Q factor — quality of oscillator: Worked examples

1.6.10 · D3 · Physics › Oscillations & Waves › Q factor — quality of oscillator

Yeh page ek drill hall hai. Parent note Q factor — quality of oscillator ne $Q$ ke teen faces build kiye the; yahan hum un formulas par har tarah ke sawaal phenkte hain — chhota $Q$ , bahut bada $Q$ , woh degenerate boundary jahan oscillation khatam ho jaati hai, ek real-world word problem, aur ek exam-twist. Pehle hum poora territory map karte hain, phir har cell ko walk karte hain.

Scenario matrix

Har $Q$ problem actually inhi cells mein se ek hoti hai. Teen "input routes" usi ek number ke teen faces hain.

Cell	Aapko kya diya gaya	Kaun sa formula	Regime
A	Har cycle mein khoi energy $Δ E / E$	$Q = 2 π \frac{E}{Δ E}$	light damping
B	$m, k, b$ (mechanical parts)	$Q = \frac{mk}{b}$	light damping
C	Resonance width $f_{0}, Δ f$	$Q = \frac{f _{0}}{Δ f}$	light damping
D	Ring-down count ( $1/ e$ tak swings)	$Q = π N_{amp}$	light damping
E	Electrical parts $R, L, C$	$Q = \frac{1}{R} L / C$	light damping
F	Scaling / "agar ek part badal doon toh"	proportionalities	light damping
G	Degenerate: $Q \leq \frac{1}{2}$	koi oscillation nahi — regime check karo	critical / over
H	Word problem (real device)	sahi face chuno	light damping
I	Exam twist (do faces combined)	do formulas chain karo	light damping

Ab hum A → I hit karte hain, har ek ka ek example (plus ek extra taaki mechanical aur electrical dono ke apne worked cases hon). Matrix ka har cell cover hoga.

Examples se pehle, ek line vocabulary ki jo hum baar baar use karenge:

Definition Symbols, seedhe alfaazon mein

Jo numbers aap compute ya read karte hain:

$Q$ — ek plain number (koi units nahi). Bada = ringy, chhota = dead-thud.
$ω_{0}$ — "swing karna chahta hai" wali angular rate, radians per second mein, $ω_{0} = k / m$ .
$f_{0}$ — ordinary frequency cycles per second (hertz) mein, $ω_{0}$ se $f_{0} = ω_{0} /2 π$ ke zariye juda ( $2 π$ radians ek cycle mein hote hain, toh divide karne se radians-per-second, cycles-per-second mein convert ho jaata hai).
$γ$ — "friction speed khaata hai" wali rate, per-seconds mein, $γ = b / m$ .
$Δ E / E$ — energy ka woh fraction jo ek poore aage-peeche ke baad gaya.
$Δ f$ — resonance hump ki width, wahan measure ki jaati hai jahan power apne peak ki aadhi ho.

Physical parts (hardware):

$m$ — mass (kg); $k$ — spring stiffness (N/m); $b$ — mechanical damping coefficient (kg/s). Yeh mass–spring–damper ke teen parts hain.
$R$ — resistance (Ω); $L$ — inductance (H); $C$ — capacitance (F). Yeh electrical twins hain: $L$ , $m$ ki tarah kaam karta hai, $1/ C$ , $k$ ki tarah, $R$ , $b$ ki tarah (dekho RLC Circuits).

Counts (kitne swings):

$N_{amp}$ — full oscillations ki sankhya jab tak amplitude $1/ e$ tak gire; $Q / π$ ke barabar.
$N_{energy}$ — full oscillations ki sankhya jab tak energy $1/ e$ tak gire; $Q /2 π$ ke barabar ( $N_{amp}$ ka aadha, kyunki energy $\propto$ amplitude $^{2}$ do guna tez decay karti hai).

Yeh sab parent mein derive kiye gaye the; yahan hum sirf use karte hain.

Yahan poora terrain ek nazar mein — wohi $Q$ jo char alag doors se reach kiya gaya.

