Mechanical waves — transverse and longitudinal: Worked examples

1.6.13 · D3 · Physics › Oscillations & Waves › Mechanical waves — transverse and longitudinal

Shuru karne se pehle, ek reminder un symbols ka jo hum baar baar use karte hain (sab parent mein banaye gaye hain):

$A$ = amplitude, wo sabse bada displacement jo ek particle reach karta hai (metres mein).
$ω$ = angular frequency (radians per second), aur $f = ω /2 π$ (cycles per second, hertz).
$k$ = angular wave number (radians per metre), aur $λ = 2 π / k$ (metres per cycle).
$v = ω / k = f λ$ = speed jis par pattern move karta hai.
$v_{p} = \partial y / \partial t$ = speed jis par ek single particle upar-neeche move karta hai.
$φ$ = phase constant (radians): ek fixed head-start jo sine ke andar add hoti hai, $y = A sin (ω t - k x + φ)$ . Ye sirf shift karta hai ki $t = 0$ par wave kahan baith rahi hai; ye kabhi $A, ω, k,$ ya $v$ nahi badalta.
$T$ = tension ek string mein (newtons): string ke saath-saath pull jo restoring force ka kaam karti hai. (Period ke saath confuse mat karo; is page par $T$ ka matlab hamesha tension hai.)
$μ$ = string ki linear mass density (kilograms per metre): mass per unit length, wo inertia jise restoring pull move karna hota hai.
Bulk media ke liye: $Y$ = Young's modulus, $B$ = bulk modulus (dono pascals mein, $Pa$ ), $ρ$ = mass density ( $kg/m^{3}$ ) — solid/fluid ke elastic aur inertial properties.

Dekho Mechanical waves — transverse and longitudinal ki in mein se har ek kaisa bana, aur Simple Harmonic Motion ke liye ki sine aata hi kyun hai.

Scenario matrix

Cell	Case class	Kya tricky hai	Example
C1	$+ x$ wave equation padhna	har quantity extract karo, travel ka sign sahi rakho	Ex 1
C2	$- x$ wave equation padhna	sign flip $sin (ω t + k x)$ → direction reverse	Ex 2
C3	Particle velocity aur slope link	$v_{p}$ ko $v$ se confuse mat karo; kisi chosen instant par $v_{p}$ ka sign	Ex 3
C4	Speed medium se (string)	$v = T / μ$ , aur tension $\times$ karne par scaling	Ex 4
C5	Speed medium se (solid vs fluid)	$Y / ρ$ ya $B / ρ$ choose karo; kaun se wave types allowed hain	Ex 5
C6	Degenerate / zero inputs	$A = 0$ , $f = 0$ , $T = 0$ , $k = 0$ — kaun sa "wave" bachta hai?	Ex 6
C7	Limiting behaviour	$f \to \infty$ ( $λ \to 0$ ), $μ \to \infty$ ( $v \to 0$ )	Ex 7
C8	Real-world word problem	echo / distance, pehle model banao phir compute karo	Ex 8
C9	Exam-style twist	graph + given direction se equation banao	Ex 9
C10	Nonzero phase constant $φ$	$y = A sin (ω t - k x + φ)$ padhna/banana	Ex 10

Neeche har numeric answer machine-checked hai verify block mein.

Ex 1 — C1 · Ek wave padhna jo $+ x$ mein move kar rahi hai

Worked example Statement

Ek wave hai $y = 0.05 sin (400 t - 8 x)$ (SI units). $A, ω, f, k, λ, v$ nikalo, aur travel ki direction batao.

Forecast: Andaaza lagao — kya $λ$ 1 metre se bada hoga ya chhota? Kya wave left ja rahi hai ya right?

Step 1. $0.05 sin (400 t - 8 x)$ ko standard form $A sin (ω t - k x)$ se compare karo. Ye step kyun? Standard form ek template hai: har slot ka ek physical meaning hai, to unhe align karne se ek baar mein har quantity mil jaati hai. $A = 0.05 m, ω = 400 rad/s, k = 8 rad/m$

Step 2. Frequency: $f = \frac{ω}{2 π} = \frac{400}{2 π} = 63.7 Hz$ . Ye step kyun? $ω$ radians per second count karta hai; $2 π$ (radians per full cycle) se divide karne par cycles per second milte hain.

