Harmonics and overtones — on strings and in pipes: Visual walkthrough

1.6.19 · D2 · Physics › Oscillations & Waves › Harmonics and overtones — on strings and in pipes

Step 1 — Wave hoti kya hai, aur woh ek equation jis par hum rely karte hain

WHAT. Ek wave ek chalti hui ripple hai. Do numbers uski shape aur speed describe karte hain:

Wavelength $λ$ — ek crest se agale identical crest tak ki doori. Socho "ek full wiggle kitni lambi hoti hai."
Frequency $f$ — kitne full wiggles ek fixed point se har second guzarte hain, hertz (Hz) mein measure hota hai.

WHY this equation. Agar ek wiggle $λ$ metres lambi hai, aur $f$ wiggles har second guzarti hain, toh ripple $f \times λ$ metres per second aage badhti hai. Wahi speed $v$ hai:

$v = wiggles per second f \times metres per wiggle λ \Rightarrow f = \frac{v}{λ} .$

Yeh aakhri rearrangement is poore page ki puri strategy hai: medium $v$ fix karta hai (dekho Speed of a wave on a string aur Speed of sound in gases); boundaries $λ$ fix karengi; phir hum simply divide karke $f$ nikal lenge. Isse zyada kuch nahi chahiye.

PICTURE. Ek wiggle ek ruler par rakhi hai, uski length $λ$ label ki gayi hai, ek arrow dikhata hai ki woh right taraf speed $v$ par slide kar rahi hai.

Figure — Harmonics and overtones — on strings and in pipes

Step 2 — Ek trapped wave kyun ek standing pattern ban jaati hai

WHAT. Dono ends clamp kar do. Ab wave escape nahi kar sakti — woh boundary se takraati hai aur reflect hokar wapas aati hai (yeh exactly Reflection of waves at boundaries hai). Forward wave aur uska reflection overlap karke ek saath add hote hain — yeh adding Standing waves and superposition hai.

WHY it matters. Jab tum ek wave ko uski khud ki reflection se add karte ho, kuch points kabhi nahi hilaate aur kuch points sabse zyada jhoolte hain. Nateeja ab travel nahi karta — bas jagah par khada hokar saans leta hai. Ise hum standing wave kehte hain.

Do named spots dikhte hain, aur baad ke har step inhi par depend karte hain:

Ek node — ek aisa point jo bilkul still rehta hai (dono waves wahan hamesha cancel hoti hain).
Ek antinode — ek aisa point jo maximum amplitude se jhoolata hai (dono waves wahan hamesha reinforce hoti hain).

PICTURE. Blue forward wave + yellow reflected wave; unka sum green mein still points dikhata hai (red dots = nodes) aur badi-swing wale points (green arrows = antinodes).

Step 3 — Standing wave ke andar chupi do rulers

WHAT. Kisi bhi cheez ko length $L$ mein fit karne se pehle, hume standing-wave shape se do distances chahiye.

WHY. Boundaries demand karengi "yahan node, wahan antinode." Yeh count karne ke liye ki kitne patterns fit hote hain, hume pata hona chahiye ki yeh features kitni door hote hain.

Wave shape se seedha padho: ek full wiggle $λ$ hai; woh har wiggle mein do baar zero cross karti hai (ek node), isliye consecutive nodes aadhi wiggle door hain. Peak (antinode) bilkul beech mein baith jaata hai, har node se quarter-wiggle ki doori par.

PICTURE. Ek standing-wave loop jisme $λ /2$ bracket do red nodes ke beech hai aur $λ /4$ bracket ek node se green antinode tak hai.

Step 4 — Case 1: string fixed at both ends (node–node)

WHAT. Guitar string dono ends par clamp hai, isliye dono ends nodes hone chahiye (ek clamped point hil nahi sakta — dekho Reflection of waves at boundaries).

