Exercises — Mechanical waves — transverse and longitudinal

Kya check karna hai: particle motion aur wave travel ke beech ka angle. Particles perpendicular move karte hain (upar/neeche) travel ke direction se (rightward) → transverse. Features: crests (sabse uunche points) aur troughs (sabse neeche points). Kyun: transverse ⇔ perpendicular ⇔ crest/trough language; longitudinal ⇔ parallel ⇔ compression/rarefaction language.

Nahi. Woh curve pressure (ya displacement) versus position plot karta hai, particles ka actual sideways path nahi. Sound longitudinal hai: air particles travel direction ke along oscillate karte hain (aage-peeche), compressions aur rarefactions banate hain. Air mein koi shear rigidity nahi hai, isliye yeh apne bulk mein transverse mechanical waves carry nahi kar sakta. Dekho Sound waves.

$y=A\sin (\omega t-kx)$ se term-by-term compare karo:

• $A=0.01\text{m}$ — aage wala multiplier maximum displacement hai.
• $\omega =200\text{rad/s}$ — $t$ ko multiply karne wala number.
• $k=4\text{rad/m}$ — $x$ ko multiply karne wala number. Itna easy kyun hai: standard form hi dictionary hai — positions match karne se symbols decode ho jaate hain.

Tool: $v=f\lambda$ (ek period mein wave ek wavelength aage badhti hai). $\lambda$ ke liye solve karo: $\lambda =\frac{v}{f}=\frac{343}{256}=1.340\text{m}$ $v$ upar kyun hai: $v$ air se fix hoti hai, fork $f$ fix karta hai, toh $\lambda$ ko $v/f$ ke hisaab se adjust karna padta hai.

Tool: transverse string speed $v=\sqrt{T/\mu }$ = √(restoring ÷ inertia). $v=\sqrt{\frac{50}{2.0\times 10^{-3}}}=\sqrt{25000}=158.1\text{m/s}$ Square root kyun hai: yeh tension (jo ek bent element ko seedha kheenchti hai — restoring) aur uski mass (jo acceleration resist karti hai — inertia) ke beech ek tug-of-war hai.

$\omega =628\text{rad/s}$, $k=2\text{rad/m}$ read karo.

• $f=\frac{\omega }{2\pi }=\frac{628}{2\pi }=99.95\text{Hz}\approx 100\text{Hz}$.
• $\lambda =\frac{2\pi }{k}=\frac{2\pi }{2}=3.1416\text{m}$.
• $v=\frac{\omega }{k}=\frac{628}{2}=314\text{m/s}$ ($f\lambda =100\times 3.1416$ se match karta hai). Kyun dono routes agree karte hain: $v=\omega /k=f\lambda$ ek hi relation hai alag-alag form mein.

$\omega =400\text{rad/s}$, $k=5\text{rad/m}$, $A=0.05\text{m}$. (a) $v=\frac{\omega }{k}=\frac{400}{5}=80\text{m/s}$ — pattern ki speed. (b) Particle velocity $v_p=\frac{\partial y}{\partial t}=A\omega \cos (400t-5x)$; iska maximum magnitude hai $v_{p,\max }=A\omega =0.05\times 400=20\text{m/s}.$ Yahan $v=80>v_{p,\max }=20$. Koi contradiction nahi: yeh alag-alag cheezein measure karte hain — shape kitni fast sideways travel karti hai vs ek single particle kitni fast upar-neeche wiggle karta hai. Inhe equal ya kisi particular order mein hona zaroori nahi.

Tool: slope–velocity link $v_p=-v\frac{\partial y}{\partial x}$ (dono same $y$ ko differentiate karne se aate hain; figure dekho). $v_p=-(60)(+0.03)=-1.8\text{m/s}.$ Negative → particle neeche move karta hai. Minus sign kyun hai: right ki taraf move karti wave ke liye, ek particle woh le leta hai jo uske left neighbour ke paas abhi abhi tha. Figure dekho: ek positive (upward) slope pe, thoda left wala point neeche hai, toh thodi der baad yeh particle bhi neeche hoga → yeh neeche ja raha hai.

Frequency source se set hoti hai aur boundary cross karne pe change nahi hoti (boundary particle source ki rate pe driven hota hai) → $f=500\text{Hz}$ dono mein.

• Air: ${\lambda }_1=\frac{v_1}{f}=\frac{340}{500}=0.680\text{m}$.
• Water: ${\lambda }_2=\frac{v_2}{f}=\frac{1480}{500}=2.960\text{m}$. Water mein $\lambda$ kyun barta hai: faster medium, same $f$, toh $\lambda =v/f$ stretch ho jaati hai.

(a) $\mu =\frac{m}{L}=\frac{9.6\times 10^{-3}}{1.2}=8.0\times 10^{-3}\text{kg/m}$. (b) $v=\sqrt{\frac{T}{\mu }}=\sqrt{\frac{120}{8.0\times 10^{-3}}}=\sqrt{15000}=122.5\text{m/s}$. (c) $\lambda =\frac{v}{f}=\frac{122.5}{210}=0.583\text{m}$. Yeh order kyun: mass/length speed mein jaata hai; source ka $f$ phir $\lambda$ set karta hai $v=f\lambda$ ke through.