Figure — Q factor — quality of oscillator

Intuition Terrain figure kaise padhein (Cell → door map)

Centre mein amber box single number $Q$ hai. Char arrows iske andar point karte hain, har corner se ek — har corner ek alag cheez hai jo aapko di gayi hai aur jisse aap wohi $Q$ compute kar sakte hain:

Top-left door → Cell A: aapko energy lost per cycle $Δ E / E$ di gayi thi, toh aap $Q = 2 π E /Δ E$ se enter karte ho.
Top-right door → Cells B aur E: aapko physical parts di gayi thi — mechanical $m, k, b$ (Cell B) ya unke electrical twins $R, L, C$ (Cell E) — toh aap $Q = mk / b$ (ya $\frac{1}{R} L / C$ ) se enter karte ho.
Bottom-left door → Cell C: aapko resonance width $f_{0}, Δ f$ di gayi thi, toh aap $Q = f_{0} /Δ f$ se enter karte ho.
Bottom-right door → Cell D: aapko ek ring-down count $N$ diya gaya tha, toh aap $Q = π N$ se enter karte ho.

Picture ka point: char bilkul alag measurements, ek destination. Jab aap neeche har example milein, pehle poochho "main kaun si door par khada hoon?" — woh ek sawaal hi aapka formula select kar deta hai. (Cells F, G, H, I nayi doors nahi hain; woh in charon doors ko aur tricky tareekon se re-use karte hain: ek part scale karna, formula refuse karna, real device, aur do doors ko chain karna.)

Cell A — energy lost per cycle diya gaya

Worked example A · Ek gong jo har swing mein 1% khota hai

(Yeh upar terrain figure ki top-left door hai.) Ek struck gong oscillate karta hai aur har full period mein apni stored energy ka $1%$ khota hai. $Q$ nikalo.

Forecast: Har cycle mein sirf ek sowa hissa khona bahut kanjoos lagta hai — guess karo ki $Q$ ka value $6$ , $60$ , ya $600$ ke paas kahan hoga, aage padhne se pehle.

Step 1 — fraction likho. $\frac{Δ E}{E} = 0.01$ . Yeh step kyun? Yahi $Q$ ke energy face ka raw input hai; abhi kuch compute nahi karna.

Step 2 — energy definition apply karo. $Q = 2 π \cdot \frac{E}{Δ E} = 2 π \cdot \frac{1}{0.01} = 200 π \approx 628.$ Yeh step kyun? Definition $Q = 2 π E /Δ E$ ratio ko invert karti hai: ek chhota loss fraction denominator mein jaata hai, toh $Q$ bada ho jaata hai — exactly wahi "kanjoos = high $Q$ " wala idea.

Verify: Sanity — har cycle mein $1%$ khona parent ke $2%$ pendulum ( $Q \approx 314$ ) se das guna kanjoos hai, aur sach mein $628 = 2 \times 314$ . Loss fraction aadhi karne se $Q$ double ho jaata hai. Units: pure number ✓.

Cell B — mechanical parts $m, k, b$ diye gaye

Worked example B · Ek car suspension element

(Yeh terrain figure ki top-right door hai — "parts in, $Q$ out".) $m = 2.0 kg$ , $k = 800 N/m$ , $b = 8.0 kg/s$ . $Q$ nikalo aur batao ki yeh oscillate karta hai ya nahi.

Forecast: Car suspensions jaanbujhkar near-dead rakhi jaati hain taaki bounce na ho. Guess karo: $Q$ bada hoga ya chhota?

Step 1 — natural frequency. $ω_{0} = k / m = 800/2 = 400 = 20 rad/s .$ Yeh step kyun? $ω_{0}$ "swing karna chahta hai" wali rate hai; har mechanical $Q$ ko yeh chahiye.

Step 2 — damping rate. $γ = b / m = 8.0/2.0 = 4.0 s^{- 1} .$ Yeh step kyun? $γ$ friction-eats-speed rate hai, $Q = ω_{0} / γ$ ka denominator.

Step 3 — combine karo. $Q = \frac{ω _{0}}{γ} = \frac{20}{4.0} = 5.0.$ Yeh step kyun? Boxed formula do rates ko quality number mein baadlti hai.

