Harmonics and overtones — on strings and in pipes: Worked examples

1.6.19 · D3 · Physics › Oscillations & Waves › Harmonics and overtones — on strings and in pipes

Woh tools jo hum reuse karenge (use se pehle earn kiye hue)

Definition Is page par sirf yahi symbols hain

$L$ — string ya air column ki length, metres mein. Socho ek ruler uske saath rakh do.
$v$ — speed jis par wave us medium mein travel karti hai, metres per second mein. Ek string par yeh tension se aati hai (dekho Speed of a wave on a string); hawa mein yeh speed of sound hai.
$λ$ (Greek letter "lambda") — wavelength: wave shape ke ek complete repeat ki length, metres mein. Socho ek wave-crest se doosre crest tak ki distance.
$f$ — frequency: har second mein kitne full wobbles hote hain, hertz (Hz) mein measure ki jaati hai.
$n$ — ek saada counting number $1, 2, 3, \dots$ jo batata hai hum kaunsa allowed pattern mean kar rahe hain.
$T$ — string mein tension: use kitni zor se kheencha ja raha hai, newtons (N) mein measure kiya jaata hai. Socho do haath string ko kheench rahe hain. (Sirf Ex 8 mein use hoga.)
$μ$ (Greek letter "mu") — linear mass density: string ke har metre mein kitne kilograms hain, kg/m mein. Ek moti bhaari string ka $μ$ bada hota hai. (Sirf Ex 8 mein use hoga.)
$r$ — pipe ki opening ka radius, metres mein. Socho hole ke centre se rim tak ki distance. (Sirf Ex 9 mein use hoga.)

Scenario matrix

Yeh un classes ki poori list hai jo is topic mein tumhare samne aa sakti hain. Har worked example neeche us cell ke saath tagged hai jo woh fill karta hai, taaki tum check kar sako ki poora grid cover ho gaya.

Formula Is topic mein har case

Cell	System / condition	Tricky kya hai	Example
A	String, dono ends fixed — all harmonics	baseline node–node	Ex 1
B	Pipe open both ends — all harmonics	antinode–antinode, same math	Ex 2
C	Pipe closed one end — odd only	overtone ≠ harmonic number	Ex 3
D	Degenerate: shortest possible fit ( $n = 1$ )	kya $n = 1$ sach mein sabse chhota hai?	Ex 4
E	Zero / limiting input ( $v \to 0$ , $L \to \infty$ )	edges par kya hota hai	Ex 5
F	Real-world word problem	words ko $L, v$ mein translate karo	Ex 6
G	Exam twist: "shared harmonic?" trap	odd-vs-all numbering	Ex 7
H	Inverse problem: given $f$ , find $L$ or $v$	formula ko ulta chalao	Ex 8
I	End correction (real pipe)	$L_{eff} = L + 0.6 r$	Ex 9

Ab hum har cell fill karte hain.

Ex 1 — Cell A · String fixed at both ends

Figure — Harmonics and overtones — on strings and in pipes

Figure s01 — Ek steel string dono ends par clamp ki gayi hai. Red curve fundamental standing wave hai: beech mein ek single hump (antinode) aur dono black walls par ek-ek node. Double-headed arrow dikhata hai ki poori length half a wavelength ke barabar hai, $L = λ /2$ .

Worked example Statement

$L = 0.80$ m lambi ek steel string $v = 480$ m/s par transverse waves carry karti hai. Yeh dono ends par clamp ki gayi hai. Fundamental $f_{1}$ aur third harmonic $f_{3}$ nikalo.

Forecast (pehle guess karo!): Kya $f_{3}$ exactly triple hoga $f_{1}$ ka, ya kuch aur messier? Ek number likh lo.