Har symbol build karo:

• $A=0.030\text{m}$.
• $\omega =2\pi f=2\pi (25)=157.08\text{rad/s}$.
• $\lambda =\frac{v}{f}=\frac{50}{25}=2.0\text{m}\Rightarrow k=\frac{2\pi }{\lambda }=\pi =3.1416\text{rad/m}$ (ya $k=\omega /v=157.08/50=3.1416$, same). $+x$ form $y=A\sin (\omega t-kx)$ mein assemble karo: $\boxed{y(x,t)=0.030\sin (157.08t-3.1416x)\text{(SI)}}$ Sign kyun: $-kx$ door wale points ko lag karta hai, jo exactly $+x$ mein move karti wave hai. $-x$ travel ke liye $+kx$ use karte.

Same $T$ hai, toh $v=\sqrt{T/\mu }\Rightarrow v\propto 1/\sqrt{\mu }$. (a) $\frac{v_2}{v_1}=\sqrt{\frac{{\mu }_1}{{\mu }_2}}=\sqrt{\frac{1}{4}}=\frac{1}{2}$ → speed half ho jaati hai. (b) $f$ source se fix hai aur join ke across change nahi hoti. $\lambda =v/f$ ke saath: $\frac{{\lambda }_2}{{\lambda }_1}=\frac{v_2}{v_1}=\frac{1}{2}$ → wavelength half ho jaati hai. $f$ invariant kyun hai: junction particle shared hai; yeh sirf ek rate pe oscillate kar sakta hai — driving rate.

$A=0.02\text{m}$, $\omega =300\text{rad/s}$, $k=15\text{rad/m}$. Particle: $v_p=\partial y/\partial t=A\omega \cos (\varphi )$, acceleration $a_p={\partial }^{2}y/\partial t^2=-A{\omega }^2\sin (\varphi )=-{\omega }^{2}y$, jahan $\varphi =300t-15x$. (a) $a_{p,\max }=A{\omega }^2=0.02\times 300^2=1800{\text{m/s}}^2$. (b) Crest pe $y=+A$, $\sin \varphi =1$ toh $\cos \varphi =0$: $v_p=A\omega \cos \varphi =0$ (momentarily rest pe), aur $a_p=-{\omega }^{2}A=-1800{\text{m/s}}^2$ (max magnitude, equilibrium ki taraf wapas point karti hai, yaani neeche). (c) $y=0$ pe, $\sin \varphi =0$ toh $\cos \varphi =\pm 1$: $v_p=\pm A\omega =\pm 6\text{m/s}$ (maximum speed), aur $a_p=-{\omega }^2(0)=0$. Yeh sirf SHM kyun hai: $a_p=-{\omega }^{2}y$ SHM ka signature hai — extreme displacement ⇔ max acceleration & zero speed; equilibrium ⇔ zero acceleration & max speed. Dekho Simple Harmonic Motion.

(a) $B=\gamma P$ ko $v=\sqrt{B/\rho }$ mein substitute karo: $v=\sqrt{\gamma P/\rho }$. (b) $v=\sqrt{\gamma P/\rho }$. Agar $P\to 2P$ aur $\rho \to 2\rho$, toh $P/\rho$ unchanged hai, toh $v$ unchanged hai. (Isliye sound speed temperature pe depend karti hai, na ki is baat pe ki tum fixed $T$ pe kitna squeeze karte ho.) (c) Fixed $P$ pe $\rho \to 0$ hone par, $P/\rho \to \infty$, toh $v\to \infty$ (unbounded). Physically ideal-gas model isse bahut pehle break down ho jaata hai — bahut rarefied gas disturbance ko particle-to-particle relay nahi kar sakti. Yeh degenerate limit hai: koi medium nahi ($\rho \to 0$) matlab "wave" ka concept khud fail ho jaata hai.

Tool: $v_p=-v\cdot (\partial y/\partial x)$, wave right move karti hai toh $v>0$.

• Point $P$ (rising front): yahan snapshot slope $\partial y/\partial x>0$ hai (curve upar jaati hai jaise $x$ badhta hai). Toh $v_p=-v(\text{positive})<0$ → $P$ neeche move karta hai? — dhyan se: figure check karo. Rightward bump ke leading (right) edge pe slope negative hai (curve aage flat string ki taraf neeche aa rahi hai). Toh $v_p=-v(\text{negative})>0$ → upar: front rise karta hai jaise pulse arrive karti hai. ✓
• Point $Q$ (back/left edge): wahan slope positive hai (curve bump mein upar climb kar rahi hai), toh $v_p=-v(\text{positive})<0$ → neeche: trailing edge fall karti hai jaise pulse wahan se leave karti hai. ✓
• Zero particle speed: pulse ke peak pe slope $=0$ hai, toh $v_p=0$ — woh particle momentarily apne highest point pe hai, instantaneously rest mein. Edges opposite kyun behave karte hain: leading edge lift ho rahi hai jaise shape uski taraf advance karti hai; trailing edge lower ho rahi hai jaise shape wahan se move off karti hai. Peak turning point hai. Yeh L5.1 ki crest/equilibrium logic se match karta hai.

L1 ::: Wave type pehchano aur standard-form symbols read karo. L2 ::: Single-formula application: $v=f\lambda$, $v=\sqrt{T/\mu }$, $y(x,t)$ decode karo. L3 ::: Particle vs wave velocity; slope–velocity link; $f$ media ke across fixed. L4 ::: Relations chain karo $v$, $\lambda$, aur complete $y(x,t)$ build karne ke liye. L5 ::: SHM acceleration link, $v=\sqrt{\gamma P/\rho }$ limits, pulse geometry & degenerate cases.