Doppler effect — all cases - source moving, observer moving, both: Question bank

1.6.21 · D5 · Physics › Oscillations & Waves › Doppler effect — all cases - source moving, observer moving

Shuru karne se pehle, ek refresher taaki yahan har symbol earned lage:

Formula ko ek picture se build karo (taaki yahan kuch bhi black box na ho)

Is page ki har cheez ek equation pe tikhi hai, toh test karne se pehle aao isse visually earn karein.

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

Sirf observer (spacing $λ = v / f$ pe fixed, arrival speed $v + v_{o}$ ho jaati hai): $f^{'} = \frac{v + v _{o}}{λ} = \frac{v + v _{o}}{v / f} = f \frac{v + v _{o}}{v}$

Sirf source — s02 dekho. Ek period $T$ mein source $v_{s} T$ teri taraf badhta hai agla crest release karne se pehle, toh har naya crest pichle se $v_{s} T$ nazdeek paida hota hai: $λ^{'} = λ - v_{s} T = \frac{v}{f} - \frac{v _{s}}{f} = \frac{v - v _{s}}{f}, f^{'} = \frac{v}{λ ^{'}} = f \frac{v}{v - v _{s}}$

Woh ek line hi hai jo neeche ke har question mein probe hoti hai.

Oblique motion — general case (kyun $cos θ$ aata hai)

Master formula assume karta tha ki sab source aur observer ko join karne wali line ke saath seedha chalte hain. Agar ek car tumse offset road pe guzre toh?

Intuition Sirf toward-component matter karta hai

s03 mein source velocity $v_{s}$ road ke saath point karti hai, lekin crests jo tumhare paas pahunche woh sirf is baat ki parwah karte hain ki source tumhare ear ki line of sight kitni tezi se close kar raha hai. $θ_{s}$ use karte hue (source ki velocity aur us line of sight ke beech ka angle, jaise upar define kiya gaya), woh closing rate projection $v_{s} cos θ_{s}$ hai. Similarly, agar observer angle $θ_{o}$ pe line of sight se move kare, toh sirf $v_{o} cos θ_{o}$ hi crest-arrival rate change karta hai. Kuch bhi sideways na crests ko compress karta hai na stretch karta hai jo tumhari taraf aa rahi hain. $f^{'} = f \frac{v + v _{o} c o s θ _{o}}{v - v _{s} c o s θ _{s}}$ Head-on matlab $θ_{s} = θ_{o} = 0$ , $cos 0 = 1$ → full master formula. Straight across (closest approach ka instant) matlab relevant angle $9 0^{\circ}$ hai, $cos 9 0^{\circ} = 0$ → us instant mein koi shift nahi, chahe speed kitni bhi zyada ho. Neeche har "perpendicular = no shift" claim yahi $cos 9 0^{\circ} = 0$ hai.

Formula $v_{s} \geq v$ pe kyun mar jaata hai — Mach cone

Intuition Crests ek wall mein pile ho rahe hain

s04 trace karo. Har dot woh jagah hai jahan source tha jab usne ek still-expanding circular crest emit ki. Jab $v_{s} < v$ (left) circles neat nest karte hain aur aage ek listener sirf high, finite $f^{'}$ pata hai. Jab $v_{s} = v$ (middle) har crest source ki current position ko touch karti hai — sab ek flat front pe pile ho jaate hain, toh $λ^{'} \to 0$ aur formula ka denominator $v - v_{s} \to 0$ , $f^{'} \to \infty$ force karta hai. Woh divergence maths ka cheekh ke kehna hai ki ordinary Doppler ab apply nahi hota. Jab $v_{s} > v$ (right) crests kabhi aage nahi jaa sakti; unka common tangent ek cone hai. $α$ us cone ka half-angle hai — source ke travel direction aur cone ke edge ke beech ka angle, s04 mein marked hai. Kyun $sin α = v / v_{s}$ : time $t$ mein shuru mein emit ki gayi ek crest outward distance $v t$ (uski radius) tak phail gayi hai, jabki source apne path mein $v_{s} t$ advance kar chuka hai. Woh do distances ek right triangle banate hain — crest ki radius $v t$ half-angle $α$ ke opposite side hai, aur source ka travel $v_{s} t$ hypotenuse hai (emission point se source ki current position tak). Toh $sin α = opposite / hypotenuse = v t / (v_{s} t) = v / v_{s}$ . Yeh sonic boom ka Mach cone hai. Is speed ke baad tumhe $f^{'} = f v / (v - v_{s})$ bilkul abandon karna hoga.