Verify: One-shot form se cross-check karo $Q = mk / b = 2 \cdot 800 /8 = 1600 /8 = 40/8 = 5.0$ ✓. Kya yeh underdamped hai? $Q = 5 > \frac{1}{2}$ , toh haan yeh oscillate karta hai — lekin sirf mildly ringy, jo ek suspension ke liye sahi hai.

Cell C — resonance width diya gaya

Worked example C · Ek tuning-fork resonance curve

(Yeh terrain figure ki bottom-left door hai — "curve width in, $Q$ out".) Ek tuning fork $f_{0} = 440 Hz$ par resonate karta hai; uska response $439.6 Hz$ aur $440.4 Hz$ par half-power tak gir jaata hai. $Q$ nikalo.

Forecast: Peak bahut razor-thin hai (sirf $\pm 0.4$ Hz). Sharp peak ⇒ high ya low $Q$ ?

Step 1 — full width read karo. $Δ f = 440.4 - 439.6 = 0.8 Hz .$ Yeh step kyun? Bandwidth do half-power points ke beech ki total width hai, half-width nahi — ek classic galti.

Step 2 — bandwidth face apply karo. $Q = \frac{f _{0}}{Δ f} = \frac{440}{0.8} = 550.$ Yeh step kyun? Bandwidth definition "kitna narrow" ko directly "kitna ringy" mein convert karti hai. Narrow $Δ f$ divide hokar bada $Q$ deta hai.

Verify: Ek tuning fork ko lamba ring karna chahiye — $Q = 550$ matlab approximately $Q / π \approx 175$ audible swings amplitude ke $1/ e$ tak girne se pehle ✓, consistent hai ek fork ke ek ache fraction of a second tak humming karne ke saath.

Cell D — ring-down count diya gaya (ek figure cell)

Worked example D · Decay trace par swings count karna

(Yeh terrain figure ki bottom-right door hai — "swing count in, $Q$ out".) Ek oscilloscope ek freely decaying oscillation dikhata hai jiska amplitude apne start ke $1/ e$ tak 20 full oscillations ke baad gir jaata hai. $Q$ estimate karo.

Forecast: Marne se pehle zyada swings = higher ya lower $Q$ ?

Neeche figure mein envelope dekho — dashed amber curve $A_{0} e^{- γ t /2}$ hai, aur horizontal amber line $1/ e$ level mark karti hai.

Step 1 — count ko $Q$ se link karo. Parent ne dikhaya tha ki amplitude approximately $N_{amp} = Q / π$ oscillations ke baad $1/ e$ tak pahunchta hai (yeh $N_{amp}$ upar vocabulary block mein define kiya gaya count hai). Yeh step kyun? Woh relation exactly "ring-down" face hai; hum ise invert karte hain $Q$ solve karne ke liye.

Step 2 — invert karo. $Q = π N_{amp} = π \times 20 \approx 62.8.$ Yeh step kyun? Counted swings ko $π$ se multiply karna $Q / π$ relation ko undo karta hai.

Verify: Energy (jo amplitude $^{2}$ ki tarah jaati hai) ko $1/ e$ tak do guna tez pahunchna chahiye: $N_{energy} = Q /2 π = 62.8/6.28 = 10$ oscillations ke baad. Das, bees ka aadha hai ✓ — consistent hai kyunki energy, amplitude rate ke do guna rate se decay karti hai.

Cell E — electrical analogue $R, L, C$

Worked example E · Ek RLC resonant filter

(Cell B jaisi hi top-right door — "parts in, $Q$ out" — lekin ab parts electrical hain.) Ek series RLC circuit mein $L = 50 μ H$ , $C = 200 pF$ , $R = 5 Ω$ hai. $Q$ aur resonant frequency $f_{0}$ nikalo.

Forecast: Radio front-ends ko sharp tuning chahiye. $Q$ expect karo tens mein, hundreds mein, ya thousands mein?

