Wave equation — derivation for string: Worked examples

1.6.15 · D3 · Physics › Oscillations & Waves › Wave equation — derivation for string

Yeh page Wave equation — derivation for string ka workout room hai. Parent note ne yeh equation banai thi: $\frac{\partial ^{2} y}{\partial t ^{2}} = v^{2} \frac{\partial ^{2} y}{\partial x ^{2}}, v = \frac{T}{μ} .$ Yahan hum isme har tarah ke sawaal daalenge — normal numbers, doubling aur halving, ek value jo zero ho jaati hai, ek degenerate string, ek real-world word problem, aur ek exam-style twist. Exam mein kuch bhi naya nahi lagega iske baad.

Shuru karne se pehle, symbols ka ek simple-sa reminder, taaki pehli line bina kisi background ke samajh aa sake:

Scenario matrix

Is topic mein jo bhi sawaal aa sakte hain, woh in cells mein se koi ek hain. Neeche ke examples mein har example ke saath uski cell tag ki gayi hai.

Cell	Kya test karta hai	Kis example mein
A. Direct value	$T, μ$ ko $v = T / μ$ mein plug karo	Ex 1
B. Scaling up	$T$ (ya $μ$ ) ko kisi factor se multiply karo — $v$ kaise badlega?	Ex 2
C. Scaling down	Kisi quantity ko divide/halve karo — ulti direction	Ex 3
D. Zero / degenerate input	$T \to 0$ , ya $μ \to 0$ , ya ek slack string	Ex 4
E. Limiting behaviour	Kya hota hai jab $T \to \infty$	Ex 4
F. Given a wave, find $v$	$v = ω / k$ se $y = A sin (k x - ω t)$ se $v$ padho	Ex 5
G. Left- vs right-mover / sign	Kya $f (x - v t)$ vs $g (x + v t)$ dono solve karte hain? $v$ ka sign	Ex 6
H. Real-world word problem	Guitar / rope with a hanging mass	Ex 7
I. Exam twist (combined)	Do effects ek saath (mass aur tension change) + time-of-travel	Ex 8

Worked examples

Ex 1 — Cell A: Direct value

Ex 2 — Cell B: Tension ko scale up karna

Ex 3 — Cell C: Density ko scale down karna (ulti direction)

Worked example Statement

Ex 1 jaisi tension ( $T = 64 N$ ), lekin ab ek-chauthai mass density wali string use karo: $μ = 0.01 kg/m$ . Nayi speed?

Forecast: Halki string — faster ya slower? Kitni?

Ab $v \propto 1/ μ$ (jab $T$ fix ho). Yeh step kyun? $μ$ fraction ke neeche aur root ke neeche hai. Denominator chota hone se result bada hota hai; root is growth ko thoda control karta hai.
Density $\times \frac{1}{4}$ hone ka matlab $1/ μ$ ka $4 = 2$ se badhna: $v_{new} = 2 v$ . Yeh step kyun? $1/ 1/4 = 4 = 2$ . Halki string wiggles ko zyada tezi se le jaati hai kyunki same tug ke liye accelerate karne ke liye kam mass hai (parent ka $F = ma$ : chota $m$ , bada $a$ ).
$v_{new} = 2 \times 40 = 80 m/s$ .

Verify: $64/0.01 = 6400 = 80 m/s$ ✓. Dhyan do ki Ex 2 aur Ex 3 ne ek hi $80$ diya: $T$ ko chaar guna karna aur $μ$ ko chauthai karna equivalent hain kyunki dono ratio $T / μ$ ko $\times 4$ se change karte hain.

Ex 4 — Cells D & E: Zero, degenerate, aur limiting inputs

Figure — Wave equation — derivation for string

Worked example Statement

$v = T / μ$ ke extremes explore karo: (a) Bilkul slack string, $T \to 0$ . (b) Infinitely rigid pull, $T \to \infty$ . (c) Massless idealised string, $μ \to 0$ .

Forecast: Kis case mein wave rukti hai, aur kis mein infinitely fast ho jaati hai?

(a) $T \to 0$ : $v = 0/ μ = 0$ . Yeh step kyun? Bina tension ke koi restoring tug nahi hota (parent Step 2: vertical force $T \times$ curvature hai — $T$ ko zero karo aur force zero ho jaata hai). Ek bent slack rope bas sag karti hai; kuch propagate nahi hota. Figure mein red curve dekho jo $T \to 0$ par axis se chipakti hai.
(b) $T \to \infty$ : $v = T / μ \to \infty$ . Yeh step kyun? Bada tug ⇒ bada restoring force ⇒ faster response. Curve bina kisi bound ke upar jaata hai. (Real materials mein physically capped hai, lekin equation $\infty$ kehti hai.)
(c) $μ \to 0$ : $v = T / μ \to \infty$ yahan bhi. Yeh step kyun? Zero mass ka matlab $F = ma$ mein kisi bhi force ke liye infinite acceleration — "shape" instantly react karta hai. Isliye ek idealised massless string ko tension instantly transmit karte hue treat kiya jaata hai.