Step 3. Wavelength: $λ = \frac{2 π}{k} = \frac{2 π}{8} = 0.785 m$ . Ye step kyun? $k$ radians per metre count karta hai; $2 π$ radians ek full spatial cycle hai = ek wavelength.

Step 4. Speed: $v = \frac{ω}{k} = \frac{400}{8} = 50 m/s$ . Ye step kyun? Ek crest wahan hota hai jahan phase $ω t - k x$ fixed rehta hai; usse fixed rakhne se $x$ ko $ω / k$ ki speed par aage badhna padta hai.

Step 5. Direction: form $ω t - k x$ hai (ek minus), to wave $+ x$ mein move karti hai. Ye step kyun? Phase constant rakhne ke liye jaise $t$ badhta hai, $x$ bhi badhna chahiye — pattern $+ x$ direction mein slide karta hai.

Verify: cross-check $v = f λ = 63.7 \times 0.785 = 50 m/s$ ✓ — Step 4 se match karta hai. Units: $(Hz) (m) = m/s$ ✓.

Ex 2 — C2 · Ek wave padhna jo $- x$ mein move kar rahi hai

Worked example Statement

$y = 0.01 sin (200 t + 5 x)$ (SI). $v$ aur direction of travel nikalo.

Forecast: Ex 1 ke form se sirf ek plus sign ka change hai. Kya ye speed, direction, ya dono badalta hai?

Step 1. $A sin (ω t + k x)$ se compare karo: $ω = 200$ , $k = 5$ . Ye step kyun? Same template idea — lekin ab phase $ω t + k x$ hai.

Step 2. Speed magnitude: $v = \frac{ω}{k} = \frac{200}{5} = 40 m/s$ . Ye step kyun? Speed hamesha $∣ ω / k ∣$ hoti hai; sign direction mein rehta hai, magnitude mein nahi.

Step 3. Direction: phase $ω t + k x$ ko constant rakhne ke liye jaise $t$ badhta hai, $x$ ghatna chahiye, to wave $- x$ mein move karti hai. Ye step kyun? Plus sign matlab "delay ulti taraf run karta hai" — pattern decreasing $x$ ki taraf travel karta hai.

Verify: plus sign ne sirf direction badla, speed nahi ( $40 m/s$ , ek plain magnitude). Units of $ω / k = (rad/s) / (rad/m) = m/s$ ✓.

Ex 3 — C3 · Particle velocity aur slope link

Figure — Mechanical waves — transverse and longitudinal

Ye figure Ex 1 wave ka ek snapshot dikhata hai $t = 0$ par freeze kiya hua (white curve, displacement $y$ page ke upar vs position $x$ uske across). Yellow dot particle ko $x = 0$ par mark karta hai, rest line $y = 0$ par baitha hua. Pink arrow seedha upar point karta hai ussi se — ye particle velocity $v_{p}$ hai, ye dikha raha hai ki ye particle us instant mein climb kar raha hai. Alag blue arrow $+ x$ ki taraf point karta hai — ye wave velocity hai, pura pattern sideway slide kar raha hai. Do alag arrows, do alag motions: inhe is tarah padho "dot pink ke saath bob karta hai; shape blue ke saath march karta hai."

Worked example Statement

$y = 0.05 sin (400 t - 8 x)$ (Ex 1 wave) ke liye, (a) maximum particle speed nikalo, aur (b) $x = 0, t = 0$ par particle velocity nikalo.

Forecast: Wave $50 m/s$ par move kar rahi hai (Ex 1). Guess karo: kya ek particle ki upar-neeche speed usse badi hogi ya chhoti?

