Beats — derivation, applications: Worked examples

1.6.20 · D3 · Physics › Oscillations & Waves › Beats — derivation, applications

Yeh page Beats ka drill-ground hai. Parent note ne woh ek formula diya jo sab kuch chalata hai:

Woh trap jo har jagah chhupa hota hai: $∣ f_{1} - f_{2} ∣$ sign ko throw away kar deta hai. Yeh batata hai ki do frequencies kitni dur hain ek doosre se, lekin kaun si badi hai nahi batata. Neeche ke aadhe examples usi lost sign ko recover karne ke liye hain.

Scenario matrix

Neeche woh har distinct tarah ka question hai jo yeh topic pooch sakta hai. Baad ke har example ko us cell ke saath tag kiya gaya hai (C1, C2, …) jise woh demolish karta hai.

Cell	Case class	Kya alag karta hai ise	Example
C1	Plain difference	Dono frequencies known → bas subtract karo	Ex 1
C2	Sign-hidden (load with wax)	$f_{?}$ upar ya neeche ho sakta hai; mass add karna ⇒ $f$ girti hai	Ex 2
C3	Sign-hidden (file/tension up)	Wahi, lekin fix badhata hai $f$ ko	Ex 3
C4	Degenerate: zero beats	$f_{1} = f_{2}$ → beats gayab (tuning ka goal)	Ex 4
C5	Limiting: difference too big	$	f_1-f_2
C6	Time/period reading	Beat rate ↔ silences ke beech ka time convert karo	Ex 6
C7	Real-world word problem (Doppler radar)	Beat = transmitted vs reflected frequency	Ex 7
C8	Exam twist: do beats do clues dete hain	Do measurements combine karke exact value pin karo	Ex 8

Example 1 — Plain difference (cell C1)

Worked example Do forks, dono known

$f_{1} = 512$ Hz aur $f_{2} = 517$ Hz ke forks ko saath bajaya gaya. Beat frequency aur suni gayi pitch nikalo.

Forecast: Aage padhne se pehle dono numbers guess karo. Beats per second = ? Pitch = ?

Beats ke liye subtract karo: $f_{beat} = ∣512 - 517∣ = 5$ Hz. Yeh step kyun? Parent derivation ne dikhaya ki loudness peaks exactly frequency difference par hoti hain — yahi poora result hai $f_{beat} = ∣ f_{1} - f_{2} ∣$ .
Pitch ke liye average karo: $f_{pitch} = \frac{512 + 517}{2} = 514.5$ Hz. Yeh step kyun? Product-of-cosines mein fast factor mean frequency par chalta hai; wahi note hai jo tumhara kaan name karta hai.

Verify: Ek second mein $517$ Hz fork exactly $5$ zyada cycles complete karta hai $512$ Hz wale se — toh woh $5$ baar re-synchronise hote hain ⇒ $5$ loud swells. Step 1 se match karta hai. ✓

Example 2 — Wax fork ko lower karta hai (cell C2)

Worked example Classic sign resolver

Ek unknown fork $480$ Hz standard ke against 6 beats/s deta hai. Unknown fork par wax ka chhota sa blob chipkaya gaya; ab beats 8/s ho gayi. Unknown frequency nikalo.

Forecast: $∣ f_{?} - 480∣ = 6$ matlab $f_{?}$ ya to $474$ hai ya $486$ . Kaun sa, aur kyun?

Do candidates list karo: $f_{?} = 480 - 6 = 474$ Hz ya $f_{?} = 480 + 6 = 486$ Hz. Yeh step kyun? $∣ f_{1} - f_{2} ∣$ symmetric hai, toh dono algebraically valid hain; experiment ko tie break karni hogi.
Wax mass add karta hai ⇒ frequency girti hai. Simple Harmonic Motion se, $f = \frac{1}{2 π} k / m$ , toh bada $m$ chota $f$ deta hai. Waxing ke baad unknown ki frequency decreases hoti hai. Yeh step kyun? Humein ek directional nudge chahiye yeh expose karne ke liye ki hum $480$ ke kis side par hain.
Dono candidates ko downward nudge ke under trace karo:
- Agar $f_{?} = 486$ (480 se upar): ise girane se woh $480$ ki taraf jata hai ⇒ gap shrinks ⇒ beats decrease hoti hain.
- Agar $f_{?} = 474$ (480 se neeche): ise girane se woh $480$ se dur jata hai ⇒ gap grows ⇒ beats increase hoti hain.
Observation se match karo: beats $6 \to 8$ gayi (increased) ⇒ hum " $474$ " case mein hain.

Answer: $f_{?} = 474$ Hz.

Verify: Maano waxing ne $474$ ko $2$ Hz se girakr $472$ kar diya: naya gap $= ∣472 - 480∣ = 8$ Hz. ✓ Exactly observed $8$ beats/s.

Example 3 — Filing fork ko raise karta hai (cell C3)

Worked example Wahi trick, ulti direction

Ek unknown fork $256$ Hz standard ke saath 4 beats/s deta hai. Unknown fork ko file kiya gaya (prongs se metal hata di), aur beats 2/s ho gayi. Unknown nikalo.