Verify: Ek concrete tiny case se sanity check: $T = 64, μ = 0.0001 \Rightarrow v = 640000 = 800 m/s$ — actually bahut bada hai jab $μ$ ghata ✓. Aur $T = 0.01, μ = 0.04 \Rightarrow v = 0.25 = 0.5 m/s$ — aslmein bahut slow jab $T$ ghata ✓.

Ex 5 — Cell F: Wave di hui hai, $v$ nikalo

Worked example Statement

Ek string par wave hai $y (x, t) = 0.02 sin (5 x - 20 t)$ (SI units). Uski speed nikalo. Agar $μ = 0.05 kg/m$ ho, toh is string mein kaunsi tension hai?

Forecast: Formula mein kaunse do numbers speed set karte hain — $0.02$ , $5$ , ya $20$ ?

Standard travelling wave $y = A sin (k x - ω t)$ se match karo: amplitude $A = 0.02$ , wavenumber $k = 5 rad/m$ , angular frequency $ω = 20 rad/s$ . Yeh step kyun? $k$ , $x$ ka coefficient hai (har metre mein kitne radians ki wiggle) aur $ω$ , $t$ ka coefficient hai (har second mein kitne radians ki wiggle). Amplitude $A$ kabhi speed ko affect nahi karta — parent ki mistake box yaad karo. Dekho Travelling wave function y = A sin(kx - omega t).
$v = ω / k$ use karo (Dispersion relation omega = vk): $v = 20/5 = 4 m/s$ . Yeh step kyun? Parent ke Ex 3 ne dikhaya tha ki sines wave equation solve karte hain tabhi jab $ω^{2} = v^{2} k^{2}$ , yaani $v = ω / k$ . Yeh woh tool hai jo ek likhi hui wave ko speed mein convert karta hai.
Ab $v = T / μ$ ko invert karo $\Rightarrow T = μ v^{2} = 0.05 \times 4^{2}$ . Yeh step kyun? Dono sides ko square karne se root hat jaata hai; $μ$ se multiply karne par $T$ akela reh jaata hai.
$T = 0.05 \times 16 = 0.8 N$ .

Verify: $T / μ = 0.8/0.05 = 16 = 4 m/s = ω / k$ ✓. Amplitude $0.02$ kabhi aaya hi nahi — jaisa promise tha.

Ex 6 — Cell G: Left-mover vs right-mover ( $v$ ka sign)

Worked example Statement

Parent ne dikhaya tha ki $y = f (x - v t)$ (ek shape right ki taraf sliding) equation solve karta hai. Kya $y = g (x + v t)$ (baayi taraf left sliding) bhi solve karta hai? Concrete pulse $y (x, t) = \frac{1}{1 + ( x + 3 t ) ^{2}}$ lo aur check karo; yahan $3$ speed ka role play karta hai.

Forecast: Kya $+$ sign equation ko tod dega, ya equation direction ki parwah nahi karta?

Andar wala part ek variable maano $u = x + v t$ jahan $v = 3$ , toh $y = \frac{1}{1 + u ^{2}}$ . Yeh step kyun? Argument ko bundle karne se hum $u$ ke respect mein ek baar differentiate kar sakte hain aur reuse kar sakte hain — parent ke forecast section wala chain rule trick.
Time derivatives: $\frac{\partial u}{\partial t} = + v$ , isliye $\frac{\partial ^{2} y}{\partial t ^{2}} = y^{''} (u) \cdot v^{2}$ . Yeh step kyun? Chain rule ke do applications; aur sabse important baat yeh ki $+ v$ square ho jaata hai, isliye uska sign gayab ho jaata hai. $(+ v)^{2}$ aur $(- v)^{2}$ identical hain — yahi toh poora point hai.
Space derivatives: $\frac{\partial u}{\partial x} = 1$ , isliye $\frac{\partial ^{2} y}{\partial x ^{2}} = y^{''} (u)$ . Yeh step kyun? $x$ mein koi hidden $v$ nahi hai, isliye second $x$ -derivative sirf $y^{''} (u)$ hai.
Compare karo: $\frac{\partial ^{2} y}{\partial t ^{2}} = v^{2} y^{''} (u) = v^{2} \frac{\partial ^{2} y}{\partial x ^{2}}$ ✓. Yeh step kyun? Yeh $y_{tt} = v^{2} y_{xx}$ se identically match karta hai, chahe $v$ ka sign kuch bhi ho. Toh dono directions solve karte hain — isliye general solution $f (x - v t) + g (x + v t)$ hai: ek right-mover aur ek left-mover, exactly wahi jo reflect hoke Standing waves on a string banaata hai.

