Shock waves — Mach number, Mach cone — - CRITICAL for rockets -: Worked examples

1.6.22 · D3 · Physics › Oscillations & Waves › Shock waves — Mach number, Mach cone — - CRITICAL for rocke

Yeh Mach-cone topic ka drill-yard hai. Parent note ne ek hi formula banaya jo is poore page ko chalata hai:

Neeche sab kuch sirf yahi equation hai, har jagah use karke. Kuch bhi solve karne se pehle, chaliye har tarah ke questions list karte hain jo yeh produce kar sakta hai.

The scenario matrix

Cell	Case class	Kya unknown hai	Trap / feature	Example
A	Forward: $M$ diya, $θ$ nikalo	$θ$	seedha substitution	Ex 1
B	Inverse: $θ$ diya, pehle $M$ phir $v_{obj}$ nikalo	$M$ , $v_{obj}$	sine ko invert karna hoga	Ex 2
C	Degenerate $M = 1$	$θ$	cone flat hokar $θ = 9 0^{\circ}$ ho jaata hai	Ex 3
D	Forbidden $M < 1$	(koi nahi)	$sin θ > 1$ → koi cone nahi	Ex 3
E	Limiting $M \to \infty$ (hypersonic)	$θ$ trend	cone $\to$ kagaz jaisa patla	Ex 4
F	Real-world word problem: boom delay	time $t$	geometry $x = h / tan θ$	Ex 5
G	Changing medium ( $v_{s}$ altitude ke saath badalta hai)	naya $M$ , naya $θ$	same $v_{obj}$ , alag $v_{s}$	Ex 6
H	Exam twist: two observers / find the height	height $h$	geometry ulti karo	Ex 7
I	Sanity/limit check: kya "boom at Mach 0.9" possible hai?	haan/nahi	conceptual + numeric	Ex 8

Yeh nau cells cover karti hain: forward map, uska inverse, dono boundary values ( $M = 1$ aur $M < 1$ ), runaway limit, ek full word problem, variable-medium case, ek inverted-geometry exam question, aur ek pure conceptual trap. Is topic mein jo bhi pucha ja sakta hai, woh is table ke bahar nahi hai.

Do geometries jo baar baar use hongi

Examples se pehle, do pictures samajh lo. Har example inhi mein se kisi ek ki taraf point karega.

Picture 1 — the cone triangle (jahan $sin θ = 1/ M$ janam leta hai):

Figure — Shock waves — Mach number, Mach cone — - CRITICAL for rockets -

Object apex $B$ par baitha hai. Ek ripple jo usne $t$ time pehle emit ki thi woh $v_{s} t$ radius ki sphere hai (yeh right triangle ki opposite side hai). Usi time mein object $v_{obj} t$ chala (hypotenuse $A B$ ). Flight path aur cone wall ke beech ka angle $B$ par $θ$ hai, aur "opposite over hypotenuse" bilkul $sin θ = v_{s} t / (v_{obj} t) = 1/ M$ hai.

Picture 2 — the ground geometry (jahan $x = h / tan θ$ janam leta hai):

Jet height $h$ par level fly kar raha hai. Uski cone wall flight line se $θ$ angle par peeche jhukti hai. $h$ length ka vertical daalo; wall ground se horizontal distance $x$ peeche milti hai jet se. Us patli triangle mein, $tan θ = h / x$ (opposite $h$ over adjacent $x$ ), isliye $x = h / tan θ$ . Yeh trailing distance $x$ hi wajah hai ki aap boom jet ke overhead jaane ke baad sunte ho.

Worked examples

Worked example Ex 2 — Cell B: inverse,

$θ$ diya speed nikalo Ek shell ke shock cone ka half-angle $θ = 2 5^{\circ}$ hai, air mein $v_{s} = 340 m/s$ ke saath. Shell kitni fast ja rahi hai?

