Doppler effect — all cases - source moving, observer moving, both: Worked examples

1.6.21 · D3 · Physics › Oscillations & Waves › Doppler effect — all cases - source moving, observer moving, observer moving

Formula use karne se pehle, chalte hain isko ek picture se build karte hain, taaki yeh ek black box rehna band kare.

Formula ko wavefronts se build karna

Doppler effect ka dil ek sawaal hai: har second kitne crests tumhare kaan se takraate hain? Suni gayi frequency ka exactly yahi matlab hai — crests per second.

Neeche ki picture dekho. Left mein source still baitha hai; woh crests ko evenly spaced rings ke set ke roop mein paint karta hai, har ek $λ = v / f$ ki doori par. Right mein source move kar raha hai: har naya crest pichle spot se thoda aage se janam leta hai, isliye rings aage bunch up hoti hain aur peeche spread out hoti hain.

Figure s02 — do ring pictures side by side. Left: ek still source jiske evenly spaced blue circles hain (spacing = wavelength). Right: ek moving source jiske circles aage bunch up hain (red) aur peeche spread out hain (green). Caption poochta hai: har second kitne crests tumhare kaan tak pahunchte hain?

Figure — Doppler effect — all cases - source moving, observer moving, both

Ab chalte hain har knob ko ek term mein badal dete hain.

Observer term (numerator). Maano rings $λ$ spacing ke saath fixed hain aur hawa mein speed $v$ par travel karti hain. Agar tum still khade ho, rings tumhare paas se speed $v$ par guzarti hain, isliye tum unse $v / λ = f$ per second milte ho — koi shift nahi. Agar ab tum unki taraf $v_{o}$ speed par chalte ho, rings tumhare paas combined speed $v + v_{o}$ par aati hain (kisi cheez ki taraf teri closing speed jo tumhare paas aa rahi hai dono speeds ka sum hai — dekho Relative velocity). Ab tum $(v + v_{o}) / λ$ crests per second milte ho. Observer ki speed closing speed mein add hoti hai, isliye woh upar $v + v_{o}$ ke roop mein baithti hai.

Figure s03 — observer term. Vertical blue crest lines fixed spacing $λ$ ke saath right par $v$ par drift karti hain; ek orange observer right par $v_{o}$ par left (unki taraf) daurta hai; ek green box batata hai ki closing speed $v + v_{o}$ hai, jo numerator ban jaata hai.

Source term (denominator). Ab observer ko still rakho aur source ko unki taraf $v_{s}$ par move karo. Ek period $T = 1/ f$ mein source ek crest emit karta hai, phir agle emit karne se pehle $v_{s} T$ aage travel karta hai. Isliye agla crest $v_{s} T$ pehle se closer shuru hota hai: aage spacing $λ = v T$ se ghatakar $λ^{'} = v T - v_{s} T = (v - v_{s}) T = \frac{v - v _{s}}{f}$ ho jaati hai. Rings phir bhi $v$ par travel karti hain (medium, source nahi, wave speed set karta hai — dekho Wave speed in a medium). Isliye tum $v / λ^{'} = f \cdot v / (v - v_{s})$ crests per second milte ho. Source ka forward motion denominator ko ghata deta hai $v - v_{s}$ tak.

Figure s04 — source term. Source "emit 1" par (orange) $v_{s} T$ distance "emit 2" (red) tak move kar chuka hai; usne jo do crest fronts paint kiye hain woh sirf $λ^{'} = (v - v_{s}) T$ apart hain (green double-arrow), jo denominator ban jaata hai.

Inhe stack karna. Observer closing speed change karta hai (upar); source spacing change karta hai (neeche). Kyunki yeh do knobs ratio $\frac{closing speed}{spacing}$ ke alag-alag hisson par act karte hain — ek numerator par, ek denominator par — dono ko ek saath ghoomana simply dono histon ko ek saath replace karta hai: $f^{'} = \frac{closing speed of crests}{spacing you meet} = \frac{v + v _{o}}{( v - v _{s} ) / f} = f \frac{v + v _{o}}{v - v _{s}} .$ Yeh step kyun? Hum allowed hain dono corrected pieces ko single fraction $\frac{closing speed}{spacing}$ mein substitute karne ke liye precisely kyunki woh independent hain: observer ki motion kabhi ring spacing ko touch nahi karti, aur source ki motion kabhi closing speed ko touch nahi karti. Independent effects usi ratio mein multiply/divide hote hain bina interfere kiye, isliye hum observer-corrected numerator ( $v + v_{o}$ ) aur source-corrected denominator ( $(v - v_{s}) / f$ ) ko ek saath plug in kar sakte hain. Yahi approach ka master formula hai. Ya to motion reverse karo aur uska sign flip ho jaata hai — jo general form deta hai aur, crucially, humein batata hai kyun sign rule kaam karta hai.