Step 1 — shortest fit draw karo. Yeh step kyun? Dono ends clamp hain (nodes), toh sabse simple standing wave ek hump hai do nodes ke beech mein. Figure s01 dekho: red curve ko apni ungli se trace karo — yeh left black wall par pinned start hoti hai, ek single bulge tak uthti hai, aur right wall par wapas zero par aati hai. Jinne zero-crossings wall ko touch karti hain unhe gino: exactly do. Woh single hump ek node-to-node distance span karta hai, jo picture rule hume batata hai $λ /2$ hai. Isliye $L = \frac{λ _{1}}{2} \Rightarrow λ_{1} = 2 L = 1.60 m .$

Step 2 — general pattern banao (WHY $f_{n} = n f_{1}$ ). Yeh step kyun? Hume ladder justify karna hai, sirf assert nahi karna. $n$ -th mode banane ke liye, dono end-nodes rakho aur aur humps pack karo. Har extra hump ek aur node-to-node stretch of $λ /2$ add karta hai, toh $n$ humps ek length $n \cdot λ_{n} /2$ fill karte hain. Kyunki yeh $L$ ke barabar hona chahiye, $L = n \frac{λ _{n}}{2} \Rightarrow λ_{n} = \frac{2 L}{n}, n = 1, 2, 3, \dots$ Har $λ_{n}$ ko master relation se feed karo: $f_{n} = \frac{v}{λ _{n}} = \frac{n v}{2 L} = n f_{1}, f_{1} = \frac{v}{2 L} .$ Toh har integer multiple allowed hai — yahi poori wajah hai $f_{n} = n f_{1}$ ki.

Step 3 — numbers plug in karo. Yeh step kyun? Ab hum general formulas ko un do specific pitches mein badlenge jo question pooch raha hai. $f_{1} = \frac{v}{2 L} = \frac{480}{1.60} = 300 Hz, f_{3} = 3 f_{1} = 900 Hz .$

Verify: Units: $\frac{m/s}{m} = 1/s = Hz$ ✓. Sanity: teen humps same length mein fit hone chahiye, toh har hump ek tehai jitna lamba hoga, toh $λ$ ek tehai aur $f$ triple — matches.

Ex 2 — Cell B · Pipe open at both ends

Figure s02 — Dono ends par open pipe (patli black tube walls). Red curve fundamental hai: dono open ends par ek-ek antinode aur beech mein ek node. Double-headed arrow dikhata hai $L = λ /2$ , exactly wahi spacing jaisi clamped string mein.

Worked example Statement

Ek flute-jaisa pipe jo dono ends par open hai $L = 0.34$ m lamba hai; sound speed $v = 340$ m/s. $f_{1}$ aur first overtone nikalo.

Forecast: String jaisa hi formula, ya alag kyunki yeh air hai string nahi?

Step 1 — shortest fit draw karo. Kyun? Dono ends open $\Rightarrow$ antinodes. Figure s02 dekho: red curve dono ends par sabse bada hai (open mouths, jahan air freely swing karti hai) aur beech mein ek baar zero cross karti hai — woh single central node hi ek pinned point hai. Har end par antinode ke saath sabse chhoti standing wave isliye antinode–node–antinode hai. Antinode se agle antinode tak bhi $λ /2$ hai, toh spacing string jaisi hi hai: $L = \frac{λ _{1}}{2} \Rightarrow λ_{1} = 2 L = 0.68 m .$

Step 2 — general pattern banao. Yeh step kyun? Ex 1 jaisi hi node-counting argument: har extra half-wave add karne se dono open ends par antinodes rahti hain, toh $n$ half-waves length fill karte hain, $L = n \frac{λ _{n}}{2} \Rightarrow λ_{n} = \frac{2 L}{n}, f_{n} = \frac{n v}{2 L} = n f_{1} .$ Har integer allowed hai, exactly string jaisa.

