Visual walkthrough — Wave equation — derivation for string

1.6.15 · D2 · Physics › Oscillations & Waves › Wave equation — derivation for string

---

Step 0 — Symbols se pehle ke words

Kisi bhi formula se pehle, string ki ek picture fix karte hain aur do cheezein name karte hain jo ise describe karti hain.

Yahaan abhi koi derivative nahi hai — bas ek picture hai char labels ke saath: $x$, $y$, $T$, $\mu$.

---

Step 1 — Ek tiny piece par zoom in karo

KYA. Hum string ka ek bahut chhota hissa kaat lete hain, position $x$ se position $x+dx$ tak. Symbol $dx$ ka matlab hai "x ka ek tiny extra bit" — itna chhota ki woh piece almost straight ho.

KYUN. Newton's second law, $F=ma$, ek object ke baare mein law hai. Ek poori wiggly string bahut complicated hai. Lekin itni chhoti piece jo almost straight ho woh ek simple object hai jise hum push kar sakte hain. Agar yeh law har tiny piece ke liye sahi hai, toh poori string ke liye bhi sahi hai.

PICTURE.

---

Step 2 — Ends par do pulls

KYA. Tension piece ko dono cut ends par khenchti hai, string ke saath baahir ki taraf.

KYUN. String ka ek piece apne aap ke middle se kuch feel nahi karta — sirf unse jo iske do ends ko kheench rahe hain. Wahi do khinchaav hain jo ise accelerate kar sakte hain. Isliye humein exactly unhe dekhna hai.

PICTURE. Figure mein notice karo ki string har end par tilted hai, aur dono tilts same nahi hain — yahi difference poora game hai.

Har pull ka size $T$ hai lekin yeh string ke saath point karta hai, jo slanted hai. Hum har pull ko ek horizontal part aur ek vertical (sideways) part mein split karte hain. Tilt angle $\theta$ kehte hain — string aur horizontal ke beech ka angle. Tab:

Right end par string upar-daayein tilts karti hai, isliye iska sideways pull $+T\sin \theta (x+dx)$ (upar ki taraf) hai. Left end par string neeche-baayein tilts karti hai, isliye iska sideways pull $-T\sin \theta (x)$ (neeche ki taraf) hai. Net sideways pull unka sum hai: $F_y=T\sin \theta (x+dx)-T\sin \theta (x).$

---

Step 3 — Horizontal pulls kyun cancel hote hain (ek zaroori side-trip)

KYA. Hum horizontal direction check karte hain aur dikhate hain ki woh kuch contribute nahi karta.

KYUN. Agar piece sideways bhi khich rahi hoti aur $x$ ke saath bhi drag ho rahi hoti, toh problem two-dimensional aur horrible hoti. Hum prove karna chahte hain ki horizontal parts ek doosre ko cancel kar dete hain taaki hum unhe bhool sakein.

PICTURE.

Small tilts ke liye, $\cos \theta \approx 1$, isliye har end par horizontal part $T\cos \theta \approx T$ hai — dono ends par same size lekin opposite directions mein. Opposite aur equal ⇒ cancel ho jaate hain.

---

Step 4 — Angle ko slope mein badlo

KYA. Hum angle $\theta$ ko string ki slope se replace karte hain.

KYUN slope? Hum angle measure kar sakte hain, lekin uspar easily calculus nahi kar sakte. Slope, doosri taraf, directly $y$ ki derivative hai — jo hume Partial derivatives and curvature ki machinery laane deta hai. Isliye hum geometry ko ek aisi derivative mein translate karte hain jo hum compute kar saken.

Yeh particular tool kyun — $\tan$? Right triangle mein, $\tan \theta =\frac{\text{opposite}}{\text{adjacent}}=\frac{\text{rise}}{\text{run}}$. Woh ratio hi string ki steepness hai, aur $y$ ki $x$ ke muqable steepness exactly derivative $\partial y/\partial x$ hai. Isliye $\tan$ woh ek trig function hai jo slope ke barabar hota hai.

PICTURE.

Small tilts ke liye, angle ka sine aur tangent almost equal hote hain (figure mein dekho ki $\theta =0$ ke paas dono curves kaise saath rehte hain), isliye:

Ab vertical force ban jaati hai, $y^{\prime }\equiv \partial y/\partial x$ ke saath: $F_y=T[\underset{\text{right end par slope}}{\underbrace{y^{\prime }(x+dx)}}-\underset{\text{left end par slope}}{\underbrace{y^{\prime }(x)}}].$

---

Step 5 — Slope ka change curvature hai

KYA. Hum "right ki slope minus left ki slope" ko ek cheez ke roop mein rewrite karte hain: slope kitni fast change hoti hai.