Formula Is page par sirf ek formula

$f^{'} = f \frac{v \pm v _{o}}{v \mp v _{s}}$

$f$ = true source frequency (crests emitted per second), $f^{'}$ = frequency heard.
$v$ = wave speed medium mein (still air $\approx 340$ m/s). Medium referee hai: saari motion iske against measure hoti hai.
$v_{o}$ = observer ki speed, $v_{s}$ = source ki speed.
$\pm$ aur $\mp$ padhna: dono symbols independently choose kiye jaate hain — top sign sirf is baat se fix hota hai ki observer kya karta hai, bottom sign sirf is baat se ki source kya karta hai. Woh lock together nahi hain; tum har ek ko apne aap neeche di gayi do lines use karke set karte ho.
Top sign (observer, $v_{o}$ ): $+$ agar observer source ki taraf move kare, $-$ agar door.
Bottom sign (source, $v_{s}$ ): $-$ agar source observer ki taraf move kare, $+$ agar door.
Ek rule jo dono produce karta hai: har sign is tarah choose karo ki saath-saath aana pitch raise kare. Upar, "toward" add karta hai ( $+ v_{o}$ , bada numerator). Neeche, "toward" subtract karta hai ( $- v_{s}$ , chhota denominator). Dono $f^{'}$ raise karte hain, opposite algebraic reasons se.

"Top" aur "bottom" matlab numerator aur denominator — dividing line ke upar aur neeche ke numbers. Yehi poora vocabulary hai joh humein chahiye.

Recall Aside: agar motion

head-on nahi ho? (off-axis case) Upar sab kuch assume karta tha ki motion source–observer line ke straight along hai. Agar observer ya source kisi angle par move kare, toh sirf woh component jo unhe join karne wali line ke along ho squeezing karta hai.

Figure s05 — off-axis angles. Ek source aur observer do dots par baithte hain jo ek dashed gray "line of sight" se jude hain. Har ek ka velocity arrow us line se kisi angle par hai: source ki velocity line se angle $φ$ banati hai, observer ki angle $θ$ banati hai. Component line ke along — source ke liye $v_{s} cos φ$ , observer ke liye $v_{o} cos θ$ — dashed line par ek chhote arrow ke roop mein draw hai; sideways (perpendicular) component faint draw hai, labelled "no Doppler shift."

$v_{o} \to v_{o} cos θ$ aur $v_{s} \to v_{s} cos φ$ replace karo, jahan (figure dekho) $θ$ observer ki velocity aur dono ko join karne wali line ke beech ka angle hai, aur $φ$ source ki velocity aur usi line ke beech ka angle hai. Head-on motion matlab velocity seedha line ke along point karti hai, isliye $θ = 0$ (ya $φ = 0$ ) aur $cos = 1$ — poori speed count hoti hai. Purely sideways motion matlab velocity perpendicular hai, $θ = 9 0^{\circ}$ , isliye $cos 9 0^{\circ} = 0$ aur koi first-order shift nahi hai. Hum poora off-axis treatment defer karte hain; is page par har example head-on hai isliye $cos = 1$ .

Scenario matrix

Har Doppler problem neeche di gayi table ki ek row hai. Last column us example ka naam batata hai jo use solve karta hai.

#	Kaun move karta hai?	Direction	Humein kya expect karna chahiye	Solved in
A	Sirf Observer	source ki taraf	pitch up	Example 1
B	Sirf Observer	source se door	pitch down	Example 2
C	Sirf Source	observer ki taraf	pitch up	Example 3
D	Sirf Source	observer se door	pitch down	Example 4
E	Dono	ek dusre ki taraf	sabse bada up	Example 5
F	Dono	source bhaage, observer peechha kare	chhota net down	Example 6
G	Zero / degenerate	koi nahi move karta ( $v_{o} = v_{s} = 0$ )	koi shift nahi	Example 7
H	Real-world word problem	ambulance paas se guzri (pehle aur baad)	pehle up phir down	Example 8
I	Hawa chal rahi hai	medium khud drift karta hai	$v$ shift karta hai	Example 9
J	Limiting / exam twist	$v_{s} \to v$	formula breaks → shock wave	Example 10

Sab examples mein hum reuse karte hain: $f = 400$ Hz, $v = 340$ m/s, jab tak problem kuch aur na kehe. Numbers constant rakhna rows ko seedhe compare karne deta hai — exams ke liye yaad rakhne layak trick.