True ya false — justify karo

Source ki motion medium mein wavelength change karti hai

True. Source crests emit karne ke beech

v_{s} T

travel karta hai, toh successive crests paas ke (aage) ya zyada door (peeche) paida hote hain:

λ^{'} = (v \mp v_{s}) / f

Observer ki motion medium mein wavelength change karti hai

False. Observer already laid down crests ko touch nahi karta;

λ

fixed rehti hai. Observer sirf woh speed change karta hai jis par crests guzarte hain,

v \pm v_{o}

Ek moving source air mein sound waves ki actual speed change karta hai

False. Medium wave speed

v

set karta hai; source sirf set karta hai ki har crest kahan se emit hoti hai. Speed

v

rehti hai, spacing change hoti hai.

30

m/s pe approach karta source wahi

f^{'}

deta hai jaise observer

30

m/s pe approach kare

False. Alag mechanisms hain — source denominator mein hota hai (wavelength), observer numerator mein (arrival speed). Woh sirf low speeds pe approximately agree karte hain.

Agar dono source aur observer ek doosre ki taraf move karein, pitch shift akele kisi bhi effect se bada hoga

True. Numerator badhta hai (

+ v_{o}

) aur denominator ghatता hai (

- v_{s}

), toh fraction akele kisi bhi change se zyada badhta hai.

Jab source aur observer same velocity pe saath move karein, koi Doppler shift nahi hoga

True. Unke beech ki distance kabhi change nahi hoti, toh crests na pile up hote hain na stretch hote hain;

f^{'} = f

Sound ka Doppler formula source aur observer ke beech symmetric hai, bilkul light ki tarah

False. Sound ka ek medium referee hai, toh source aur observer motions asymmetrically enter karte hain. Light ka koi medium nahi — sirf relative velocity matter karti hai (aur relativity enter hoti hai) — Doppler effect of light symmetric hai.

Agar observer wave speed se tezi se door jaaye, toh woh kuch nahi sunेगा

True. Crests kabhi ek

v_{o} > v

pe bhagte observer ko catch nahi kar sakti; numerator

v - v_{o}

negative ho jaata hai, jo physically matlab hai ki koi wavefronts kabhi reach nahi karte.

Jaise source ki speed wave speed ke paas jaati hai, suni gayi frequency infinity ki taraf jaati hai

True. Denominator

v - v_{s} \to 0

, toh

f^{'} \to \infty

: sab crests ek front pe pile ho jaate hain — shock wave ki shuruat, Sonic boom and shock waves dekho.

Source se observer ki taraf blow karne wali tailwind (

w > 0

) sunी gayi pitch badhati hai

False. Wind dono numerator aur denominator mein

v \to v + w

shift karta hai, toh still source aur observer ke liye factor cancel ho jaata hai aur pitch unchanged rehti hai. Wind mostly sound ko bend aur delay karta hai, uski pitch nahi.

Error pakdo

"Observer ek still source se

v_{o}

pe door bhaag raha hai, toh

f^{'} = f \frac{v}{v - v _{o}}

Galat line. Observer ki speed hamesha numerator mein hoti hai:

f^{'} = f \frac{v - v _{o}}{v}

v_{o}

ko denominator mein daalna source ka slot hai.