Step 3 — frequency mein convert karo. Yeh step kyun? Hum allowed wavelength tak pahunch gaye lekin question ek pitch poochh raha hai, aur master relation $f = v / λ$ hi ek aisi length se — jo hum draw kar sakte hain — frequency tak ki ek bridge hai — jo hum sun sakte hain — toh hum ise yahan apply karte hain. $f_{1} = \frac{v}{λ _{1}} = \frac{340}{0.68} = 500 Hz .$

Step 4 — first overtone. Kyun? "Overtone" = $f_{1}$ ke upar agla allowed frequency. Yahan saari harmonics exist karti hain, toh agla hai $f_{2} = 2 f_{1} = 1000$ Hz. 1st overtone IS the 2nd harmonic.

Verify: Wahi $λ = 2 L$ math jaisi the string parent ke claim ko confirm karta hai ki string aur open pipe formulas share karte hain. Units pehle jaisi check ✓.

Ex 3 — Cell C · Pipe closed at one end (odd only)

Figure s03 — Left par closed pipe (moti black wall) aur right par open. Red curve fundamental hai: closed wall par ek node jo open end par ek full antinode tak grow karta hai. Double-headed arrow dikhata hai sirf ek quarter wavelength fit hoti hai, $L = λ /4$ — ek lambi wave, isliye low pitch.

Worked example Statement

Us hi $L = 0.34$ m pipe ( $v = 340$ m/s) ka ek end band karo. $f_{1}$ aur uska first overtone nikalo.

Forecast: Jab hum ek end plug karte hain toh kya $f_{1}$ upar jaega, neeche, ya same rahega?

Step 1 — shortest fit draw karo. Kyun? Closed end = node, open end = antinode. Figure s03 dekho: notice karo red curve moti left wall ke against flat pinned hai (ek node) lekin open right mouth par apne fullest tak swing karti hai (ek antinode). Double-headed arrow follow karo — red curve tube mein sirf ek quarter of a full wave shape complete karti hai: $L = \frac{λ _{1}}{4} \Rightarrow λ_{1} = 4 L = 1.36 m .$ Notice karo $λ_{1}$ open pipe se do guna lamba hai — longer wave ka matlab lower pitch.

Step 2 — odd-only ladder. Yeh step kyun? Node→antinode se shuru karke, har agla mode ek poora half-wave add karta hai (node→antinode→node→antinode), toh sirf odd quarter-waves fit hote hain: $L = (2 n - 1) \frac{λ _{n}}{4} \Rightarrow λ_{n} = \frac{4 L}{2 n - 1}, f_{n} = (2 n - 1) \frac{v}{4 L} = (2 n - 1) f_{1} .$

Step 3 — frequency. Yeh step kyun? Ab hum allowed wavelength tak pahunch gaye lekin question ek pitch poochh raha hai, toh hum bridge $f = v / λ$ ek baar phir cross karte hain us length ko ek frequency mein badlne ke liye. $f_{1} = \frac{v}{λ _{1}} = \frac{340}{1.36} = 250 Hz .$ Exactly open-pipe answer (500 Hz) ka aadha — ek octave neeche, jaisa promise kiya tha.

Step 4 — first overtone. Kyun? Allowed pitches hain $250, 750, 1250, \dots$ (odd multiples). $250$ ke upar pehla hai $f_{1st overtone} = 3 f_{1} = 750 Hz = 3rd harmonic.$

Verify: $f_{1}^{closed} = 250 = \frac{1}{2} (500) = \frac{1}{2} f_{1}^{open}$ ✓. Even harmonic $500$ Hz simply odd list mein appear hi nahi karta — parent ka "even harmonics forbidden" match karta hai.

Ex 4 — Cell D · Degenerate: kya $n = 1$ sach mein sabse chhota hai?

Figure s04 — Kyun $n = 1$ sabse chhota closed-pipe mode hai. Red solid curve allowed quarter wave hai (closed wall par node, open end par antinode). Black dotted curve ek chhota shape hai jo closed wall par ek bulge rakhega — forbidden, kyunki woh wall node honi chahiye.

Worked example Statement

Ek closed pipe mein $L = 0.10$ m, $v = 340$ m/s hai. Kya yeh possibly half a quarter-wave ke saath vibrate kar sakta hai — yaani $n = 1$ se bhi chhota pattern? Aur uska actual lowest note kya hai?