Forecast: Candidates hain $252$ ya $260$ . Filing mass remove karti hai — $f$ kis direction mein jata hai, aur kaun sa candidate bachta hai?

Do candidates: $f_{?} = 256 \pm 4 = 252$ ya $260$ Hz. Kyun? Wahi symmetric ambiguity jaise pehle.
Filing mass remove karti hai ⇒ frequency badhti hai. $f = \frac{1}{2 π} k / m$ mein chota $m$ ⇒ bada $f$ . Unknown ki frequency increases hoti hai. Kyun? Yeh waxing ke exact opposite nudge hai — argument ka sign flip ho jata hai.
Dono ko upward nudge ke under trace karo:
- Agar $f_{?} = 252$ (256 se neeche): ise badhane se woh $256$ ki taraf jata hai ⇒ beats decrease hoti hain.
- Agar $f_{?} = 260$ (256 se upar): ise badhane se woh $256$ se dur jata hai ⇒ beats increase hoti hain.
Match karo: beats $4 \to 2$ gayi (decreased) ⇒ " $252$ " case.

Answer: $f_{?} = 252$ Hz.

Verify: Agar filing ne $252$ ko $2$ Hz se badhakar $254$ kar diya: gap $= ∣254 - 256∣ = 2$ Hz. ✓ Observed 2 beats/s se match karta hai.

Example 4 — Degenerate case: zero beats (cell C4)

Worked example Silence par tuning karna

Ek guitarist ek string ko $440$ Hz reference ke against tune karti hai. Woh $3$ beats/s sunti hai. Woh string ko thoda tight karti hai aur ab 0 beats/s sunti hai. String ki frequency kya hai, aur pehle kya thi?

Forecast: Zero beats physically kya matlab rakhti hai? Aur kya tightening ne pitch badhaya ya ghataया?

Zero beats ⇒ exact match. $f_{beat} = ∣ f_{string} - 440∣ = 0 \Rightarrow f_{string} = 440$ Hz. Yeh step kyun? Yeh degenerate limit hai $f_{1} = f_{2}$ : slow envelope $cos (2 π \frac{f _{1} - f _{2}}{2} t)$ ban jata hai $cos (0) = 1$ , ek constant — loudness kabhi dip nahi karti. Beats ki khamoshi tuning ka goal hai.
Tightening frequency badhata hai (higher tension ⇒ string par higher pitch). Kyunki tightening ne 3-beat gap remove kiya, string pehle $440$ se neeche thi. Kyun? Below-reference string ko uthane se gap zero ho jata hai.
Original frequency: $440 - 3 = 437$ Hz.

Answer: Ab $440$ Hz; pehle $437$ Hz tha.

Verify: $∣437 - 440∣ = 3$ Hz ✓ (pre-tuning beats se match karta hai), aur $∣440 - 440∣ = 0$ ✓ (silent match).

Figure — Beats — derivation, applications

Figure dekho: amber envelope shrink hoti hai jab do frequencies converge hoti hain, aur top par (equal frequencies) woh flat ceiling ban jati hai — bilkul koi throb nahi. Woh flat line wahi hai jo "zero beats" lagti hai.

Example 5 — Limiting case: gap too large (cell C5)

Worked example Jab koi beats sunne ko nahi hoti

$200$ Hz fork aur $260$ Hz fork ko saath bajaya gaya. "Beat frequency" kya hai, aur listener actually kya perceive karta hai?

Forecast: Formula mein plug karo — phir poochho: kya kaan use throb ki tarah sun sakta hai?

Formula phir bhi number deta hai: $∣200 - 260∣ = 60$ Hz. Yeh step kyun? Algebra kabhi fail nahi hoti — product-of-cosines hamesha valid hai.
Lekin kaan ise resolve nahi kar sakta. Beat tabhi sunai deti hai jab envelope slow ho — roughly $∣ f_{1} - f_{2} ∣ ≲ 10$ Hz. $60$ Hz par "swells" itni tezi se aate hain ki kaan alag nahi kar sakta; do tones simply do alag notes ki tarah sunai dete hain (ek musical interval, yahan roughly minor third). Yeh step kyun? Parent note ki condition ki $∣ f_{1} - f_{2} ∣$ chota hona chahiye optional decoration nahi hai — yahi cheez phenomenon ko audible banati hai.

Answer: Mathematically $60$ Hz; perceptually koi beats nahi — tum do alag pitches sunte ho.

Verify: $60$ Hz $> 10$ Hz threshold ⇒ envelope kaan ke liye too fast ⇒ beating ki tarah perceive nahi hota. ✓ Definition ki small-difference requirement ke saath consistent.

Example 6 — Silences ke beech ka time padhna (cell C6)

Worked example Beat rate se stopwatch tak

$340$ Hz aur $348$ Hz ke do pipes saath bajte hain. (a) Beat frequency? (b) Do successive moments of silence ke beech kitna time? (c) $5$ seconds mein kitni silences?