Step 1. Particle velocity hai $v_{p} = \frac{\partial y}{\partial t}$ . $y = A sin (ω t - k x)$ ko $t$ ke respect mein differentiate karo ( $x$ fixed rakhte hue): $v_{p} = A ω cos (ω t - k x)$ Ye step kyun? "Ek particle kitni tez upar-neeche move karta hai?" exactly uske displacement ki time mein rate of change hai — ye time-derivative hai, ek particle at a time.

Step 2. Maximum particle speed = $A ω$ kyunki $cos$ maximum 1 tak jaata hai: $v_{p, m a x} = A ω = 0.05 \times 400 = 20 m/s$ Ye step kyun? Cosine 1 se exceed nahi kar sakta; sabse badi speed tab hoti hai jab particle apni rest line se guzarta hai (crest/trough wahan hain jahan wo momentarily rukta hai).

Step 3. $x = 0, t = 0$ par: $v_{p} = A ω cos (0) = 20 m/s$ (upar move kar raha hai, positive). Ye step kyun? Wo instant plug in karo. Figure mein yellow dot dekho: is instant par particle $y = 0$ par baitha hai aur climb kar raha hai — maximum upward speed.

Step 4 (sanity, slope link — yahan derived, assumed nahi). Usi $y = A sin (ω t - k x)$ ko $x$ ke respect mein differentiate karo ( $t$ fixed rakhte hue) snapshot ka slope nikalne ke liye: $\frac{\partial y}{\partial x} = - A k cos (ω t - k x)$ Compare karo ise time-derivative $\partial y / \partial t = A ω cos (ω t - k x)$ se Step 1 se. Ek ko doosre se divide karo: $\frac{\partial y / \partial t}{\partial y / \partial x} = \frac{A ω cos}{- A k cos} = - \frac{ω}{k} = - v$ . Rearrange karo: $v_{p} = \frac{\partial y}{\partial t} = - v \frac{\partial y}{\partial x}$ Ye step kyun? Ye "slope link" hai: ek particle ki upar-neeche speed frozen shape ki local steepness ka $- v$ guna hoti hai. Minus sign physics hai — jahan snapshot right ki taraf neeche slope karta hai, wahan wave slide hoti hua us point mein particle ko upar push karti hai. Wapas plug karne par $v_{p} = A ω cos$ milta hai, to dono routes agree karte hain.

Verify: $v_{p, m a x} = 20 m/s$ wave speed $50 m/s$ se chhoti hai yahan — lekin dhyan do ki ye alag cheezein measure karte hain (upar-neeche vs pattern). Units: $(m) (rad/s) = m/s$ ✓.

Ex 4 — C4 · Speed string se set hoti hai, aur scaling

Worked example Statement

Ek string ki linear mass density $μ = 4 \times 1 0^{- 3} kg/m$ hai aur tension $T = 100 N$ . (a) $v$ nikalo. (b) Agar tension quadruple hokar $400 N$ ho jaaye, to naya $v$ nikalo.

Forecast: Tension quadruple karo — kya $v$ quadruple hoga, double hoga, ya kuch aur?

Step 1. Transverse string speed: $v = \frac{T}{μ}$ . Ye step kyun? Speed ek tug-of-war hai: restoring pull $T$ (tension, upar) vs inertia $μ$ (mass per length, neeche). Square root unke units ko balance karke m/s deta hai.

Step 2. $v = \frac{100}{4 \times 1 0 ^{- 3}} = 25000 = 158 m/s$ . Ye step kyun? Diye gaye medium properties ka direct substitution.

Step 3. $T$ quadruple karo: $v_{new} = \frac{400}{4 \times 1 0 ^{- 3}} = 100000 = 316 m/s$ . Ye step kyun? Kyunki $v \propto T$ , $T$ ko $4$ se multiply karne par $v$ ko $4 = 2$ se multiply kiya jaata hai.

Verify: $316/158 = 2$ exactly ✓ — ne $\times 4$ ko $\times 2$ mein badal diya. Units: $(N) / (kg/m) = m^{2} / s^{2} = m/s$ ✓.