Step 1 — parts se quality. $Q = \frac{1}{R} \frac{L}{C} = \frac{1}{5} \frac{50 \times 1 0 ^{- 6}}{200 \times 1 0 ^{- 12}} .$ Yeh step kyun? Yeh $Q = mk / b$ ka electrical twin hai: $L$ , $m$ ki jagah leta hai, $1/ C$ , $k$ ki jagah, $R$ , $b$ ki jagah. Dekho RLC Circuits.

Step 2 — ratio crunch karo. $\frac{L}{C} = \frac{50 \times 1 0 ^{- 6}}{200 \times 1 0 ^{- 12}} = 2.5 \times 1 0^{5}$ , toh $\dots = 500$ . Phir $Q = 500/5 = 100.$ Yeh step kyun? Ten ke powers ko explicit rakho taaki square root clean ho.

Step 3 — resonant frequency (bonus). $ω_{0} = 1/ L C = 1/ 50 \times 1 0^{- 6} \cdot 200 \times 1 0^{- 12} = 1/ 1 0^{- 14} = 1 0^{7} rad/s$ , toh $f_{0} = ω_{0} /2 π \approx 1.59 MHz .$ Yeh step kyun? $ω_{0} = 1/ L C$ electrical $k / m$ hai — zaroorat padegi agar baad mein bandwidth $Δ f = f_{0} / Q$ chahiye.

Verify: Bandwidth hogi $Δ f = f_{0} / Q = 1.59 MHz /100 \approx 15.9 kHz$ — ek plausibly narrow AM-band filter ✓. $L / C$ ke units: $Ω \cdot s \div s /Ω = Ω$ , ohms mein $R$ cancel ho jaata hai, ek pure number bachta hai ✓.

Cell F — scaling / "agar main ek part badal doon"

Worked example F · Stiffness teen guna karo

Ek lightly damped mass–spring ka $Q_{1} = 30$ hai. $m$ aur $b$ fix rakhte hue, aap spring ko teen guna stiffer ( $k \to 3 k$ ) se replace karte ho. Naya $Q_{2}$ nikalo.

Forecast: Stiffer spring tez swing karta hai. Kya isse woh zyada ringy (higher $Q$ ) ho jaata hai ya nahi?

Step 1 — $k$ -dependence isolate karo. $Q = mk / b$ se, sirf $k$ badalta hai; $m, b$ constant hain. Yeh step kyun? Formula ko proportionality $Q \propto k$ mein turn karne se hum $m$ aur $b$ dobara compute karna skip kar sakte hain.

Step 2 — ratio apply karo. $\frac{Q _{2}}{Q _{1}} = \frac{3 k}{k} = 3, Q_{2} = 303 \approx 51.96.$ Yeh step kyun? Do $Q$ s ko divide karne se sab kuch unchanged cancel ho jaata hai, sirf stiffness ratio root ke andar bachta hai.

Verify: $3 \approx 1.732$ , aur $30 \times 1.732 = 51.96$ ✓. Reasonableness: stiffer ⇒ tez swing ⇒ friction (unchanged) ke jeette se pehle zyada swings ⇒ higher $Q$ , forecast se match karta hai jo "higher" hona chahiye tha.

Cell G — degenerate case (koi oscillation nahi)

Worked example G · Jab formula refuse karna zaroori hai

$m = 1.0 kg$ , $k = 4.0 N/m$ , $b = 5.0 kg/s$ . Ek student $Q$ compute karta hai aur report karta hai "mass approximately $Q / π$ swings tak ring karta hai." Kya galat hai?

Forecast: Chhote $k$ ke paas bada $b$ — kya aap koi bouncing hone ki expect karte ho?

Step 1 — number waise bhi compute karo. $Q = mk / b = 1 \cdot 4 /5 = 2/5 = 0.4.$ Yeh step kyun? Hume number ko danger threshold cross karte dekhna chahiye pehle flag karne se pehle.