Forecast: $2 5^{\circ}$ , $M = 2$ ke $3 0^{\circ}$ se chhota hai, isliye shell Mach 2 se faster hai — expect karo $M$ thoda $2$ se upar aur $v_{obj}$ $680 m/s$ se kaafi zyada.

$sin θ = 1/ M$ se, invert karo: $M = \frac{1}{sin θ} = \frac{1}{sin 2 5 ^{\circ}} = \frac{1}{0.4226} = 2.366$ . Yeh step kyun? Hume angle pata hai, speed ratio nahi, isliye pehle $M$ ke liye formula solve karte hain.
Definition $M = v_{obj} / v_{s}$ se speed recover karo: $v_{obj} = M v_{s} = 2.366 \times 340 = 804.5 m/s$ . Yeh step kyun? $M$ akela ek pure number hai; $v_{s}$ se multiply karne par physical speed m/s mein milti hai.

Verify: $sin 2 5^{\circ} \approx 0.4226$ , aur $1/2.366 = 0.4226$ ✓. $M = 2.366 > 2$ aur $804 m/s > 680 m/s$ , forecast se match karta hai. Units: $(dimensionless) \times (m/s) = m/s$ ✓.

Worked example Ex 3 — Cells C & D: boundary

$M = 1$ aur forbidden $M < 1$ (a) $M = 1$ par exactly $θ$ kya hai? (b) Formula $M = 0.8$ par kya kehta hai, aur physically uska matlab kya hai?

Forecast: $M = 1$ par object apne ripples ke saath pace karta hai — guess hai cone "poori tarah khul" jaayegi $9 0^{\circ}$ tak. Usse neeche, guess hai "koi cone nahi."

(a) $sin θ = 1/1 = 1 \Rightarrow θ = sin^{- 1} (1) = 9 0^{\circ}$ . Yeh step kyun? $sin θ = 1$ sine ki sabse badi value hai, isliye $θ$ apne maximum par hai — cone wall flight path ke perpendicular khadi ho jaati hai. "Cone" ek flat plane ban jaata hai: sound barrier.
(b) $sin θ = 1/0.8 = 1.25$ . Yeh step kyun? Sub-sonic value seedhe plug in karo aur result dekho.
Sine kabhi $1$ se zyada nahi ho sakti, isliye $sin θ = 1.25$ ka koi real solution nahi — koi angle $θ$ nahi milta. Yeh step kyun? Mathematics khud cone banane se mana kar deta hai; physically source abhi bhi apne expanding ripples ke andar hai, isliye kuch pile up nahi hota. $M < 1$ par koi shock cone nahi banta.

Verify: $sin 9 0^{\circ} = 1 = 1/1$ ✓. (b) ke liye: $1/0.8 = 1.25 > 1$ , sine ki range ke bahar ✓ — "no real cone" ka verdict confirm hota hai.

Worked example Ex 4 — Cell E: hypersonic limit

$M \to \infty$ Ek re-entry capsule $M = 25$ tak pahunchta hai. $θ$ nikalo, aur batao $M$ badhte badhte kya hota hai.

Forecast: Bahut zyada speed ⇒ cone almost needle jaisi honi chahiye, $θ$ sirf kuch degrees, $0^{\circ}$ ki taraf jaati hui.

$sin θ = 1/25 = 0.04 \Rightarrow θ = sin^{- 1} (0.04) = 2.2 9^{\circ}$ . Yeh step kyun? Direct forward substitution, Ex 1 jaisa, lekin extreme $M$ par limiting shape dekhne ke liye.
Jab $M \to \infty$ , $1/ M \to 0$ , isliye $sin θ \to 0$ aur $θ \to 0^{\circ}$ . Yeh step kyun? Hum ek single value nahi, behaviour padh rahe hain: cone flight path ke saath chipak kar kagaz jaisi patli sliver ban jaati hai — hypersonic flow ki physical picture.