WHY shapes quantised hain. Do nodes ke beech tumhe poori gannat ke node-to-node gaps fit karne hain, har ek $λ /2$ lamba. Tum dedh gaps nahi fit kar sakte — aadha gap chhod dega ek hilta hua point (antinode) jo clamp par baithega, jo forbidden hai. Isliye:

$L = number of loops n \times one loop’s length \frac{λ _{n}}{2}, n = 1, 2, 3, \dots$

Allowed wavelengths ke liye solve karo, phir har ek ko Step 1 ke $f = v / λ$ mein daalo:

$λ_{n} = \frac{2 L}{n} ⟹ f_{n} = \frac{v}{λ _{n}} = \frac{v}{2 L / n} = \frac{n v}{2 L} = n f_{1}, f_{1} = \frac{v}{2 L} .$

Term by term: $L$ fixed string length hai; $n$ loops count karta hai (aur harmonic number bhi hai); $v$ tension aur mass se set hota hai — $v = T / μ$ , dekho Speed of a wave on a string; $f_{1}$ woh sabse gehri note hai jo string baja sakti hai. Har integer $n$ kaam karta hai, isliye saare harmonics maujood hain.

PICTURE. Teen stacked shapes $n = 1, 2, 3$ ke liye, red nodes dono walls par pinned, loops gine hue.

Step 5 — Case 2: pipe open at both ends (antinode–antinode)

WHAT. Ek open-ended pipe mein, hawa har open end par freely andar-bahar aati hai, isliye dono ends antinodes hone chahiye.

WHY math string jaisa hi hai. Step 3 par wapas jaao: antinode → agla antinode bhi $λ /2$ hai (yeh bas pattern hai jo quarter-wave shift ho gaya hai; same-type features ke beech spacing unchanged hai). Isliye fitting rule same equation hai same $λ /2$ brick ke saath:

$L = n \frac{λ _{n}}{2} ⟹ λ_{n} = \frac{2 L}{n} ⟹ f_{n} = \frac{n v}{2 L} = n f_{1} .$

Yahan $v$ Speed of sound in gases hai string speed ki jagah — yahi ek physical farq hai. Saare harmonics maujood hain, bilkul string ki tarah.

PICTURE. Open pipe ek tube ki tarah draw ki gayi, dono mouths par green antinodes, andar $n = 1, 2, 3$ modes.

Step 6 — Case 3: pipe closed at one end (node–antinode)

WHAT. Ek end band karo. Hawa cap par hil nahi sakti → woh end ek node hai. Open end abhi bhi freely jhoolata hai → antinode. Ab dono ends alag types ke hain.

WHY sirf quarter-wavelengths hain — aur sirf odd wale. Sabse chota shape jo ek end par node aur doosre end par antinode rakhta hai woh ek single node-to-antinode span hai, jo Step 3 batata hai $λ /4$ hai. Agla legal shape paane ke liye tumhe ek aur poora node-to-antinode-to-node stretch ( $λ /2$ ) add karna hoga taaki ends apne required types maintain karein. Baar baar $λ /2$ add karne se $λ /4$ par $\frac{1}{4} λ, \frac{3}{4} λ, \frac{5}{4} λ, \dots$ milta hai — odd quarter-wavelengths:

$L = 1, 3, 5, \dots (2 n - 1) \frac{λ _{n}}{4}, n = 1, 2, 3, \dots$

Solve karo aur $f = v / λ$ se convert karo:

$λ_{n} = \frac{4 L}{2 n - 1} ⟹ f_{n} = \frac{( 2 n - 1 ) v}{4 L} = (2 n - 1) f_{1}, f_{1} = \frac{v}{4 L} .$

$(2 n - 1)$ factor sirf $1, 3, 5, \dots$ generate karta hai — even harmonics forbidden hain. Aur kyunki fundamental sirf $λ /4$ fit karta hai (open pipe ke $λ /2$ se lamba $λ$ ), $f_{1}$ half ho jaata hai: ek closed pipe same length ke open pipe se ek octave neeche bajaata hai.

PICTURE. Closed pipe: capped end par red node, open mouth par green antinode, $n = 1$ ke modes (1st harmonic) aur $n = 2$ (jo 3rd harmonic hai), even shape crossed out.

Step 7 — Degenerate & edge cases (reader ko kabhi adhura mat chhodna)

WHAT tod sakta hai? Chalo corners sweep karte hain taaki koi bhi scenario surprise na kare.