Step 2 — regime test karo. Underdamped (oscillating) ke liye $γ < 2 ω_{0}$ , yaani $Q > \frac{1}{2}$ chahiye. Yahan $Q = 0.4 < 0.5$ . Yeh step kyun? Poori ring-down interpretation underdamping assume karti hai. Decay roots hain $r = - \frac{γ}{2} \pm \frac{γ ^{2}}{4} - ω_{0}^{2}$ ; us root ke andar ki quantity, $\frac{γ ^{2}}{4} - ω_{0}^{2}$ , discriminant hai. Jab yeh positive ho toh do roots real hain (koi cosine nahi ⇒ koi swings nahi); jab negative ho toh complex hain (genuine oscillation).

Step 3 — discriminant evaluate karo. $ω_{0} = 2$ , $γ = 5$ ; discriminant $\frac{γ ^{2}}{4} - ω_{0}^{2} = \frac{25}{4} - 4 = 6.25 - 4 = 2.25 > 0$ . Positive ⇒ real roots ⇒ overdamped. Yeh step kyun? Yeh Damped Harmonic Motion ka direct check hai: positive discriminant ka matlab hai koi imaginary part nahi, isliye koi cosine nahi, isliye koi swings nahi.

Verify: Toh $Q = 0.4$ number ke taur par galat nahi hai, lekin "rings for $Q / π \approx 0.13$ swings" meaningless hai — zero swings hain. Sahi statement: overdamped, mass bina zero cross kiye sluggishly rest par waapas aata hai ✓. Yeh woh boundary cell hai jo har exam pasand karta hai.

Cell H — ek real-world word problem

Worked example H · Wine glass kitni der gaata hai?

Ek crystal wine glass tap karo; woh $f_{0} = 1000 Hz$ par ring karta hai aur aap ise fade hone se pehle approximately $t_{1/ e} = 2.0 s$ tak sun sakte ho (amplitude ka $1/ e$ tak). $Q$ estimate karo.

Forecast: Hazaar hertz par do poore second ki ringing bahut zyada cycles hain. Rough $Q$ ?

Step 1 — us time mein oscillations count karo. $N_{amp} = f_{0} t_{1/ e} = 1000 \times 2.0 = 2000$ full oscillations. Yeh step kyun? Frequency times time = cycles ki sankhya; yeh audible duration ko swing count mein convert karta hai, jo ring-down face ka input hai.

Step 2 — count ko $Q$ mein convert karo. $Q = π N_{amp} = π \times 2000 \approx 6283.$ Yeh step kyun? Wohi $Q = π N_{amp}$ ring-down relation jaise Cell D mein — humne bas ise ek screen se read kiye count ki jagah clock se derive kiye count se feed kiya.

Verify: Yahan usi number tak pahunchne ka ek independent rasta hai, vocabulary block mein di gayi do definitions se built.

Amplitude ka $1/ e$ -time woh time hai jab $e^{- γ t /2}$ , $e^{- 1}$ tak pahunche, yaani $γ τ /2 = 1$ , toh $τ = 2/ γ$ . (Factor of 2 isliye aata hai kyunki amplitude envelope apne exponent mein $γ /2$ carry karta hai, $γ$ nahi — energy $γ$ use karta aur aadha time deta.)
$Q = ω_{0} / γ$ aur $τ = 2/ γ$ combine karo: doosre se $γ = 2/ τ$ solve karo, phir pehle mein substitute karo — $Q = \frac{ω _{0}}{γ} = \frac{ω _{0}}{2/ τ} = \frac{ω _{0} τ}{2}$ . Yeh chaining kyun kaam karti hai: dono formulas mein wohi $γ$ hai, toh $γ$ eliminate karne se $Q$ directly measured ring-time $τ$ se link ho jaata hai.
$ω_{0} = 2 π f_{0} = 2 π (1000)$ aur $τ = 2.0 s$ plug in karo: $Q = \frac{ω _{0} τ}{2} = \frac{( 2 π \cdot 1000 ) ( 2.0 )}{2} = 2 π \cdot 1000 \approx 6283$ ✓ — Step 2 jaisa hi. $Q \sim 6000$ wala crystal glass realistic hai — ache crystal mein bahut kam internal friction hota hai.