Neeche ka map poori sign logic ko ek card par collect karta hai — cells ke through kaam karte waqt ise view mein rakho.

Figure s01 — sign map. Master formula upar baitha hai; ek blue arrow numerator ko OBSERVER word se jodta hai (toward $\to +$ ), ek orange arrow denominator ko SOURCE se jodta hai (toward $\to -$ ), ek green box master rule batata hai "moving together raises pitch," aur ek red line validity boundary $v_{s} < v$ mark karti hai.

Worked examples

Cell A — Sirf Observer, taraf move karta hua

Worked example 1 — Tum ek parked horn ki taraf sprint karte ho

Horn rest par, $f = 400$ Hz, $v = 340$ m/s. Tum uski taraf $v_{o} = 20$ m/s par daurte ho.

Forecast: Source still hai, isliye wavelength untouched hai. Tum crests mein daudh ke jaate ho, isliye tum unse tezi se milte ho → zyada pitch. 400 aur, kaho, 430 Hz ke beech ka number guess karo.

Step 1 — Decide karo kaun si speeds non-zero hain. Source move nahi karta: $v_{s} = 0$ , isliye bottom sirf $v$ hai. Yeh step kyun? Pehle zero terms hatana sign mistakes rokta hai aur dikhata hai ki sirf observer knob active hai.

Step 2 — Observer sign chuno. Taraf jaana → pitch rise honi chahiye → upar $+ v_{o}$ choose karo. Yeh step kyun? Derivation se, rings mein daurna unki closing speed mein add karta hai ( $v + v_{o}$ ).

Step 3 — Substitute karo. $f^{'} = 400 \cdot \frac{340 + 20}{340} = 400 \cdot \frac{360}{340} = 423.5 Hz$ Yeh step kyun? Ab yeh pure arithmetic hai; physics decisions pehle hi ho chuki hain.

Verify: $423.5 > 400$ ✓ (pitch up, jaise forecast kiya). Units: dono sides par Hz kyunki $v$ -ratio dimensionless hai ✓.

Cell B — Sirf Observer, door bhaagta hua

Worked example 2 — Tum usi horn se bhaag jaate ho

Wahi horn, lekin ab tum $v_{o} = 20$ m/s par bhagte ho.

Forecast: Tum crests ko outrun kar rahe ho, isliye woh tumhe slower pakad rahi hain → pitch drop. 400 Hz se thoda kam expect karo.

Step 1 — Zero terms. $v_{s} = 0$ , bottom plain $v$ hai. Yeh step kyun? Source abhi bhi move nahi kar raha, isliye woh wavelength touch nahi kar sakta; sirf observer knob live hai, exactly jaise Example 1 mein.

Step 2 — Observer sign. Door jaana pitch kam karta hai → upar $- v_{o}$ choose karo. Yeh step kyun? Rings se bhaagna unki closing speed subtract karta hai ( $v - v_{o}$ ); yeh Example 1 ka mirror image hai.

Step 3 — Substitute karo. $f^{'} = 400 \cdot \frac{340 - 20}{340} = 400 \cdot \frac{320}{340} = 376.5 Hz$ Yeh step kyun? Chosen sign ko observer-only formula mein plug karo; physics settle ho gayi hai, sirf arithmetic bachi hai.

Verify: $376.5 < 400$ ✓. Notice karo Ex 1 mein up-shift $+ 23.5$ tha aur yahan down-shift $- 23.5$ hai — observer case ke liye size mein equal, kyunki $v_{o}$ linearly upar enter karta hai. Yeh symmetry observer ke liye special hai; source ise tod dega (Examples 3 & 4 dekho).

Cell C — Sirf Source, taraf move karta hua

Worked example 3 — Ek train horn tumhari taraf daudta hai (tum still khade ho)

Train $f = 400$ Hz sound karti hai jabki tumhari taraf $v_{s} = 30$ m/s par move karti hai. Tum stationary ho.

Forecast: Source apni khud ki waves ka peecha karta hai, aage wavelength squeeze karta hai. Fixed wave speed par chhoti wavelength → zyada pitch. Usi speed par move karne wale observer se bada shift expect karo, kyunki denominator shrink hota hai.

