Exercises — Wave equation — derivation for string

KYA chahiye: $v$. YEH formula kyun: string par speed sirf medium se fix hoti hai (tension aur density), isliye hum $v=\sqrt{T/\mu }$ use karte hain — koi bhi wave shape ki zaroorat nahi. $v=\sqrt{\frac{T}{\mu }}=\sqrt{\frac{64}{0.04}}=\sqrt{1600}=40\text{m/s}.$ Units check: $\sqrt{\text{N}÷(\text{kg/m})}=\sqrt{({\text{kg⋅m/s}}^2)\cdot (\text{m/kg})}=\sqrt{{\text{m}}^2/{\text{s}}^2}=\text{m/s}.$ ✓

(a), (b), (d) hamesha solutions hain. Wave equation linear hai, isliye ek right-mover plus ek left-mover phir bhi ek solution hai — yahi (d) hai, general solution (dekho Travelling wave function y = A sin(kx - omega t)). (c) generally solution NAHI hai: ek arbitrary product $f(x)g(t)$ sirf special separated forms ke liye kaam karta hai (jaise $\sin (kx)\cos (\omega t)$ wali standing waves), har product ke liye nahi. Dekho Standing waves on a string. KYU: parent note mein humne verify kiya tha ki $u=x\pm vt$ se $y_{tt}=v^2{y}_{xx}$ automatically satisfy hoti hai; ek free product mein koi aisa built-in $v$ nahi hota.

$y$ ek length hai (m), $x$ ek length hai (m), $t$ ek time hai (s). $\left[\frac{{\partial }^{2}y}{\partial x^2}\right]=\frac{\text{m}}{{\text{m}}^2}={\text{m}}^{-1},\left[\frac{{\partial }^{2}y}{\partial t^2}\right]=\frac{\text{m}}{{\text{s}}^2}.$ YEH kyun matter karta hai: equation $y_{tt}=v^2{y}_{xx}$ ke liye phir $[v^2]=\frac{{\text{m/s}}^2}{{\text{m}}^{-1}}={\text{m}}^2/{\text{s}}^2$ chahiye — ek speed squared. Units force karte hain ki $v$ ek speed ho.

KYA: $v=\sqrt{T/\mu }$ ko $T$ ke liye solve karo. KYU: teen mein se do quantities pata hain; algebra teesri deta hai. Square root hatane ke liye dono sides square karo: $v^2=\frac{T}{\mu }\Rightarrow T=\mu v^2=0.02\times 120^2=0.02\times 14400=288\text{N}.$

Same material + same thickness-per-material matlab $\mu \propto$ area, toh ${\mu }_B=2{\mu }_A$. Tension equal. $\frac{v_B}{v_A}=\frac{\sqrt{T/{\mu }_B}}{\sqrt{T/{\mu }_A}}=\sqrt{\frac{{\mu }_A}{{\mu }_B}}=\sqrt{\frac{1}{2}}=\frac{1}{\sqrt{2}}\approx \mathrm{0.707.}$ KYU: length ke hisaab se zyada bhaari ($\mu \uparrow$) → hilane ke liye zyada inertia → dheemi wave. Moti string patli wali ki speed ka lagbhag $70.7\%$ par chalti hai.

Step 1 — wave se speed. Non-dispersive string par ek sine ke liye Dispersion relation omega = vk deta hai $v=\frac{\omega }{k}=\frac{200}{5}=40\text{m/s}.$ KYU: parent note ke Example 3 mein, sine ko wave equation mein plug karne par ${\omega }^2=v^2{k}^2$ force hua, yaani $v=\omega /k$. Step 2 — medium se tension. Ab $v=\sqrt{T/\mu }$ ulta use karo: $T=\mu v^2=0.05\times 40^2=0.05\times 1600=80\text{N}.$

Neeche wala figure dekho.

Shape kya batati hai: crest par string ek pahaad ki tarah bend karti hai — woh neeche ki taraf curve karti hai, isliye ${\partial }^{2}y/\partial x^2<0$ (slope baayein taraf se positive aur daayein taraf se negative hai, yaani woh ghatt raha hai). Isse force kyun milti hai: parent note se net vertical force hai $F_y=T\frac{{\partial }^{2}y}{\partial x^2}dx.$ ${\partial }^{2}y/\partial x^2<0$ ke saath, force $F_y<0$ hai — neeche ki taraf. Ek crest hamesha axis ki taraf wapas khinchi jaati hai. YEH KAISA LAGTA HAI: chote se pahaad ke donon siron par tension arrows thoda sa neeche-aur-bahar ki taraf point karte hain; unke baache hue vertical parts ek neeche ki taraf tug mein add ho jaate hain (figure mein red arrow).

