Wave equation — derivation for string: Question bank

1.6.15 · D5 · Physics › Oscillations & Waves › Wave equation — derivation for string

Neeche do figures tumhare mental anchors hain: Figure 1 ek tiny curved element ka free-body picture dikhata hai (kyun pulls cancel nahi hoti); Figure 2 ek small-slope wave (rigidly glides) aur ek large-slope wave (distorts) ko contrast karta hai. Jab tum items work karo toh inhe refer karte raho.

Sach ya jhooth — justify karo

Ek zyada tez (badi amplitude wali) wave string ke along tezi se travel karti hai?

Jhooth. Speed

v = T / μ

hai, jisme koi amplitude term nahi hai; amplitude energy carried set karti hai, na ki shape kitni tezi se move karta hai — dekho Energy carried by a wave.

Ek higher-frequency wave usi ideal string par lower-frequency wali se aage nikal jaati hai?

Jhooth. Ideal string ke liye dispersion relation

v = ω / k =

constant deta hai har frequency ke liye, toh saari frequencies usi speed par saath travel karti hain.

Tension double karne se wave speed double ho jaati hai?

Jhooth. Kyunki

v \propto T

hai,

T

double karne se

v

sirf

2 \approx 1.41

se multiply hota hai; speed double karne ke liye tumhe tension chaar guna karni padegi.

Wave equation sirf Newton's second law hai jo string ke ek tiny piece par apply kiya gaya hai?

Sach.

F_{y} = d m a_{y}

jahan

F_{y} = T y_{xx} d x

aur

d m = μ d x

directly

y_{tt} = (T / μ) y_{xx}

deta hai — yeh Newton's Second Law disguise mein hai.

Agar string ek segment ke upar bilkul seedhi hai, toh kya us segment ko zero net transverse force feel hota hai?

Sach. Seedha matlab slope constant hai, toh

\partial^{2} y / \partial x^{2} = 0

; zero curvature ke saath dono end-tensions ek hi line ke along pull karti hain aur cancel ho jaati hain — koi leftover tug nahi (Figure 1, straight case dekho).

Kya ek nonzero vertical tension component ek end par net transverse force guarantee karta hai?

Jhooth. Dono ends par usually nonzero vertical components hote hain; net force unka difference hota hai, jo zero ho jaata hai agar dono ends ki slope equal ho (seedhi string).

Kya derivation mein tension ke horizontal components ko simply ignore kiya ja sakta hai?

Jhooth. Inhe ignore nahi kiya jaata — yeh dikhaya jaata hai ki yeh dono ends par equal hain (

\approx T cos θ \approx T

) aur isliye cancel ho jaate hain, jo precisely yahi hai jo

T

ko string ke along uniform rakhta hai.

Kya

y = f (x + v t)

bhi ek valid solution hai, sirf

f (x - v t)

nahi?

Sach. Dono

y_{tt} = v^{2} y_{xx}

satisfy karte hain;

f (x - v t)

ek shape hai jo speed

v

par right slide karta hai,

g (x + v t)

ek left slide karta hai — general solution unka sum hai.

Kya derived speed

v

negative ho sakta hai kyunki wave left move kar sakti hai?

Jhooth.

v = T / μ

ek positive magnitude hai; leftward motion argument ke andar ke sign

x + v t

se capture hoti hai, negative

v

se nahi.

Kya small-angle assumption ek mathematical convenience hai jisme koi physical content nahi hai?

Jhooth. Yeh small slopes ke liye ek real physical restriction hai (roughly

θ ≪ 1

rad, yaani

∣ \partial y / \partial x ∣ ≲ 0.1

); isse drop karo aur

sin θ \neq = tan θ

, equation nonlinear ho jaati hai, aur pulse travel karte waqt distort hoti hai (Figure 2, right).

Error dhundo

Har item mein pehli line flawed statement hai aur ::: ke baad reveal kehta hai ki kya inspect karein aur kaise fix karein. Us symbol ya sign ko dhundho jo Figure 1 flag karega.

"Net force

= T sin θ (x + d x) + T sin θ (x)

, dono vertical pulls ko add karke."

Error $+$ sign mein hai. Left end neeche-aur-left pull karta hai, toh uska vertical component

- T sin θ (x)

hai; sahi net force difference

T [sin θ (x + d x) - sin θ (x)]

hai, sum nahi.

sin θ \approx tan θ = \frac{\partial y}{\partial t}

string ki slope hai."

Error $\partial t$ mein hai. Slope ek spatial rate hai:

tan θ = \partial y / \partial x

, na ki

\partial y / \partial t

. Time derivative transverse velocity hai, bilkul alag quantity.

"Net force

T \frac{\partial y}{\partial x} d x

hai, slope ke proportional."

Error first derivative use karne mein hai. Net force curvature

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

ke proportional hai; sirf slope matlab string seedhi-par-tilted hai aur koi net transverse force feel nahi karti — tumhe slope ka change chahiye.

"Kyunki

d m = μ d x

right side par appear karta hai aur

d x

left par, result depend karta hai hum ne

d x

kitna chota choose kiya."