KYUN. Bracket $[y^{\prime }(x+dx)-y^{\prime }(x)]$ do nearby jagahon par slope ka difference hai. "(Value yahaan $+dx$ par) minus (value yahaan par)" ko $dx$ se divide karna ek derivative hai — yahi derivative ki definition hai. Slope $y^{\prime }$ par yeh karna slope ki derivative deta hai, yaani second derivative.

PICTURE. Figure piece ke upar string dikhata hai: jahan slope turn hoti hai (string bend hoti hai), dono end-slopes differ karte hain. Ek straight piece (koi bend nahi) ke equal end-slopes hote hain aur isliye zero net force.

Ab net sideways pull sirf curvature se likhi jaati hai: $\boxed{F_y=T\frac{{\partial }^{2}y}{\partial x^2}dx}$ Ise zyaaz se padho: net sideways force = tension × curvature × length. Zyaada bend ⇒ zyaada force. Ek straight piece par koi net pull feel nahi hoti — exactly figure ka point yahi hai.

---

Step 6 — Piece par Newton's second law

KYA. Hum force ko mass times acceleration ke barabar set karte hain.

KYUN. Yeh Newton's Second Law hai — woh akela law jo force ko motion se connect karta hai. Humne Step 1 mein left side (force) aur mass build kiya; ab acceleration chahiye.

Acceleration yeh hai ki sideways position $y$ time mein kitni fast speed up hoti hai, jo second time-derivative ${\partial }^{2}y/\partial t^2$ hai. (Pehli time-derivative = velocity; doosri = acceleration — same idea jaise position→speed→acceleration.)

PICTURE.

---

Step 7 — $dx$ cancel karo aur speed padho

KYA. Dono sides ko $dx$ se divide karo; woh gayab ho jaata hai.

KYUN yeh matter karta hai ki $dx$ cancel ho. Humne jo piece choose ki uska size arbitrary tha. Agar answer abhi bhi $dx$ par depend karta, toh hamari derivation bakwaas hoti. Yeh cancel ho jaata hai — isliye law har piece ke liye same hai, aur isliye poori string ke liye bhi.

$\frac{{\partial }^{2}y}{\partial t^2}=\frac{T}{\mu }\frac{{\partial }^{2}y}{\partial x^2}.$

Ise standard shape $y_{tt}=v^2{y}_{xx}$ se compare karo. Curvature ko multiply karne wala number ek speed squared hona chahiye:

Yahi $v$ Wave speed on a string — v = sqrt(T over mu) mein use hota hai aur dispersion relation ke peeche hai.

---

Step 8 — Degenerate cases (reader ko kabhi stranded mat chodo)

KYA. Extreme inputs test karo taaki sure ho sake ki picture complete hai.

Curvature ka har sign piece ko straight ki taraf wapas bend karta hai — wahi restoring tug exactly oscillation banata hai, aur oscillations ki ek travelling chain wave hai (dekho Travelling wave function y = A sin(kx - omega t)).

---

Ek-picture summary

Poori derivation logic ka ek arrow hai:

$\underset{{\partial }^{2}y/\partial x^2}{\underbrace{\text{curvature}}}\stackrel{\times T}{\to }\underset{F_y}{\underbrace{\text{net force}}}\stackrel{÷dm}{\to }\underset{{\partial }^{2}y/\partial t^2}{\underbrace{\text{acceleration}}}⟹v=\sqrt{\frac{T}{\mu }}$

Recall Feynman: poora walkthrough kisi dost ko batao

Ek lamba tight rope lo. Usme se ek teeny piece kaat lo, itna chhota ki woh straight lagey. Dono taraf ki rope iske do ends ko apni-apni taraf kheenchti hai. Agar woh chhota piece bent hai — chahe thoda sa bhi — toh do tugs thodi-thodi-different directions mein point karte hain, isliye woh fully cancel nahi hote aur ek chhota leftover sideways pull reh jaata hai. Kitna leftover pull? Yeh badhta hai jitni tightly rope kheenchi ho ($T$) aur jitna sharply piece bent ho (uski curvature). Woh leftover pull, piece kitna heavy hai ($\mu$ times uski length) se divide karke, batata hai ki woh kitni tezi se upar ya neeche accelerate karta hai. Ise likho aur piece ki length cancel ho jaati hai — proof ki yeh poori rope ke liye sach hai. Tugs ko time ke saath line karo aur tumhe ek shape milti hai jo ek fixed speed $v=\sqrt{T/\mu }$ par rope ke saath chalti hai: tighter aur lighter matlab faster. Bend, tug, shake — poori wave equation bas yahi hai.

---

Connections

• Wave equation — derivation for string (parent)
• Newton's Second Law
• Partial derivatives and curvature
• Wave speed on a string — v = sqrt(T over mu)
• Travelling wave function y = A sin(kx - omega t)
• Dispersion relation omega = vk
• Standing waves on a string
• Transverse vs Longitudinal Waves
• Energy carried by a wave