Ex 5 — C5 · Solid vs fluid: sahi elasticity choose karo

Worked example Statement

(a) Ek steel rod: Young's modulus $Y = 2.0 \times 1 0^{11} Pa$ , density $ρ = 8000 kg/m^{3}$ . Longitudinal wave speed nikalo. (b) Paani: bulk modulus $B = 2.2 \times 1 0^{9} Pa$ , $ρ = 1000 kg/m^{3}$ . Sound speed nikalo. (c) In mein se kaun sa medium transverse bulk wave carry kar sakta hai?

Forecast: Steel paani se kahin zyada stiff hai. Guess: kya steel ki speed ~2×, ~5×, ya ~10× paani ki hai?

Step 1. Solid rod Young's modulus $Y$ use karta hai: $v = Y / ρ$ . Ye step kyun? Apni length ke saath stretch hua ek rod $Y$ se resist karta hai; ye longitudinal rod wave ke liye restoring elasticity hai. $v_{steel} = \frac{2.0 \times 1 0 ^{11}}{8000} = 2.5 \times 1 0^{7} = 5000 m/s$

Step 2. Fluid bulk modulus $B$ use karta hai: $v = B / ρ$ . Ye step kyun? Fluid mein koi shear stiffness nahi hoti, sirf compression ka resistance hota hai — wahi $B$ hai. $v_{water} = \frac{2.2 \times 1 0 ^{9}}{1000} = 2.2 \times 1 0^{6} = 1483 m/s$

Step 3. Transverse bulk waves: sirf steel unhe carry kar sakta hai. Ye step kyun? Transverse motion ke liye shear rigidity chahiye. Solids mein hoti hai; fluid (water bulk) mein nahi — bilkul parent ka rule. Dekho Sound waves ki fluid sound longitudinal kyun hai.

Verify: $5000/1483 = 3.37$ — steel ~3.4× faster hai ✓. Dono ke units: $(Pa) / (kg/m^{3}) = m/s$ ✓.

Ex 6 — C6 · Degenerate / zero inputs

Worked example Statement

Discuss karo (a) $A = 0$ , (b) $f = 0$ , (c) string par tension $T = 0$ , (d) $k = 0$ . Kaun sa "wave" bacha rehta hai?

Forecast: In charon mein se kaun sa abhi bhi kuch move karne deta hai?

Step 1. $A = 0$ : $y = 0 \cdot sin (ω t - k x) = 0$ sabhi $x, t$ ke liye. Ye step kyun? Amplitude poore displacement ko scale karta hai; zero amplitude = flat medium = koi wave hi nahi. Har particle ghar mein hi rehta hai.

Step 2. $f = 0$ : to $ω = 2 π f = 0$ , isliye $y = A sin (- k x)$ , time mein frozen. Ye step kyun? Zero frequency matlab time mein kuch oscillate nahi hota — ek static bent shape milti hai, travelling wave nahi. Koi energy propagate nahi hoti.

Step 3. $T = 0$ : $v = T / μ = 0 = 0$ . Ye step kyun? Koi tension nahi to koi restoring force nahi, to ek disturbance aage nahi badh sakti — speed zero ho jaati hai; ek "kink" wahan ka wahan baith jaata hai.

Step 4. $k = 0$ : to $λ = 2 π / k \to \infty$ aur $y = A sin (ω t)$ — koi spatial variation nahi, har particle ek saath (in phase) oscillate karta hai. Ye step kyun? Zero wave number matlab "wave" ki infinite wavelength hai — $x$ ke saath kahin koi crest ya trough nahi, to poora medium ek single SHM oscillator ki tarah bob karta hai. Ye $f = 0$ ke opposite degenerate limit hai: yahan time vary karta hai lekin space nahi. Kuch propagate nahi hota (points ke beech koi phase lag nahi), isliye ye bhi travelling wave nahi hai.

Verify: chaar degeneracies, propagation ki chaar distinct failures — $A \to 0$ (flat), $ω \to 0$ (frozen in time), $v \to 0$ (carry nahi ho sakta), $k \to 0$ (koi spatial pattern nahi). $0 = 0$ ✓ aur $2 π / k \to \infty$ as $k \to 0$ ✓.