Error yeh sochne mein hai ki $d x$ survive karta hai.

d x

ki single power dono sides se cancel ho jaati hai, toh wave equation element size se independent hai — exactly isliye yeh poori string ke liye valid hai.

"Units:

T / μ = N / (kg \cdot m)

, jo ek speed nahi hai."

Error $μ$ ke units mein hai.

μ

kg per metre (

kg/m

) hai, toh

T / μ = (kg\cdotm/s^{2}) / (kg/m) = m^{2} / s^{2}

; square root m/s hai — ek genuine speed.

y = A sin (k x - ω t)

wave equation ko

kisi bhi ω aur k

ke liye solve karta hai."

Error 'kisi bhi' mein hai. Yeh tab hi kaam karta hai jab

ω = v k

; plug in karne par

- ω^{2} y = - v^{2} k^{2} y

milta hai, jo dispersion relation

ω = v k

force karta hai.

Why questions

Kyun curvature — slope nahi — transverse force create karti hai?

Constant slope matlab dono end-tensions ek hi line ke along point karti hain aur cancel ho jaati hain; sirf changing slope (curvature) dono pulls ko alag-alag point karwata hai, ek net sideways tug chodta hai (Figure 1).

Kyun humein ordinary

d / d x

ki jagah partial derivatives

\partial y / \partial x

aur

\partial y / \partial t

use karni chahiye?

y (x, t)

do independent variables par depend karta hai;

\partial / \partial x

time ko freeze karta hai shape padhne ke liye,

\partial / \partial t

position ko freeze karta hai motion padhne ke liye — dekho Partial derivatives and curvature.

Kyun wave speed medium (

T, μ

) par depend karti hai lekin sound-related "loudness" nahi karti?

v

force-per-curvature (

T

) versus inertia-per-length (

μ

) ke balance se aata hai — yeh dono string khud ki properties hain, tumne use kaise shake kiya usse independent.

"Tug versus sluggishness" view se, kyun

T / μ

natural combination hai?

T

restoring pull hai (bada

T

⇒ snappier return ⇒ faster),

μ

motion resist karne wali inertia hai (bada

μ

⇒ slower); speed

T

ke saath badhti hai aur

μ

ke saath ghatti hai, aur square root units sahi banata hai — Wave speed on a string — v = sqrt(T over mu).

Kyun do opposite-moving solutions ko add karke standing waves bana sakte hain?

Wave equation linear hai, toh solutions ka koi bhi sum ek solution hai; ek right-mover aur ek equal left-mover fixed nodes aur antinodes mein interfere karte hain — Standing waves on a string.

Kyun guitar string ki pitch raise karne ke liye thodi si gain ke liye bahut zyada tightening chahiye?

Pitch

\propto v \propto T

hai, toh yeh sirf tension ke square root ke saath badhti hai; frequency ko 2 factor se raise karne ke liye string ko chaar guna tightly kheenchna padega.

Kyun same

F = ma

derivation ki parwah nahi ki wave transverse hai ya generally longitudinal?

Method (spatial variation se net force, phir

ma

) universal hai, lekin physical restoring quantity differ karti hai — yahan tension versus kahin compression; contrast karo Transverse vs Longitudinal Waves mein.

Edge cases

Agar displacement small nahi hai (large amplitude) toh derivation ka kya hoga?

sin θ \approx tan θ

fail ho jaata hai aur

cos θ \neq \approx 1

, toh tension uniform nahi rehti; equation nonlinear ho jaati hai aur pulses rigidly glide karne ki jagah steepen ya distort hote hain (Figure 2, right panel).

Agar

μ \to 0

(ek massless string) finite tension ke saath ho, toh

v

kya predict karta hai?

v = T / μ \to \infty

; ek inertia-free string disturbances ko instantaneously transmit karti — ek signal ki idealisation break down ho gayi, kyunki koi real string massless nahi hoti.

Agar

T \to 0

(ek slack string) ho, toh wave propagation ka kya hoga?

v = T / μ \to 0

; koi restoring pull nahi hai toh koi bent piece ko snap back karne wala nahi, toh transverse waves travel nahi kar sakti — medium unhe support nahi kar sakta.

Ek bilkul flat, horizontal string at rest (

y \equiv 0

) — kya yeh wave equation ka solution hai?

Haan, trivial wala: dono

y_{tt}

aur

y_{xx}

zero hain, toh

0 = v^{2} \cdot 0

hold karta hai; equilibrium hamesha ek valid (agar boring) solution hai.

Ek sine wave ke exact crest par, slope zero hai — kya wahan transverse force bhi zero hai?

Nahi. Slope zero matlab shape ki peak hai, lekin force curvature

y_{xx}

follow karta hai, jo crest par magnitude mein maximum hai (sabse sharply bent) — isliye strongest downward tug.

Waveform ke ek inflection point par, jahan curvature

y_{xx} = 0

ho, net transverse force kya hai?

Us instant zero, kyunki force

\propto y_{xx} = 0

; wahan string momentarily seedhi hai, toh woh element accelerate nahi ho raha chahe woh move kar raha ho.