Net force $\propto$ curvature $y_{xx}$. Yeh wahan zero hai jahan curvature sign change karti hai — inflection points, jahan bump upar se curve karna (crest ke paas) se neeche curve karna (tails par) switch karta hai. Do baar differentiate karo: $y=h(1+s^2{)}^{-1},s=x/a.$ $\frac{dy}{ds}=-2hs(1+s^2{)}^{-2},\frac{d^{2}y}{d{s}^2}=h\frac{6s^2-2}{(1+s^2{)}^3}.$ Set numerator zero: $6s^2-2=0\Rightarrow s^2=\frac{1}{3}\Rightarrow s=\pm \frac{1}{\sqrt{3}}$, yaani $x=\pm \frac{a}{\sqrt{3}}\approx \pm 0.577a.$ KYU: un do points par string momentarily seedhi hai (curvature $=0$), isliye tension pulls exactly cancel ho jaate hain — wahan koi transverse force nahi, chahe string move kar rahi ho.

Woh step: humne $\sin \theta$ ko $\tan \theta =\partial y/\partial x$ se replace kiya aur $\cos \theta$ ko $1$ se. Linear kyun: $\partial y/\partial x$ ek linear function hai $y$ ka ($y$ double karo, slope double ho jaati hai). Usse seedha $F=ma$ mein daalne par har term $y$ ya uski derivatives ke proportional rehti hai — koi $y^2$ nahi, koi products nahi. Ek linear equation superposition obey karta hai: do solutions add karke ek solution milta hai (isliye Problem 1.2(d) kaam karta hai). Kya tootta hai: exact vertical force $T\sin \theta$ hai, aur $\sin \theta =\frac{y_x}{\sqrt{1+y_x^2}}$. Woh $\sqrt{1+y_x^2}$ $y_x$ mein nonlinear hai. Bade slopes ke liye isse ab nahi chhoda ja sakta; superposition fail ho jaata hai aur pulses steepen/distort ho jaate hain. Isliye small-angle step ek sachchi physical assumption hai (small amplitude), koi shortcut nahi.

Speeds (same $T$, alag $\mu$): $v_1=\sqrt{\frac{90}{0.01}}=\sqrt{9000}\approx 94.87\text{m/s},v_2=\sqrt{\frac{90}{0.04}}=\sqrt{2250}\approx 47.43\text{m/s}.$ Ratio: $\frac{v_1}{v_2}=\sqrt{\frac{{\mu }_2}{{\mu }_1}}=\sqrt{\frac{0.04}{0.01}}=\sqrt{4}=2.$ Halka half wave ko exactly do guna tezi se carry karta hai. Wave type: string element string ke perpendicular move karta hai ($y$-direction mein) jabki wave $x$ ke along chalti hai — yeh ek transverse wave hai. (Transverse vs Longitudinal Waves mein, transverse = displacement ⟂ propagation.)

Relations chain karo. Dispersion relation omega = vk se, $v=\omega /k$. Definitions substitute karo: $v=\frac{\omega }{k}=\frac{2\pi f}{2\pi /\lambda }=f\lambda .$ Aur $v=\sqrt{T/\mu }$, isliye $\boxed{f\lambda =\sqrt{\frac{T}{\mu }}}.$ Numbers: pehle $v=\sqrt{100/0.01}=\sqrt{10000}=100\text{m/s}$. Phir $f=\frac{v}{\lambda }=\frac{100}{0.5}=200\text{Hz}.$ YEH powerful kyun hai: yeh ek shape property ($\lambda$), ek timing property ($f$), aur ek medium property ($T,\mu$) ko ek hi line mein weld karta hai. Koi bhi ek change karo aur baaki adjust ho jaate hain taaki product $=v$ rahe.

Overlap ke dauran displacement sum $y=f(x-vt)+f(x+vt)$ hai — pulses ek doosre se guzar jaate hain, har point par add hote hain (superposition). Yahi building block hai jisse Standing waves on a string arise hoti hain (equal opposite travelling waves). Confirm karo ki equation satisfy hoti hai: wave equation linear hai, aur humne pehle se har piece alag verify kiya hai: $\frac{{\partial }^2}{\partial t^2}[f(x-vt)]=v^2\frac{{\partial }^2}{\partial x^2}[f(x-vt)],$ aur identically $f(x+vt)$ ke liye. Do sachi equations add karne par ek sachi equation milti hai, toh sum $y_{tt}=v^2{y}_{xx}$ satisfy karta hai. YEH kyun hold karta hai: kisi bhi step ne $y^2$ jaisa product introduce nahi kiya, isliye additivity kharaab nahi hoti.