Forecast: Silences complete-destructive dips hain. Woh kitni baar aati hain?

Beat frequency: $f_{beat} = ∣340 - 348∣ = 8$ Hz. Kyun? Hamesha ki tarah frequencies ka difference.
Beat period = ek loud–soft–loud cycle ke liye time: $T_{beat} = \frac{1}{f _{beat}} = \frac{1}{8} = 0.125$ s. Yeh step kyun? Frequency aur period reciprocals hain: $T = 1/ f$ . Har beat cycle mein ek silence hoti hai, toh silences $T_{beat}$ se spaced hoti hain.
5 s mein count karo: $5 \times 8 = 40$ silences. Kyun? $5$ seconds ke liye $8$ per second.

Answers: (a) $8$ Hz, (b) $0.125$ s, (c) $40$ silences.

Verify: $0.125 s \times 8 beats/s = 1$ s beats ke har second ke liye — reciprocal self-consistent hai, aur $40 = 8 \times 5$ . ✓ Units: $[s] \times [Hz] = [s] \times [1/ s] =$ dimensionless count. ✓

Example 7 — Real-world word problem: Doppler radar (cell C7)

Worked example Speed gun as a beat detector

Ek radar ek car par $f_{0} = 10.000 000$ GHz transmit karta hai. Reflected wave Doppler-shifted hokar $10.000 002$ GHz wapas aati hai. Receiver transmitted aur reflected signals mix karta hai aur beat read karta hai. Kaun si beat frequency appear hoti hai, aur yeh kya encode karti hai?

Forecast: Do frequencies ek almost invisible amount se differ karti hain. Mixer se kaun sa tiny number nikalta hai?

Beat = do frequencies ka difference: $f_{beat} = ∣10.000 002 - 10.000 000∣ GHz = 0.000 002 GHz = 2000 Hz .$ Yeh step kyun? Exactly beats mechanism — do near-equal frequencies mix karne par slow difference ek countable tone ke roop mein bachta hai. Ise heterodyning kehte hain.
Interpret karo: $2000$ Hz beat car ki speed ke proportional hai (Doppler shift formula ke zariye). $2$ kHz ka gigahertz difference — five million mein ek part — asaani se audible-range beat ki tarah measure hota hai. Yeh step kyun? Beats as a tool ka yahi poora point hai: woh ek unmeasurably-tiny frequency gap ko ek easy-to-count rhythm mein magnify karte hain.

Answer: $f_{beat} = 2000$ Hz, jo target ki speed encode karta hai.

Verify: $10.000 002 - 10.000 000 = 0.000 002$ GHz. Convert karo: $0.000 002 \times 1 0^{9} Hz = 2000$ Hz. ✓ Units: GHz $= 1 0^{9}$ Hz, toh arithmetic aur scaling dono hold karte hain.

Example 8 — Exam twist: do clues ek value nail karte hain (cell C8)

Worked example Do beat measurements combine karna

Fork X ek $500$ Hz standard ke saath $5$ beats/s par beats karta hai. Fork X ek $508$ Hz standard ke saath bhi $3$ beats/s par beats karta hai. $f_{X}$ exactly nikalo — koi wax nahi chahiye.

Forecast: Har clue akela do candidates deta hai. Kya do lists ka overlap ek unique answer de sakta hai?

Clue 1 candidates: $∣ f_{X} - 500∣ = 5 \Rightarrow f_{X} \in {495, 505}$ . Kyun? Pehle standard se symmetric difference.
Clue 2 candidates: $∣ f_{X} - 508∣ = 3 \Rightarrow f_{X} \in {505, 511}$ . Kyun? Doosre standard se symmetric difference.
Do lists intersect karo: dono mein sirf ek value hai ${495, 505} \cap {505, 511} = {505}$ . Yeh step kyun? $f_{X}$ ko dono experiments simultaneously satisfy karne honge; sirf common value hi true frequency ho sakti hai.

Answer: $f_{X} = 505$ Hz.

Verify: $∣505 - 500∣ = 5$ ✓ aur $∣505 - 508∣ = 3$ ✓ — dono measurements ek single value se reproduce hote hain. ✓

Recall Kaun si fix kis direction mein nudge karti hai?

Wax mass add karta hai → frequency decreases. Filing mass remove karti hai (ya higher tension) → frequency increases. Beats grow after the fix ::: tumhara fork us side par tha jis side se fix tumhe dur push karta hai. Beats shrink after the fix ::: tumhara fork us side par tha jis taraf fix tumhe push karta hai.

Connections

Superposition Principle — har example displacements ke add hone par tika hai.
Interference of Waves — beats time mein interference hain.
Doppler Effect — woh frequency shift supply karta hai jo Ex 7 mein beat ki tarah read hoti hai.
Simple Harmonic Motion — wax/file ke peeche $f = \frac{1}{2 π} k / m$ argument (Ex 2–3).
Standing Waves — same superposition algebra, opposite-travelling waves.
Amplitude Modulation — envelope maths radio aur heterodyne radar mein reuse hoti hai.

Worked examples — Beats — derivation, applications