Verify: $sin (2.2 9^{\circ}) = 0.04 = 1/25$ ✓. Trend check: $M = 25$ par already $2.2 9^{\circ}$ aa gaya, $M = 2$ ke $3 0^{\circ}$ se bahut kam, yeh $0^{\circ}$ ki taraf monotone squeeze confirm karta hai.

Worked example Ex 5 — Cell F: real-world boom delay (Picture 2 use karta hai)

Ek jet $M = 1.8$ par $h = 3 km$ height par level fly kar raha hai, air mein $v_{s} = 330 m/s$ ke saath. Jet seedha overhead se guzarne ke kitne seconds baad boom sunai deta hai?

Forecast: Parent ke Mach-1.5 example se faster aur lower-cone, isliye delay kuch seconds honi chahiye — guess around 4–6 s.

Mach angle: $sin θ = 1/1.8 = 0.5556 \Rightarrow θ = 33.7 5^{\circ}$ . Yeh step kyun? Cone ka lean-back angle sirf $M$ se fix hota hai; ground geometry touch karne se pehle yeh chahiye.
Ground par trailing distance (Picture 2): $x = \frac{h}{tan θ} = \frac{3000}{tan 33.7 5 ^{\circ}} = \frac{3000}{0.6682} = 4490 m$ . Yeh step kyun? Cone wall jet se itni door peeche ground se milti hai; $tan θ = h / x$ kyunki $h$ aur $x$ us ground triangle ke do legs hain.
Jet ground speed: $v_{obj} = M v_{s} = 1.8 \times 330 = 594 m/s$ . Yeh step kyun? Boom tab sunai deta hai jab jet itna aage nikal jaata hai ki uski trailing cone aapki jagah tak pahunche — isliye jet (aur uske dragged cone) ki speed chahiye.
Delay: $t = \frac{x}{v _{obj}} = \frac{4490}{594} = 7.56 s$ . Yeh step kyun? Cone wall ka ground intercept jet se exactly $x$ peeche hota hai; jet aapki position ke upar se uss gap ko time $x / v_{obj}$ mein close karta hai.

Verify: $tan 33.7 5^{\circ} \approx 0.668$ ; $3000/0.668 \approx 4490 m$ ✓. $594 m/s$ aur $4490/594 = 7.56 s$ ✓. Units: $m / (m/s) = s$ ✓. (Mere forecast se thoda zyada — lower cone angle ne lag ko stretch kiya; Doppler-bunching ka "wall trailing far behind" intuition sahi hai.)

Worked example Ex 6 — Cell G: same speed, alag medium

Ek missile fixed $v_{obj} = 700 m/s$ par fly kar raha hai. Uski Mach cone compare karo (i) ground ke paas jahan $v_{s} = 340 m/s$ aur (ii) upar jahan thand wali patli hawa mein $v_{s} = 295 m/s$ hai.

Forecast: Same speed, lekin sound upar slower hai, isliye missile apne ripples ko zyada beat kar raha hai ⇒ higher $M$ ⇒ upar narrower cone. Guess karo altitude par $θ$ chhota hoga.

Ground: $M_{1} = 700/340 = 2.059$ , isliye $sin θ_{1} = 1/2.059 = 0.4857 \Rightarrow θ_{1} = 29.0 6^{\circ}$ . Yeh step kyun? $M$ speed hai local sound speed ke relative; har layer ke liye recompute karte hain kyunki $v_{s}$ badla hai, $v_{obj}$ nahi.
Altitude: $M_{2} = 700/295 = 2.373$ , isliye $sin θ_{2} = 1/2.373 = 0.4214 \Rightarrow θ_{2} = 24.9 3^{\circ}$ . Yeh step kyun? Same missile, thandi hawa ⇒ chhota $v_{s}$ ⇒ bada $M$ ⇒ chhota angle, bilkul forecast jaisa.