Step 1 — Zero terms. $v_{o} = 0$ , top plain $v$ hai. Yeh step kyun? Tum move nahi kar rahe, isliye rings ki closing speed sirf unki medium speed $v$ hai; sirf source knob active hai.

Step 2 — Source sign. Source taraf → pitch up → neeche $- v_{s}$ choose karo (chhota denominator = bada fraction). Yeh step kyun? Derivation se, source ka tumhari taraf aana ring spacing ko $(v - v_{s}) / f$ tak ghata deta hai.

Step 3 — Substitute karo. $f^{'} = 400 \cdot \frac{340}{340 - 30} = 400 \cdot \frac{340}{310} = 438.7 Hz$

Verify: $438.7 > 400$ ✓. Ek hypothetical observer se compare karo 30 m/s par: woh deta $400 \cdot \frac{370}{340} = 435.3$ Hz. Source shift (438.7) bada hai — proof ki do mechanisms alag hain. ✓

Cell D — Sirf Source, door bhaagta hua

Worked example 4 — Train guzar gayi; horn dur jaata hai

Wahi train, $f = 400$ Hz, ab door $v_{s} = 30$ m/s par move karti hai; tum still khade ho.

Forecast: Source ke peeche wavelength stretched hai → kam pitch. Kyunki $v_{s}$ neeche hai, drop ka size Example 3 ke rise se chhota hona chahiye (division symmetric nahi hai).

Step 1 — Zero terms. $v_{o} = 0$ , top plain $v$ . Yeh step kyun? Tum stationary ho, isliye rings medium speed $v$ par guzarti hain; sirf source knob ghoomaya ja raha hai.

Step 2 — Source sign. Door → pitch down → bottom ko bada banao → $+ v_{s}$ . Yeh step kyun? Peechha karte hue, source har crest ko pichle se zyada door paint karta hai, spacing ko $(v + v_{s}) / f$ tak stretch karta hai.

Step 3 — Substitute karo. $f^{'} = 400 \cdot \frac{340}{340 + 30} = 400 \cdot \frac{340}{370} = 367.6 Hz$ Yeh step kyun? Receding sign ko source-only formula mein insert karo; bada denominator $f^{'}$ ko 400 se neeche le jaata hai.

Verify: $367.6 < 400$ ✓. Ex 3 mein rise $+ 38.7$ tha; yahan drop sirf $- 32.4$ hai. Asymmetric, exactly kyunki $v_{s}$ denominator mein rehta hai. Isi liye ambulance ki pitch zyada suddenly girti hai jab woh approach karti hai, baad ke gentle sag se zyada.

Cell E — Dono move karte hain, ek dusre ki taraf

Worked example 5 — Head-on: tum charging train ki taraf daurte ho

$f = 400$ , $v = 340$ , observer $v_{o} = 20$ par taraf, source $v_{s} = 30$ par taraf.

Forecast: Dono effects pitch raise karte hain; woh stack hote hain. Ab tak ka sabse bada shift expect karo — Example 3 ke 438.7 Hz se upar.

Step 1 — Dono non-zero. Koi bhi speed zero nahi hai, isliye dono slots use hote hain. Yeh step kyun? Dono knobs ab ghoomaye ja rahe hain, isliye hum koi bhi term drop nahi kar sakte.

Step 2 — Signs. Observer taraf → $+ v_{o}$ (top). Source taraf → $- v_{s}$ (bottom). Dono pitch raise karne ke liye chosen. Yeh step kyun? Closing speed bhi badhti hai aur spacing bhi ghatti hai — derivation ke do mechanisms ek saath kaam karte hain.

Step 3 — Substitute karo. $f^{'} = 400 \cdot \frac{340 + 20}{340 - 30} = 400 \cdot \frac{360}{310} = 464.5 Hz$

Verify: $464.5$ dono single-cause results (423.5 aur 438.7) se zyada hai ✓. Sanity: shift $= + 64.5$ , do individual shifts ke sum ke close ( $23.5 + 38.7 = 62.2$ ) — woh combine hote hain, multiplication se ek chhota extra ke saath. ✓

Cell F — Dono move karte hain, source bhaagta hai jab observer peechha karta hai

Worked example 6 — Tum bhagte siren ka peechha karte ho

Source door $v_{s} = 30$ par move karta hai; tum uska peechha karte ho (yaani source ki taraf move karte ho) $v_{o} = 20$ par.

Forecast: Do opposing effects. Peecha karna pitch thoda raise karta hai; source ka bhaagna ise lower karta hai. Recession neeche baitha hai aur dominate karne ka trend karta hai → 400 Hz se thoda drop expect karo.

