai jab denominator sabse chhota ho, essentially $\omega \approx {\omega }_0$ par, jahan $({\omega }^2-{\omega }_0^2{)}^2=0$ aur denominator ${\gamma }^2{\omega }_0^2$ hai. Isliye $P_{\max }\propto 1/({\gamma }^2{\omega }_0^2)$. Half-power condition. Denominator ko uske minimum ka do guna set karo: $({\omega }^2-{\omega }_0^2{)}^2+{\gamma }^2{\omega }^2=2{\gamma }^2{\omega }_0^2.$ Resonance ke paas likho $\omega ={\omega }_0+\delta$ jahan $\delta$ chhota hai. Tab ${\omega }^2-{\omega }_0^2=(\omega -{\omega }_0)(\omega +{\omega }_0)\approx \delta \cdot 2{\omega }_0$, aur ${\gamma }^2{\omega }^2\approx {\gamma }^2{\omega }_0^2$. Condition ban jaati hai $(2{\omega }_0\delta {)}^2+{\gamma }^2{\omega }_0^2=2{\gamma }^2{\omega }_0^2\Rightarrow 4{\omega }_0^2{\delta }^2={\gamma }^2{\omega }_0^2\Rightarrow \delta =\pm \frac{\gamma }{2}.$ Do half-power points ${\omega }_0\pm \gamma /2$ par hain, isliye full width hai $\Delta \omega =\gamma$, jo deta hai $Q={\omega }_0/\Delta \omega$. $■$ Numbers: $\Delta \omega ={\omega }_0/Q=1000/250=4\text{rad/s}$. ✓