Wave equation — derivation for string — Hinglish notes

1.6.15 · Physics › Oscillations & Waves

HOW: derivation from first principles

Hum string ka ek tiny element $x$ aur $x + d x$ ke beech lete hain aur transverse (vertical, $y$ ) direction mein $F = ma$ apply karte hain.

Figure — Wave equation — derivation for string

Element ki length $\approx d x$ hai (small slopes ke liye) aur mass hai: $d m = μ d x .$ Yeh step kyun? Mass = (mass per length) × (length). Hum assume karte hain ki displacements chhote hain, isliye stretched length abhi bhi $\approx d x$ hai.

Step 2 — Dono ends par forces

Tension $T$ har end par string ke along kaam karti hai. Right end par string horizontal se angle $θ (x + d x)$ banati hai; left end par angle $θ (x)$ .

Tension ke vertical components:

Right end: $+ T sin θ (x + d x)$ (upar-aur-right kheenchta hai)
Left end: $- T sin θ (x)$ (neeche-aur-left kheenchta hai)

Yeh step kyun? Sirf $T$ ka vertical component transverse motion cause karta hai; horizontal components almost cancel ho jaate hain (aage aur detail hai).

Step 3 — Small-angle approximation

Chhote slopes ke liye, $θ$ bahut chhota hota hai, isliye: $sin θ \approx tan θ = \frac{\partial y}{\partial x} .$ Yeh step kyun? $tan θ$ literally string ki slope hai, jo $\partial y / \partial x$ hai. Yeh geometry ko derivatives mein convert karta hai. (Saath hi $cos θ \approx 1$ , isliye horizontal tension components equal hain aur cancel ho jaate hain — tension har jagah $\approx T$ rehti hai.)

To net vertical force: $F_{y} = T [\frac{\partial y}{\partial x}_{x + d x} - \frac{\partial y}{\partial x}_{x}] .$

Step 4 — Difference ko derivative mein badlo

$x + d x$ aur $x$ par kisi function ka difference, implicitly $d x$ se divide karne par, ek derivative ban jaata hai: $\frac{\partial y}{\partial x}_{x + d x} - \frac{\partial y}{\partial x}_{x} \approx \frac{\partial ^{2} y}{\partial x ^{2}} d x .$ Yeh step kyun? Yeh second derivative ki definition hai: slope element ke across kitna change hoti hai. Changing slope ka matlab hai string curved hai — aur curvature hi net force produce karta hai. To: $F_{y} = T \frac{\partial ^{2} y}{\partial x ^{2}} d x .$

Step 5 — Newton's second law

Transverse acceleration $\partial^{2} y / \partial t^{2}$ hai. Phir $F_{y} = d m \cdot a_{y}$ : $T \frac{\partial ^{2} y}{\partial x ^{2}} d x = (μ d x) \frac{\partial ^{2} y}{\partial t ^{2}} .$ Yeh step kyun? Element ke liye $F = ma$ . Dono sides par same $d x$ cancel ho jaata hai — result is baat par independent hai ki humne kitna chhota piece choose kiya, isliye yeh poori string ke liye hold karta hai.

Step 6 — Cancel karo aur $v$ identify karo

$\frac{\partial ^{2} y}{\partial t ^{2}} = \frac{T}{μ} \frac{\partial ^{2} y}{\partial x ^{2}} .$ Standard form $y_{tt} = v^{2} y_{xx}$ se compare karne par: $v = \frac{T}{μ}$ Yeh step kyun? Dimensional sense: $[T] = N = kg\cdotm/s^{2}$ , $[μ] = kg/m$ , to $T / μ$ ki units hain $m^{2} / s^{2}$ — ek speed squared. ✓

WHY any $f (x - v t)$ solves it (Forecast-then-Verify)

Forecast: Ek pulse jo apni shape maintain karte hue speed $v$ se right slide kare, wo hai $y = f (x - v t)$ . Aao verify karte hain ki yeh equation satisfy karta hai.

Maano $u = x - v t$ .

$\frac{\partial y}{\partial t} = f^{'} (u) \cdot (- v)$ , to $\frac{\partial ^{2} y}{\partial t ^{2}} = f^{''} (u) \cdot v^{2}$ .
$\frac{\partial y}{\partial x} = f^{'} (u)$ , to $\frac{\partial ^{2} y}{\partial x ^{2}} = f^{''} (u)$ .

Phir $y_{tt} = v^{2} f^{''} (u) = v^{2} y_{xx}$ . ✓ Verified. Koi bhi rigidly-moving shape kaam karti hai.

Worked examples

Common mistakes (Steel-man + fix)

Active recall

Recall Quick self-test (hide and answer)

String par wave speed kon se do quantities set karti hain?
Derivation mein curvature kahan enter hoti hai?
Horizontal tension components kyun disappear ho jaate hain?
Kaunsi approximation equation ko linear banati hai?

String ke liye wave equation kya hai?

\partial^{2} y / \partial t^{2} = v^{2} \partial^{2} y / \partial x^{2}

String par wave speed kya hai?

v = T / μ

, jahan

T

tension aur

μ

linear mass density hai.

Derivation mein, kaunsi physical quantity net transverse force produce karti hai?

Curvature

\partial^{2} y / \partial x^{2}

(element ke across slope ka change).

Kaunsa small-angle step use hota hai?

sin θ \approx tan θ = \partial y / \partial x

(aur

cos θ \approx 1

Horizontal tension components kyun vanish ho jaate hain?

Dono

\approx T cos θ \approx T

hain aur dono ends par equal hain, isliye cancel ho jaate hain —

T

uniform rehta hai.

Kya string wave speed amplitude ya frequency par depend karti hai?

Nahi — sirf

T

aur

μ

par (medium par).

F = ma

mein element ka mass kya hai?

d m = μ d x

1-D wave equation ka general solution form kya hai?

y = f (x - v t) + g (x + v t)

(right- aur left-moving shapes).

Agar tension chaar guna ho jaaye, to

v

ka kya hoga?

v

double ho jaayegi, kyunki

v \propto T

Recall Feynman: 12-saal ke bachche ko samjhao

Ek lamba jump-rope imagine karo. Jab rope ka kuch hissa ek chhoti hill ki tarah curve hota hai, to us hill ke dono ends par rope ko thodi alag directions mein kheeencha ja raha hai. Wo pulls perfectly cancel nahi hote, isliye ek chhota leftover tug hota hai jo rope ke us hisse ko upar ya neeche kheenchta hai. Rope jitni zyada sharply bent (curved) hogi, tug utna bada hoga — aur bada tug matlab tez upar-neeche shake. Yeh tug-vs-shake rule, saath mein rope kitni bhaari aur kitni tight hai, decide karta hai ki wiggle rope ke along kitni fast jaayegi.

Connections

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

Wave equation — derivation for string

WHAT we are deriving