Step 1 — Dono non-zero. Yeh step kyun? Observer aur source dono move karte hain, isliye dono terms survive karti hain; humein sirf unki directions chahiye.

Step 2 — Signs. Observer source ki taraf move kar raha hai → $+ v_{o}$ (top, raises). Source door move kar raha hai → $+ v_{s}$ (bottom, lowers). Yeh step kyun? Dhyan se: "source ki taraf" wahi hai jo top par $+$ define karta hai bhale hi source khud bhaag raha ho. Observer ki motion ka direction source ke relative $v_{o}$ ka sign fix karta hai; source ki apni direction $v_{s}$ ka sign fix karti hai.

Step 3 — Substitute karo. $f^{'} = 400 \cdot \frac{340 + 20}{340 + 30} = 400 \cdot \frac{360}{370} = 389.2 Hz$ Yeh step kyun? Do effects opposite lines par opposite pitch influence ke saath land karte hain; arithmetic decide karta hai kaun jeeta.

Verify: $389.2 < 400$ ✓ — chhota drop, jaise forecast kiya, kyunki bhaagta source (bottom) tumhare peechha (top) se zyada hai. Agar tum faster daurte, $v_{o}$ overtake karke $f^{'}$ ko 400 se upar lift kar sakta tha. $v_{o} = 50$ try karo: $400 \cdot \frac{390}{370} = 421.6$ Hz > 400. ✓

Cell G — Degenerate: koi move nahi karta

Cell H — Real-world word problem

Worked example 8 — Ambulance tumhare paas se guzarti hai: pehle aur baad

Ek ambulance siren $f = 400$ Hz sound karti hai, $v_{s} = 30$ m/s par straight road par drive karti hai. Tum footpath par khade ho. Jab woh approach kare tum kaun si pitch suno, aur jab woh guzar jaaye? (Woh moments lo jab woh road par bahut upar/neeche ho taaki uski motion almost straight tumhari taraf / straight door ho.)

Forecast: Approaching → high, receding → low, ek sudden fall ke saath jab woh guzarti hai. Yeh classic "neeeee-oowww" hai.

Step 1 — Approaching phase. Source taraf, observer still. Yeh exactly Cell C hai: $f_{approach}^{'} = 400 \cdot \frac{340}{340 - 30} = 438.7 Hz$ Yeh step kyun? Road par bahut upar se velocity almost straight tumhari taraf hai, isliye $cos φ \approx 1$ aur head-on source-only formula apply hota hai.

Step 2 — Receding phase. Source door, observer still. Exactly Cell D: $f_{recede}^{'} = 400 \cdot \frac{340}{340 + 30} = 367.6 Hz$ Yeh step kyun? Ek physical event do matrix cells mein split hota hai — yeh recognize karna exam skill hai.

Step 3 — Jump ka size. $Δ f = 438.7 - 367.6 = 71.1 Hz$ Yeh step kyun? Audible drama do phases ke beech ka difference hai, isliye hum approaching pitch se receding pitch subtract karte hain.

Verify: Pitch ambulance ke guzarate hi lagbhag $71$ Hz "instantly" girta hai — roughly $438.7/367.6 \approx 1.19$ ka ek musical interval, ek minor third ke close. Real siren swoop se match karta hai ✓. Note karo asli 400 Hz kabhi nahi suni jaati except us single instant par jab woh closest approach par ho (velocity tumhari taraf momentarily zero hai — off-axis $cos = 0$ moment).

Cell I — Wind: medium khud move karta hai

Worked example 9 — Ek tailwind source se observer ki taraf chalta hai

Source observer ki taraf $v_{s} = 30$ par; tum still ( $v_{o} = 0$ ); ek wind $w = 10$ m/s par source se tumhari taraf chalta hai.

Forecast: Hawa (referee) tumhari taraf drift karti hai, isliye crests hawa par sawaar hote hain aur effectively medium mein tezi se travel karte hain. Har jagah $v$ ko $v + w$ se replace karo jahan woh medium-mein-wave speed represent karta hai. Source ki frequency shift thodi soft hogi kyunki ek tez $v$ matlab hai $v_{s}$ uska ek chhota fraction hai.

Step 1 — Wave speed correct karo. Observer ki taraf wind wave speed mein add karta hai: $v_{eff} = v + w = 350$ m/s. Yeh step kyun? Medium wave speed ground ke relative set karta hai. Agar puri hawa ka body tumhari taraf $w$ par drift kare, ek crest jo uspar sawaar hai woh hawa ke through $v$ par travel karta hai plus hawa ka $w$ , isliye ground ke relative $v + w$ .