Ex 7 — C7 · Limiting behaviour

Worked example Statement

Ek fixed string par (speed $v = 200 m/s$ medium se fixed): (a) jaise $f \to \infty$ , $λ$ ka kya hoga? $f = 100$ , $f = 1000$ , $f = 10000 Hz$ par $λ$ compute karo. (b) Agar instead $μ \to \infty$ fixed $T$ par ho, to $v$ ka kya hoga?

Forecast: Kaun si quantity har limit mein zero ki taraf shrink hoti hai?

Step 1. $v = f λ$ se $v$ fixed rakhte hue: $λ = v / f$ . Ye step kyun? Medium $v$ fix karta hai; source $f$ choose karta hai; to $λ$ adjust hone par majboor hai — ye yahan dependent variable hai.

Step 2. Compute karo: $λ_{100} = \frac{200}{100} = 2 m, λ_{1000} = 0.2 m, λ_{10000} = 0.02 m$ Ye step kyun? Jaise $f \to \infty$ , $λ = v / f \to 0$ — waves tight se tight hoti jaati hain jabki unki speed kabhi nahi badlti.

Step 3. $μ \to \infty$ limit: $v = T / μ \to 0$ . Ye step kyun? Ek infinitely heavy string ki infinite inertia hoti hai, to ek fixed restoring tension use barely accelerate kar sakti hai; disturbance ever more slowly pass hoti hai, aur limit mein wave rengti reh jaati hai ( $v \to 0$ ).

Verify: har $\times 10$ in $f$ se $λ$ mein $\div 10$ mila ( $2 \to 0.2 \to 0.02$ ) ✓ — clean inverse proportionality, aur $v$ poore time $200 m/s$ raha.

Ex 8 — C8 · Real-world word problem (echo)

Ex 9 — C9 · Exam twist: snapshot se equation banao

Ye figure us wave ka ek snapshot hai jo tum abhi banana wale ho: displacement $y$ (upar) vs position $x$ (across), amplitude $A = 0.03 m$ blue vertical arrow se mark kiya gaya hai rest line se crest tak, aur wavelength $λ = 0.5 m$ pink double-arrow se mark kiya gaya hai crest to crest spanning (do yellow dots). Neeche yellow arrow required travel direction ( $- x$ ) dikhata hai. Template $A sin (ω t \pm k x)$ bharne ke liye jo kuch chahiye wo seedha is picture se padhte hain.

Worked example Statement

Ek transverse wave ki amplitude $A = 0.03 m$ , wavelength $λ = 0.5 m$ , frequency $f = 20 Hz$ hai, aur ye $- x$ direction mein travel kar rahi hai. $y (x, t)$ likho aur $v$ nikalo.

Forecast: $- x$ travel diya gaya hai — kya tumhari equation mein $- k x$ hoga ya $+ k x$ ?

Step 1. Angular frequency: $ω = 2 π f = 2 π (20) = 40 π \approx 125.66 rad/s$ . Ye step kyun? Standard form ko $ω$ chahiye, $f$ nahi; cycles/s ko radians/s mein $2 π$ se convert karo.

Step 2. Wave number: $k = \frac{2 π}{λ} = \frac{2 π}{0.5} = 4 π \approx 12.57 rad/m$ . Ye step kyun? Form ko $k$ (radians per metre) chahiye, wavelength se nikala jaata hai.

Step 3. Direction $- x$ ⇒ $ω t + k x$ use karo (Ex 2 ke rule se). To: $y (x, t) = 0.03 sin (40 π t + 4 π x) m$ Ye step kyun? Andar plus sign phase constant rakhne ke liye decreasing $x$ ki zaroorat dalta hai — yaani $- x$ ki taraf travel, jaise required hai.

Step 4. Speed: $v = f λ = 20 \times 0.5 = 10 m/s$ . Ye step kyun? Ek period mein pattern ek wavelength aage badhta hai, isliye $v = f λ$ — direction se independent.