Forecast: Kya fundamental se neeche koi allowed pattern hai?

Step 1 — boundary rule smallest fit par check karo. Kyun? Ek standing wave ko closed end par node aur open end par antinode ZAROOR rakhna chahiye. Figure s04 dekho: left wall par dono curves compare karo. Red solid curve wahan zero par hai (ek legal node); black dotted curve wahan ek bulge par chadh jaati hai (ek closed wall par illegal antinode). Sabse chhoti legal curve jo ek wall par node aur doosri par bulge rakhti hai woh red quarter wave hai — tum physically isse chhota koi draw hi nahi kar sakte. Toh $n = 1$ definition se sabse chhota hai; kuch bhi lower exist nahi karta.

Step 2 — woh fundamental compute karo. Yeh step kyun? Yeh prove karne ke baad ki quarter wave sabse chhota allowed shape hai, hum ab master relation se uski wavelength ko actual lowest pitch mein badlte hain. $λ_{1} = 4 L = 0.40 m, f_{1} = \frac{v}{λ _{1}} = \frac{340}{0.40} = 850 Hz .$

Verify: "Half-a-quarter-wave" shape (black dotted) closed end par antinode rakhega — jo node rule violate karta hai — isliye forbidden hai. Achha: degenerate case $f_{1}$ se neeche sneak nahi kar sakta. Units ✓.

Ex 5 — Cell E · Zero aur limiting inputs

Worked example Statement

Ek open pipe ke liye $f_{1} = v /2 L$ use karke describe karo ki pitch ka kya hota hai jab (a) $v \to 0$ , aur (b) $L \to \infty$ . Phir borderline number evaluate karo: ek open pipe jahan $L = 1.00$ m, $v = 340$ m/s.

Forecast: Har limit mein, kya pitch upar jaegi, neeche, ya vanish ho jaegi?

Step 1 — $v \to 0$ limit. Kyun? $f_{1} = v /2 L$ directly proportional hai $v$ ke. Agar wave speed zero tak shrink ho jaaye (socho ek aisa medium jahan kuch propagate nahi karta), toh $f_{1} = \frac{v}{2 L} v \to 0 0.$ Koi travelling wave nahi $\Rightarrow$ koi standing wave nahi $\Rightarrow$ koi note nahi. Physically sensible.

Step 2 — $L \to \infty$ limit. Kyun? $L$ denominator mein hai, toh ek huge pipe $f_{1}$ ko zero ki taraf push karta hai: infinitely long pipes infinitely deep hum karte hain (ek rumble hearing se neeche). $f_{1} = \frac{v}{2 L} L \to \infty 0.$

Step 3 — finite borderline value. Yeh step kyun? Limits hume edges batate hain; ab hum middle ko ek concrete, checkable number se anchor karte hain yeh sure karne ke liye ki formula extremes ke beech sensibly behave karta hai. $f_{1} = \frac{340}{2 ( 1.00 )} = 170 Hz .$

Verify: Dono limits $0$ Hz ki taraf push karti hain — consistent, kyunki either speed ko khatam karna ya length ko stretch karna wave ko lengthen/lose karta hai. $170$ Hz number ek low, gentle pitch hai — ek metre-long open tube ke liye plausible. Units ✓.

Ex 6 — Cell F · Real-world word problem

Worked example Statement

Ek didgeridoo player ke paas ek straight wooden tube hai jo dono ends par open hai, $L = 1.50$ m. Ek garam din par speed of sound $346$ m/s hai. Drone (fundamental) pitch kya hai, aur woh kaunsi second harmonic par overblow kar sakti hai?

Forecast: Middle-ish (~250 Hz) se upar ya neeche?