Cell I — exam twist (do faces chain karo)

Worked example I · Bandwidth se kitni der ring karta hai

Ek microwave cavity $f_{0} = 10 GHz$ par resonate karti hai aur measured bandwidth $Δ f = 1.0 MHz$ hai. Drive band karne ke baad, stored energy ke $1/ e$ tak girne se pehle kitne oscillations guzarte hain?

Forecast: GHz cavity sirf MHz width ke saath extremely sharp hai. Thousands, ya tens of thousands, cycles expect karo?

Step 1 — bandwidth face se $Q$ nikalo. $Q = \frac{f _{0}}{Δ f} = \frac{10 \times 1 0 ^{9}}{1.0 \times 1 0 ^{6}} = 1 0^{4} = 10000.$ Yeh step kyun? Measurement ek resonance width hai, toh face C (bottom-left door) natural entry point hai.

Step 2 — $Q$ ko energy ring-down count mein convert karo. Energy $N_{energy} = Q /2 π$ oscillations ke baad $1/ e$ tak pahunchti hai (vocabulary block mein defined count). $N_{energy} = \frac{10000}{2 π} \approx 1592.$ Yeh step kyun? Yahi twist hai — sawaal energy ke baare mein poochh raha hai, jiska decay amplitude se do guna tez hai, toh hum $Q /2 π$ use karte hain, $Q / π$ nahi. Inhe mix up karna hi trap hai. Hum bandwidth door se ander gaye aur ring-down door se bahar — do faces chained.

Verify: Amplitude count se cross-check karo $N_{amp} = Q / π = 3183$ ; energy ko $1/ e$ tak aadhe cycles mein pahunchna chahiye, aur $3183/2 = 1592$ ✓ — amplitude aur energy decay ke beech factor-of-two hold karta hai, confirming ki humne do faces sahi se chain kiye. Dekho Resonance & Forced Oscillations aur Energy in Oscillations.

Nau cells ko ek saath pull karna

Intuition Poora matrix ek saans mein

Nau cells mein se har ek usi ek number $Q$ tak pahuncha (ya refuse kiya). Inhe ek story ke roop mein padhein: A aapko loss fraction deta hai, B aur E physical parts dete hain (mechanical ya electrical — wohi formula, alag hardware), C resonance width deta hai, D swing count deta hai. Yeh terrain figure ke char doors hain. Phir mushkil cells un doors ko re-use karti hain: F ek part badalta hai aur proportionality par sawaar hota hai, G number compute karta hai lekin story refuse karta hai kyunki system overdamped hai ( $Q < \frac{1}{2}$ , positive discriminant, zero swings), H ring-down door ko ek real singing glass ki tarah dress karta hai, aur I ek door se andar jaata hai (bandwidth) aur doosre se bahar (ring-down), do faces chain karta hua — kabhi nahi bhulte ki energy amplitude se do guna tez marti hai. Ek sawaal master karo "mujhe kya diya gaya tha?" aur har $Q$ problem sahi door chunne tak collapse ho jaati hai.

Recall Kya har cell ko ghar mila?

Har example ko uski matrix row se match karo. A energy-per-cycle ::: gong, $Q = 200 π \approx 628$ B mechanical parts ::: suspension, $Q = 5$ C resonance width ::: tuning fork, $Q = 550$ D ring-down count (figure) ::: scope trace, $Q \approx 62.8$ E electrical parts ::: RLC filter, $Q = 100$ F scaling one part ::: $3 \times$ stiffer, $Q = 303 \approx 52$ G degenerate / no oscillation ::: overdamped, $Q = 0.4 < \frac{1}{2}$ H real-world word problem ::: wine glass, $Q \approx 6283$ I exam twist (chain two faces) ::: cavity, $N_{energy} \approx 1592$

Connections

Q factor — quality of oscillator — parent jo har formula yahan use kiye derive karta hai.
Damped Harmonic Motion — $Q > \frac{1}{2}$ underdamped test jo Cell G mein use kiya.
Simple Harmonic Motion — $ω_{0} = k / m$ baseline.
Resonance & Forced Oscillations — bandwidth face (Cells C, I).
RLC Circuits — electrical analogue (Cell E).
Energy in Oscillations — energy amplitude se do guna tez kyun decay karti hai (Cell I).