Element par exact vertical force (donon siron ka net): $F_y=T\sin \theta {|}_{x+dx}-T\sin \theta {|}_x=T\frac{\partial }{\partial x}\left(\sin \theta \right)dx.$ Ab $\tan \theta =y_x$, isliye $\sin \theta =\frac{y_x}{\sqrt{1+y_x^2}}$. Yeh exact geometry hai. Newton's second law (Newton's Second Law) mass $dm=\mu dx$ par: $T\frac{\partial }{\partial x}\left(\frac{y_x}{\sqrt{1+y_x^2}}\right)dx=\mu dx\frac{{\partial }^{2}y}{\partial t^2}.$ Dropped term: small slopes $y_x\ll 1$ ke liye, $\frac{1}{\sqrt{1+y_x^2}}\approx 1-\frac{1}{2}{y}_x^2+\cdots$ expand karo. Correction $-\frac{1}{2}{y}_x^2$ slope mein second order hai. Isse drop karne par $\sin \theta \approx y_x$ milta hai, isliye $T\frac{\partial }{\partial x}(y_x)=\mu y_{tt}\Rightarrow T{y}_{xx}=\mu y_{tt}\Rightarrow \frac{{\partial }^{2}y}{\partial t^2}=\frac{T}{\mu }\frac{{\partial }^{2}y}{\partial x^2}.$ Recover hua $v=\sqrt{T/\mu }$ ke saath. ✓ Saari approximation ek hi jagah rehti hai: $y_x^2$ correction ko throw away karna.

(a) $\mu \to 0$: $v=\sqrt{T/\mu }\to \infty$. Massless string mein koi inertia nahi hai, isliye ek disturbance instantly har jagah appear hoti hai — ek idealisation, literally kabhi realise nahi hoti. (b) $T\to 0$: $v=\sqrt{T/\mu }\to 0$. Koi tension nahi toh koi restoring pull nahi, isliye transverse disturbances propagate nahi karti — string bas flop ho jaati hai. Yeh degenerate "no wave" limit hai. (c) $y_{xx}=0$ har jagah: toh $F_y=T{y}_{xx}dx=0$, isliye $y_{tt}=0$. Koi net force nahi, koi acceleration nahi. Ek seedhi (shayad tilted) string jaisi hai waise rehti hai ya constant transverse velocity se drift karti hai — Newton's first law ke consistent. Koi curvature nahi ⇒ koi wave force nahi. (d) Sudden kink: ek kink par slope $y_x$ discontinuous hai, isliye $y_{xx}$ infinite hai (ek spike) wahan. Idealised equation ek point par infinite force predict karta hai — physically kink instantly smooth ho jaata hai aur do travelling pulses mein split ho jaata hai ($f(x-vt)$ aur $g(x+vt)$). Real strings mein finite bending stiffness hoti hai, jo ise cap karti hai. Clean wave equation smooth slopes assume karta hai, isliye ek true kink strictly uski validity ke thoda bahar hai.

Fixed-length string par fundamental frequency $f=\frac{v}{2L}=\frac{1}{2L}\sqrt{\frac{T}{\mu }}$ hai (length $\lambda =2L$ set karta hai; dekho Standing waves on a string). (i) Sirf $T$ change karo, $L,\mu$ same rakho: $f\propto \sqrt{T}$. $f$ double karne ke liye: $2=\sqrt{\frac{T_{new}}{T_{old}}}\Rightarrow \frac{T_{new}}{T_{old}}=4.$ Tension chaar guna karni padegi (×4) pitch double karne ke liye — isliye over-tightening se strings toot jaati hain. (ii) Density half karo, $T,L$ same rakho: $f\propto 1/\sqrt{\mu }$, isliye $\frac{f_{new}}{f_{old}}=\sqrt{\frac{{\mu }_{old}}{{\mu }_{new}}}=\sqrt{2}\approx \mathrm{1.414.}$ Pitch $\sqrt{2}$ factor se badhti hai (roughly ek musical tritone), 2 se nahi. Is tarike se pitch double karne ke liye $\mu$ ko one quarter karna padega.

• L1: curvature $y_{xx}$ ki units? ::: ${\text{m}}^{-1}$
• L2: $v=120$, $\mu =0.02$ ⇒ $T$? ::: $288\text{N}$
• L3: crest par $y_{xx}$ ka sign, aur force direction? ::: negative; force downward
• L4: $f\lambda =?$ batao ::: $\sqrt{T/\mu }$
• L5: pitch double karne ke liye $T$ par factor? ::: ×4