Step 1 — words ko symbols mein translate karo. Kyun? "Open both ends" $\Rightarrow$ antinode–antinode $\Rightarrow$ use karo $f_{1} = v /2 L$ saari harmonics allowed ke saath. Yahi is sentence ka poora physics content hai. $f_{1} = \frac{v}{2 L} = \frac{346}{2 ( 1.50 )} = \frac{346}{3.00} \approx 115.33 Hz .$

Step 2 — overblown note. Kyun? Overblowing agla mode excite karta hai; open pipes har integer allow karte hain, toh second harmonic hai $f_{2} = 2 f_{1} = \frac{346}{1.50} \approx 230.67 Hz .$

Verify: Real life mein ek didgeridoo drone 60–130 Hz ke aas paas hota hai, toh $\approx 115$ Hz bilkul sahi jagah hai ✓. $f_{2}$ exactly double hai $f_{1}$ ka jaisa ek open pipe demand karta hai.

Ex 7 — Cell G · Exam twist: "shared harmonic" trap

Worked example Statement

Exam par claim: "Ek open pipe aur ek closed pipe same length $L = 0.50$ m ( $v = 340$ m/s) ke dono apni harmonic series mein kahin na kahin 340 Hz tone produce karte hain." Sach hai ya jhooth?

Forecast: Reasonable lagta hai — aage padhne se pehle decide karo.

Step 1 — OPEN pipe ki allowed frequencies list karo. Kyun? Hume check karna hai ki kaunse modes exist karte hain, sirf formula plug nahi karna. Open: $f_{1} = \frac{340}{2 ( 0.50 )} = 340$ Hz, phir $680, 1020, \dots$ — 340 Hz fundamental hai, definitely present.

Step 2 — CLOSED pipe ki allowed frequencies list karo. Kyun? Closed pipes even harmonics forbid karte hain. $f_{1} = \frac{340}{4 ( 0.50 )} = 170$ Hz, phir odd multiples: $170, 510, 850, 1190, \dots$

Step 3 — compare karo. Yeh step kyun? Claim ek shared tone ke baare mein hai, toh ise settle karne ka ek hi tarika hai — poochho ki kya wahi number dono lists mein appear karta hai — set membership, koi naya formula nahi. Kya 340 Hz closed list ${170, 510, 850, 1190, \dots}$ mein hai? Nahi — 340 equals $2 \times 170$ hai, closed fundamental ka ek even multiple, jo bilkul wahi family hai jise closed pipe throw away karta hai. Sabse nazdeeqi closed tones iske dono taraf hain ( $170$ neeche, $510$ upar) lekin kabhi hit nahi karte.

Answer: Shared-tone claim FALSE hai. Dono pipes yahan koi common tone share nahi karte: open pipe ki series ${340, 680, 1020, \dots}$ hai aur closed pipe ki ${170, 510, 850, \dots}$ , aur yeh lists kabhi overlap nahi karti (open = even $\times 170$ , closed = odd $\times 170$ ).

Verify: Trap yeh assume karna hai ki dono pipes har integer fill karte hain. Kyunki closed list odd-only hai, open pipe ka fundamental gap mein land karta hai — exactly parent note ki numbering warning ✓.

Ex 8 — Cell H · Inverse problem (formula ulta chalaya)

Worked example Statement

Ek guitar ki open string $f_{1} = 110$ Hz ka fundamental sound karti hai. String length $L = 0.64$ m hai. Iske liye kaunsi wave speed $v$ chahiye, aur $v=\sqrt{T/\mu}$ ke > zariye kaunsa tension $T$ woh deta hai? (Linear density $μ = 3.0 \times 1 0^{- 3}$ kg/m lo.)

Forecast: Kya $v$ kuch sau m/s hoga (strings ke liye typical)?