"Source

v_{s}

pe recede karta hai, toh

- v_{s}

use karo:

f^{'} = f \frac{v}{v - v _{s}}

Sign ulta hai. Receding lower pitch karna chahiye, toh

+ v_{s}

chahiye:

f^{'} = f \frac{v}{v + v _{s}}

. Hamesha 'approach raises, recession lowers' se sanity-check karo.

"Wind speed

w

observer ki taraf hai, toh

v \to v + w

use karo lekin

v_{o}

ko bhi

v_{o} + w

mein change karo."

Sirf wave speed

v

wind absorb karta hai. Source aur observer speeds

v_{s}, v_{o}

ground ke relative measure kiye jaate hain aur jaise hain waise rehte hain; wind ko double count karna trap hai.

"Sound speed se do guna speed pe move karta source sirf formula se bahut high

f^{'}

deta hai."

v_{s} \geq v

ke liye formula invalid hai. Denominator negative/zero ho jaata hai — physically source apne crests ko peeche chhod deta hai aur ek Mach cone banata hai (s04); koi ordinary Doppler frequency exist nahi karti.

"Source aur observer dono same speed pe right move kar rahe hain, toh top mein

+ v_{o}

aur bottom mein

- v_{s}

plug karo."

Signs approach reflect karne chahiye, sirf motion nahi. Same direction mein equal speed pe move karna matlab hai ki na koi doosre ke paas aa raha hai: yeh ek ke liye receding-from-behind hai aur doosre ke liye approaching-from-front — woh cancel karke

f^{'} = f

dete hain, shift nahi.

"Doppler shift source aur observer ke beech ki distance pe depend karta hai."

Yeh unki line of sight ke saath relative velocity pe depend karta hai, separation pe nahi. Kisi bhi distance pe tumhari taraf

v_{s}

pe move karta source wahi

f^{'}

deta hai.

"Kareeb wali road pe guzarti car jaate waqt steadily shift karti hai."

Nahi — shift

cos θ_{s}

pe depend karta hai (s03). Yeh approach karte waqt high hota hai, closest approach pe smoothly zero se guzarta hai (

θ_{s} = 9 0^{\circ}

), phir low ho jaata hai. Change ek continuous swoop hai, step nahi.

Why questions

v_{o}

numerator mein kyun hai lekin

v_{s}

denominator mein?

Woh alag quantities pe act karte hain: observer crest-arrival speed change karta hai (

f^{'} = speed / λ

ka top), source wavelength

λ

change karta hai (bottom). Alag physics, alag line.

Medium — source nahi — wave speed

v

kyun decide karta hai?

Ek wave medium ke particles mein se guzarta disturbance hai; woh disturbance kitni tezi se travel karta hai medium ki stiffness aur density pe depend karta hai, na ki source kaise move kiya. Wave speed in a medium dekho.

Source ko tumhari taraf move karna wavelength kyun shortens karta hai?

Ek period

T

mein source

v_{s} T

tumhari taraf advance karta hai agla crest emit karne se pehle, toh crests ke beech ka gap exactly

v_{s} T

se shrink hota hai:

λ^{'} = λ - v_{s} T

(s02 dekho).

Ek moving observer air mein already laid down wavelength kyun change nahi kar sakta?

Crests emit hone ke baad space mein fixed hain; observer unhe sirf tezi se ya dheere sample karta hai. Wavelength wave pattern ki property hai, independent hai is baat se ki kaun use dekhta hai. Yeh Wavelength and frequency relation pe tika hai.

Ambulance siren paas se guzarti hai toh pitch kyun girti hai?

Approach karte waqt, toward-component

v_{s} cos θ_{s}

wavelength compress karta hai (higher

f^{'}

); jaate waqt,

cos θ_{s}

smoothly zero se guzarta hai, toh

f^{'}

true

f

se neeche glide karta hai; receding mein, component reverse ho jaata hai aur wavelength stretch hoti hai (lower

f^{'}

). Pitch continuously change hoti hai, high se low sweep karti hai bina kisi instant jump ke — sirf passing point ke paas sabse tezi se change hoti hai.