Verify: Sample point $x = 0, t = 0$ par: symbolic differentiation (VERIFY dekho) $v = 3$ ke liye $y_{tt} = v^{2} y_{xx}$ numerically deta hai ✓. Figure mein red arrow pulse ko left ki taraf move karte dikhata hai jab $t$ badhta hai, kyunki $x + 3 t = const$ ke liye $x$ ka kam hona zaroori hai.

Ex 7 — Cell H: Real-world word problem (hanging mass)

Worked example Statement

Ek rope jiska mass $0.2 kg$ aur length $2 m$ hai, horizontally ek pulley ke upar jaati hai; ek $5 kg$ ka block uske end se latka hua hai, jisse woh khicha hua hai ( $g = 10 m/s^{2}$ lo). Horizontal part mein ek pluck kitni tezi se travel karega?

Forecast: Tension hanging block se aati hai — kya tum block ka mass use karoge ya uska weight?

Tension = hanging block ka weight: $T = m g = 5 \times 10 = 50 N$ . Yeh step kyun? Block equilibrium mein latka hua hai, isliye rope ki tension uspar gravity ko exactly balance karti hai (Newton's Second Law jab $a = 0$ ho). Force $m g$ use karo, mass nahi — yeh ek classic trap hai.
Rope ki linear density: $μ = \frac{mass}{length} = \frac{0.2}{2} = 0.1 kg/m$ . Yeh step kyun? $μ$ rope ki property hai, uske khud ke mass aur length se compute hoti hai — block ka isse koi lena-dena nahi.
Speed: $v = T / μ = 50/0.1 = 500 \approx 22.4 m/s$ . Yeh step kyun? Same master formula; physics ka kaam sirf $T$ aur $μ$ sahi se dhundna tha.

Verify: $500 = 22.36 \dots$ ; wapas square karke $22.3 6^{2} \approx 500 = 50/0.1$ ✓. Units: $N / (kg/m) = m/s$ ✓.

Ex 8 — Cell I: Exam twist (do changes + travel time)

Worked example Statement

Ek wire ki $v_{1} = 100 m/s$ hai aur length $L = 5 m$ hai. Ek naye experiment mein tum double mass density wali wire use karte ho aur usse triple tension se kheenchte ho. (a) Nayi speed $v_{2}$ nikalo. (b) Ab ek pulse ko $5 m$ cross karne mein kitna time lagega?

Forecast: Do changes opposite directions mein pull kar rahe hain — kya $v$ overall upar jaayegi ya neeche?

Dono effects ko ek ratio mein combine karo. Kyunki $v = T / μ$ , $\frac{v _{2}}{v _{1}} = \frac{T _{2} / μ _{2}}{T _{1} / μ _{1}} = \frac{3 T _{1}}{2 μ _{1}} \cdot \frac{μ _{1}}{T _{1}} = \frac{3}{2} .$ Yeh step kyun? Ratio lene se original $T_{1}, μ_{1}$ cancel ho jaate hain, isliye hume sirf factors $3$ aur $2$ chahiye. Isse actual numbers kabhi jaanne ki zaroorat nahi padti.
Toh $v_{2} = 100 3/2 = 100 1.5 \approx 122.5 m/s$ . Yeh step kyun? $1.5 > 1$ , isliye tension ko triple karna density double karne se jeet jaata hai — wire end mein faster ho jaati hai. Root, phir se, change ko control karta hai.
Travel time (b): $t = \frac{L}{v _{2}} = \frac{5}{122.5} \approx 0.0408 s$ . Yeh step kyun? Distance $=$ speed $\times$ time, rearrange karo. Time aur speed inversely linked hain, isliye faster wire jaldi cross karti hai.

Verify: $v_{2}^{2} / v_{1}^{2} = (122.47)^{2} /10 0^{2} \approx 1.5 = 3/2$ ✓. Aur $v_{2} \cdot t = 122.47 \times 0.04083 \approx 5 m = L$ ✓. Dekho Energy carried by a wave ki jab $T$ aur $μ$ badlein toh speed ke alawa kya kya badalta hai.

Recall Rapid re-run (answers chhupa lo)

Ratio jab $T \times 4$ , $μ$ fixed? ::: $v \times 2$ Ratio jab $μ \times \frac{1}{4}$ , $T$ fixed? ::: $v \times 2$ $T \to 0$ se $v =$ ? ::: $0$ (koi restoring force nahi) $y = A sin (5 x - 20 t)$ se $v$ padho? ::: $v = ω / k = 20/5 = 4 m/s$ Hanging $5 kg$ se tension kitni? ::: $T = m g = 50 N$ , $5$ nahi Kya $g (x + v t)$ equation solve karta hai? ::: Haan — $v$ square ho jaata hai, sign irrelevant hai

Connections

Wave equation — derivation for string
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
Newton's Second Law
Partial derivatives and curvature
Energy carried by a wave
Transverse vs Longitudinal Waves