Verify: cross-check $v = ω / k = 40 π /4 π = 10 m/s$ ✓ — $f λ$ se match karta hai. Figure snapshot dikhata hai $λ = 0.5 m$ crest-to-crest marked ke saath.

Ex 10 — C10 · Nonzero phase constant $φ$

Worked example Statement

Ek wave hai $y = 0.04 sin (50 t - 2 x + \frac{π}{2})$ (SI). (a) $A, ω, k, v$ nikalo. (b) $x = 0, t = 0$ par $y$ kya hai, aur ye zero kyun nahi hai? (c) Is wave ko cosine use karke rewrite karo.

Forecast: Us extra $+ \frac{π}{2}$ ke saath andar, kya $A$ , $ω$ , $k$ , ya $v$ ek $φ = 0$ wave ke comparison mein badlega?

Step 1. Full form $A sin (ω t - k x + φ)$ se compare karo: read off karo $A = 0.04$ , $ω = 50$ , $k = 2$ , $φ = \frac{π}{2}$ . Ye step kyun? Phase constant $φ$ template mein bas ek aur slot hai — yahan sabse bada naya idea ye hai ki isse ek fixed head-start ke roop mein recognize karo, rate nahi.

Step 2. Speed aur geometric sab kuch untouched rehta hai: $v = \frac{ω}{k} = \frac{50}{2} = 25 m/s$ . Ye step kyun? $φ$ ek constant hai; ye tab out ho jaata hai jab tum ek crest track karte ho (phase $=$ const abhi bhi $v = ω / k$ deta hai). To $φ$ sirf kahan wave start hoti hai shift karta hai, kitni tez nahi.

Step 3. $x = 0, t = 0$ par: $y = 0.04 sin (0 - 0 + \frac{π}{2}) = 0.04 sin \frac{π}{2} = 0.04 m$ . Ye step kyun? $φ$ ke bina wave origin par $y = 0$ par hoti; $\frac{π}{2}$ head-start use quarter-cycle aage le jaata hai, isliye ye ek crest ( $y = A$ ) par start hoti hai. Exactly yehi ek phase constant karta hai — starting point set karta hai.

Step 4. Cosine form: kyunki $sin (θ + \frac{π}{2}) = cos θ$ , humein milta hai $y = 0.04 cos (50 t - 2 x)$ . Ye step kyun? $+ \frac{π}{2}$ phase shift hai hi sine aur cosine ke beech ka difference — ye dikhata hai ki "kaun sa trig function" aur "kya $φ$ " same choice hain, do alag physics nahi.

Verify: $A, ω, k, v$ bilkul $φ = 0$ case ke method se match karte hain ( $v = 25 m/s$ ); sirf $t = 0$ starting value $0$ se $0.04 m$ par chali gayi ✓. Aur $0.04 sin (50 t - 2 x + \frac{π}{2}) = 0.04 cos (50 t - 2 x)$ ek identity hai ✓.

Recall Scenario matrix — answers cover karo

Sign inside the sine minus hai ( $ω t - k x$ ) ::: wave $+ x$ mein travel karti hai Sign inside the sine plus hai ( $ω t + k x$ ) ::: wave $- x$ mein travel karti hai Maximum particle speed ::: $A ω$ Tension quadruple karne par $v$ factor se badalta hai ::: $4 = 2$ Fluid bulk kaun sa wave type carry nahi kar sakta ::: transverse (koi shear rigidity nahi) $A = 0$ , $f = 0$ , $T = 0$ , $k = 0$ mein se har ek propagation ko khatam karta hai via ::: $A \to 0$ (flat), $ω \to 0$ (frozen), $v \to 0$ (carry nahi), $k \to 0$ (koi spatial pattern nahi) Fixed $v$ par $f \to \infty$ hone par, $λ$ ::: $\to 0$ Echo: round-trip distance ko divide karo ::: 2 se Ek phase constant $φ$ badalta hai ::: sirf kahan wave $t = 0$ par baith rahi hai, $A, ω, k, v$ nahi

Related deep tools: Wave equation, Superposition and Interference, Standing waves & resonance, Doppler effect.