Step 2 — $v_{o}$ aur $v_{s}$ ko ground speeds rakho. $w$ ko $v_{o}$ ya $v_{s}$ mein mat jodo. Yeh step kyun? Wind sirf rings ko carry karti hai; woh ambulance ke wheels ya tumhare pair nahi dhakelta. Observer aur source ki speeds ground ke relative di gayi thin, aur ground move nahi kiya, isliye woh numbers same rehte hain. Sirf wave, jo drifting air mein rehti hai, wind feel karti hai — isi liye $w$ correction sirf $v$ se attach hoti hai.

Step 3 — Master formula ko $v_{eff}$ ke saath apply karo. Observer still, source taraf: $f^{'} = 400 \cdot \frac{v _{eff}}{v _{eff} - v _{s}} = 400 \cdot \frac{350}{350 - 30} = 400 \cdot \frac{350}{320} = 437.5 Hz$ Yeh step kyun? Corrected referee speed set hone ke baad, ordinary source-only formula baaki kaam karta hai.

Verify: Wind ke bina (Example 3) humein $438.7$ Hz mila; tailwind ke saath $437.5$ Hz — thoda chhota shift, jaise forecast kiya, kyunki bada $v_{eff}$ ratio ko dilute karta hai ✓.

Cell J — Limiting case: source sound ki speed tak pahunchta hai

Worked example 10 — Agar

$v_{s} \to v$ ? Formula ka breaking point Source speed ko wave speed ki taraf push karo: $v_{s} \to 340$ m/s, observer still.

Forecast: Denominator $v - v_{s} \to 0$ , isliye $f^{'} \to \infty$ . Physically infinite frequency nonsense hai — kuch tootna chahiye. Aisa hi hota hai: wavefronts ek dusre par pile ho jaate hain.

Step 1 — Denominator ko shrink hote dekho. $v_{s} = 300, 330, 339$ ke liye: $f^{'} = 400 \cdot \frac{340}{40} = 3400, 400 \cdot \frac{340}{10} = 13600, 400 \cdot \frac{340}{1} = 136000 Hz$ Yeh step kyun? Numbers blow-up ko visceral banate hain: jaise $v_{s} \to v$ , ring spacing $(v - v_{s}) / f \to 0$ , isliye $f^{'}$ infinity ki taraf rocket karta hai.

Step 2 — Physically interpret karo. $v_{s} = v$ par har crest pichle ki same forward edge par emit hota hai jahan pichla arrive kiya — woh pressure ki ek single wall mein superpose ho jaate hain. Woh wall ek shock wave hai; $v_{s} > v$ ke liye woh ek sonic boom ka cone banata hai. Yeh step kyun? Math kehti hai " $\infty$ "; physics kehti hai assumption (crests alag rings ki tarah spread hoti hain) ab hold nahi karta. Dekho Sonic boom and shock waves.

Verify: $f^{'} \to \infty$ jaise $v_{s} \to v^{-}$ ✓, aur formula $v_{s} \geq v$ ke liye simply invalid hai — tumhe models switch karne honge. Yahi woh ek row hai matrix ki jise master formula fill nahi kar sakta; woh boundary jaanna khud hi answer hai.

Recall Self-test: cell ka naam batao

"Tum ek stationary bell se door cycle karte ho." Kaun sa cell, aur pitch rise ya fall karta hai? ::: Cell B — sirf observer, door jaata hua → pitch falls ( $- v_{o}$ upar). "Ek drone seedha tumhare upar se ek tone sound karte hue guzarta hai." Kaun se cells, kis order mein? ::: Cell C phir Cell D (approach up, recede down) — Example 8 ki story. "Ek jet exactly Mach 1 par." Kaun sa cell, aur formula kya karta hai? ::: Cell J — denominator $\to 0$ , $f^{'} \to \infty$ , formula invalid; shock wave banta hai.

Connections

Parent: Doppler — all cases — woh derivation jise yeh examples exercise karte hain.
Wave speed in a medium — kyun wind $v$ change karta hai (Cell I), $v_{s}$ nahi.
Wavelength and frequency relation — $λ = v / f$ har step ke peeche.
Relative velocity — Cells A, B, F mein observer-frame crest speed.
Sonic boom and shock waves — $v_{s} \geq v$ limit (Cell J).
Doppler effect of light — symmetric, medium-free cousin.
Beats — do shifted tones mil ke kya banate hain.