Step 1 — $f_{1} = v /2 L$ ko $v$ ke liye invert karo. Kyun? Hume pitch aur length di gayi hai aur speed chahiye, toh algebraically unknown ke liye solve karo — dono sides ko $2 L$ se multiply karo: $f_{1} = \frac{v}{2 L} \Rightarrow v = 2 L f_{1} = 2 (0.64) (110) = 140.8 m/s .$

Step 2 — tension nikalo. Kyun? String par speed $v = T / μ$ se aati hai; $T$ ke liye solve karne ke liye dono sides ko square karo (square root undo karo) aur $μ$ se multiply karo: $v = T / μ \Rightarrow T = μ v^{2} = (3.0 \times 1 0^{- 3}) (140.8)^{2} \approx 59.47 N .$

Verify: $v = 140.8$ m/s ek believable low-E-ish string speed hai ✓; $T \approx 59.5$ N ek realistic guitar tension hai (kuch kilograms ka pull). Units: $kg/m \cdot (m/s)^{2} = kg\cdotm/s^{2} = N$ ✓.

Ex 9 — Cell I · End correction (real-pipe fix)

Worked example Statement

Ek open organ pipe ki physical length $L = 0.30$ m aur radius $r = 0.02$ m hai, $v = 340$ m/s. Naive fundamental compute karo, phir corrected fundamental $L_{eff} = L + 0.6 r$ per open end use karke (do open ends hain).

Forecast: Kya correction pitch ko upar uthayega ya neeche girayega?

Step 1 — naive value. Kyun? Compare karne ke liye baseline, clean open-pipe formula use karke. $f_{1}^{naive} = \frac{v}{2 L} = \frac{340}{0.60} \approx 566.67 Hz .$

Step 2 — end corrections add karo. Kyun? Antinode actually har open end ke thoda bahar bulge karta hai, toh wave aise behave karti hai jaise tube thoda lambi ho. Do open ends matlab do corrections: $L_{eff} = L + 2 (0.6 r) = 0.30 + 2 (0.6) (0.02) = 0.324 m .$

Step 3 — corrected frequency. Yeh step kyun? Correction ka poora point pitch fix karna tha, toh hum bade effective length ko wapas $f_{1} = v /2 L_{eff}$ mein feed karte hain yeh value paane ke liye jo ek real pipe actually sound karta hai. $f_{1}^{corr} = \frac{v}{2 L _{eff}} = \frac{340}{0.648} \approx 524.69 Hz .$

Verify: Ek lambi effective tube lower pitch deti hai, toh $524.69 < 566.67$ Hz ✓ — correction hamesha frequency drop karta hai, exactly jaisa parent ka mistake-box warn karta hai.

Active recall

Recall Quick self-test

Ex 3 mein, ek end band karne se fundamental aadha kyun ho jaata hai? ::: Node–antinode boundary ko sirf $λ /4$ chahiye, toh $λ = 4 L$ open pipe ke $2 L$ se double hai; double $λ$ se $f$ half ho jaata hai. Ex 7 mein, 340 Hz closed pipe ki list mein kyun nahi hai? ::: 340 Hz closed fundamental 170 Hz ka even multiple hai, aur closed pipes even harmonics forbid karte hain. Ex 5 mein, dono limits ( $v \to 0$ , $L \to \infty$ ) $f_{1}$ ko kis value par bhejti hain? ::: Zero hertz — koi note nahi. Ex 8 mein, $f_{1}$ aur $L$ se $v$ kaise nikaalte ho? ::: $f_{1} = v /2 L$ ko invert karo: $v = 2 L f_{1}$ . Ex 9 mein, kya end correction pitch ko upar uthata hai ya neeche girata hai? ::: Neeche girata hai, kyunki effective length badi ho jaati hai.

Connections

Standing waves and superposition — kyun sirf certain $λ$ survive karte hain bilkul.
Reflection of waves at boundaries — node/antinode rules jo har step mein use hote hain.
Speed of a wave on a string — Ex 1 aur Ex 8 mein $v$ supply karta hai.
Speed of sound in gases — Ex 2, 3, 6, 9 mein $v$ supply karta hai.
Beats and resonance — kaise ek driver in frequencies par lock karta hai.
Timbre and Fourier synthesis — kyun in overtones ka mix ek instrument define karta hai.