$n = 0$ ? Iska matlab hoga $L = 0$ — zero length, na pipe, na string. Yeh ek physical mode nahi hai; ladder $n = 1$ se shuru hoti hai. Koi "zeroth harmonic" nahi hota.
Closed pipe ka 2nd harmonic ( $2 f_{1}$ )? Yeh simply exist nahi karta. $λ = 2 L /2 = L$ daalne se ek aisa node aur antinode swap demand hoga jo ek boundary violate karta hai. Isliye closed pipes ke liye "1st overtone = 2nd harmonic" galat hai — 1st overtone $3 f_{1}$ par jump karta hai.
Kya equal length ka ek open aur closed pipe kabhi koi pitch share karte hain? Open allow karta hai $170, 340, 510, 680, 850, 1020 \dots$ ; closed allow karta hai $170, 510, 850, 1190 \dots$ ( $f_{1} = 170$ use karke). Woh share karte hain $170, 510, 850, \dots$ — exactly odd open-pipe harmonics. Koi bhi even open harmonic (jaise $1020$ ) closed list mein missing hai.
Bahut zyada $n$ ? Kuch naya nahi: $λ_{n} \to 0$ aur $f_{n} \to \infty$ principle mein, lekin real materials tiny wiggles ko damp kar dete hain, isliye sirf pehle kaafi saare matter karte hain jo tumhe sunne mein aate hain (unka mix timbre hai).

PICTURE. Do harmonic ladders side by side (open vs closed), shared rungs green highlight ki gayi, forbidden even closed rungs greyed out aur crossed.

Ek-picture summary

Upar ki saari cheez ek single diagram mein collapse hoti hai: boundary type → allowed feature spacing → wavelength → frequency ladder. String aur open pipe same road par chalte hain ( $λ /2$ bricks, saare $n$ ); closed pipe $λ /4$ wala turn leta hai aur har even rung drop kar deta hai.

Recall Poore walkthrough ki Feynman-style retelling

Ek rope pakdo jo wall se bandhi ho aur use hilao. Wall end hil nahi sakta, isliye woh ek dead spot hai (ek node). Ek neat standing shape banane ke liye tumhe dono dead ends ke beech poori gannat ke "loops" fit karne hain — ek hump, do humps, teen humps — kabhi bhi dedh hump nahi, kyunki adha wall par ek hilta hua spot rakh dega, jo allowed nahi hai. Har neat shape ko ek choti aur choti wiggle-length chahiye, aur choti wiggles zyada uuncha hum karti hain, isliye tum pitches ki ek ladder par chadhte ho: 1×, 2×, 3× sabse gehri note. Dono ends par open pipe ki same kahaani hai jisme hawa har mouth par freely puffs karti hai — same ladder. Lekin pipe ka ek end band karo aur tum use lopsided bana dete ho: ek end pinned (node), ek end free (antinode). Sabse chota fit ab sirf ek quarter-wiggle hai, isliye uski sabse gehri note aur gehri hai — open pipe se ek octave neeche. Aur bura yeh ki, ends ko sahi types rakhne ke liye tum sirf half-wiggles add kar sakte ho, jo tumhe sirf odd multiples par land karta hai: 1×, 3×, 5×. Har doosri pitch ki woh skipping isliye hoti hai ki ek stopped organ pipe ek flute se hollow aur alag kyun lagta hai. Poori derivation bas yahi hai: har end par node-ya-antinode decide karo, count karo ki kitne $λ /2$ ya $λ /4$ bricks fit hote hain, phir $v$ ko us length se divide karo.

Connections

Standing waves and superposition — Step 2 ka wave + reflection ka addition.
Reflection of waves at boundaries — kyun fixed/closed ends nodes force karte hain.
Speed of a wave on a string — Case 1 mein $v$ .
Speed of sound in gases — Cases 2 & 3 mein $v$ .
Beats and resonance — kaise ek driven column in frequencies par lock karta hai.
Timbre and Fourier synthesis — kyun in harmonics ka mix ek instrument define karta hai.