Off-axis source ke liye sirf

v_{s} cos θ_{s}

kyun matter karta hai?

Sirf line of sight ke saath velocity ka component tumhari taraf aane wali crests compress karta hai; perpendicular part source ko sideways slide karta hai bina tumhari line pe crest spacing change kiye (s03).

Sound aur light Doppler formulas structure mein alag kyun hain?

Sound ko ek medium chahiye absolute referee ke roop mein, toh source aur observer motions asymmetrically enter karte hain. Light ka koi medium nahi — sirf relative velocity matter karti hai (aur relativity enter hoti hai), use symmetric banata hai; Doppler effect of light se contrast karo.

Sound Doppler shift sahi paane ke liye 'relative velocity' kaafi kyun nahi hai?

Kyunki same relative velocity wali do motions (source-moves vs observer-moves) alag

λ

-vs-arrival-speed effects deti hain. Tumhe har ek ki velocity medium ke relative jaanni chahiye, Relative velocity ke according.

Do kareeb Doppler-shifted sirens throbbing sound kyun produce karte hain?

Unki slightly different heard frequencies interfere karti hain, difference frequency pe ek slow amplitude pulsation produce karti hain — woh Beats hai.

Edge cases

Source aur observer dono stationary

v_{s} = v_{o} = 0

, toh

f^{'} = f

. Medium mein koi motion nahi matlab koi compression nahi aur koi arrival-speed change nahi.

Observer wave speed pe exactly source ki taraf move kar raha hai

f^{'} = f (v + v) / v = 2 f

. Crests do guni tezi se aate hain kyunki observer double speed pe unke andar sweep karta hai — perfectly finite, source case se alag.

Observer exactly wave speed pe bhaag raha hai (

v_{o} = v

)

f^{'} = f (v - v) / v = 0

. Crests exactly pace rakhte hain aur kabhi aage nahi jaate; observer koi wavefronts nahi milता aur silence sunta hai.

Source exactly wave speed pe approach kar raha hai (

v_{s} = v

)

Denominator

\to 0

f^{'} \to \infty

. Sab crests ek air wall pe emit ho jaate hain — yeh Sonic boom and shock waves (s04) ka shock front, boundary hai.

Source velocity line of sight ke perpendicular

Us instant mein

cos θ_{s} = 0

, toh momentarily

f^{'} = f

(sound ke liye transverse case), chahe source kitni bhi tezi se move kar raha ho. Sirf along-the-line speed enter karta hai.

Dono equal speed

u

pe ek doosre ki taraf move karein

f^{'} = f \frac{v + u}{v - u} > f

: phir bhi net rise hai, aur larger hai akele kisi bhi ke move karne se, kyunki top aur bottom dono

f^{'}

upar push karte hain.

Observer faster-receding source ka peecha kar raha hai

Agar source aage hai aur

v_{s}

pe recede kar raha hai jabki tum

v_{o} < v_{s}

pe chase kar rahe ho,

f^{'} = f \frac{v + v _{o}}{v + v _{s}} < f

: recession phir bhi jeetta hai, ek small drop, kyunki tum gap puri tarah close nahi kar sakte.

Supersonic source,

v_{s} > v

Mach cone banta hai (s04) half-angle

sin α = v / v_{s}

ke saath; ordinary formula void hai — ek listener kuch nahi sunta jab tak shock front arrive nahi karta, phir ek boom.

Wind observer se source ki taraf blow kar rahi hai (

w < 0

, sound ke liye headwind)

Dono lines mein

v \to v + w = v - ∣ w ∣

replace karo. Still source aur observer ke liye factor phir bhi cancel hota hai — koi pitch change nahi — lekin effective wave speed ab slower hai, air ke referenced to jo sound ke against drift kar rahi hai.