Verify: $700/340 = 2.059$ aur $sin 29.0 6^{\circ} \approx 0.4857 = 1/2.059$ ✓; $700/295 = 2.373$ aur $sin 24.9 3^{\circ} \approx 0.4214 = 1/2.373$ ✓. Aur $θ_{2} < θ_{1}$ : altitude par cone thinner ✓. Isliye kisi bhi Mach figure ke saath local sound speed bhi batana zaroori hai.

Worked example Ex 7 — Cell H: exam twist, height back out karo

Ground crew ek supersonic drone ka boom exactly $5.0 s$ baad sunti hai jab woh overhead se guzarta hai. Drone $M = 2.0$ par fly kar raha hai, $v_{s} = 340 m/s$ ke saath. Yeh kitni height par fly kar raha hai?

Forecast: Yeh Ex 5 ulta hai — hume delay pata hai, height chahiye. $M = 2$ par cone $3 0^{\circ}$ hai; ~680 m/s par 5-second lag ka matlab kuch km trailing distance hai, isliye expect karo $h$ around 2 km.

Ground speed: $v_{obj} = M v_{s} = 2.0 \times 340 = 680 m/s$ . Yeh step kyun? Delay trailing distance mein convert hoti hai sirf jet ki ground speed se.
Delay se trailing distance: $x = v_{obj} t = 680 \times 5.0 = 3400 m$ . Yeh step kyun? Cone wall drone ke $x$ peeche hoti hai; drone aapke upar se uss gap ko measured $5.0 s$ mein cross karta hai, isliye $x = v_{obj} t$ (Ex 5 step 4 ko rearrange kiya).
Mach angle: $sin θ = 1/2 \Rightarrow θ = 3 0^{\circ}$ . Yeh step kyun? Horizontal $x$ ko vertical $h$ se link karne ke liye cone ka lean chahiye.
Picture 2 se height: $h = x tan θ = 3400 \times tan 3 0^{\circ} = 3400 \times 0.5774 = 1963 m$ . Yeh step kyun? $x = h / tan θ$ ko $h = x tan θ$ mein invert karte hain kyunki ab $h$ unknown leg hai.

Verify: $h / tan θ = 1963/0.5774 = 3400 m = x$ ✓, aur $x / v_{obj} = 3400/680 = 5.0 s$ ✓ — diye gaye delay par round-trip karta hai. $h \approx 1.96 km$ , ~2 km forecast se match karta hai. Units: $m \times (dimensionless) = m$ ✓.

Worked example Ex 8 — Cell I: conceptual + numeric sanity trap

"Ek dost claim karta hai unhone $M = 0.9$ par ek jet ke liye $θ = 4 8^{\circ}$ ka half-angle Mach cone measure kiya." Kya yeh possible hai? Agar koi real cone kisi $M$ par exist karta hai jo $θ = 4 8^{\circ}$ deta hai, toh woh $M$ kya hai?

Forecast: $M = 0.9$ subsonic hai, isliye Ex 3(b) se koi cone hona hi nahi chahiye — claim galat lagta hai. Phir ek genuine $4 8^{\circ}$ cone ko koi supersonic $M$ chahiye; guess karo $1$ se thoda upar.

Claim test karo: $M = 0.9$ par, $sin θ = 1/0.9 = 1.111 > 1$ — impossible. Koi cone nahi banta; dost ne measure nahi kiya hoga. Yeh step kyun? Hum check karte hain ki required $sin θ$ $[- 1, 1]$ mein hai ya nahi; subsonic $M$ isse $1$ se upar push kar deta hai, koi real angle possible nahi.
Ab honest question — $θ = 4 8^{\circ}$ kaun sa $M$ deta hai? $sin 4 8^{\circ} = 0.7431$ , isliye $M = 1/0.7431 = 1.346$ . Yeh step kyun? Real $4 8^{\circ}$ cone ke liye $sin θ = 1/ M$ mein $sin θ < 1$ chahiye, jo force karta hai $M > 1$ ; nikalne ke liye invert karte hain.

Verify: $1/0.9 = 1.111 > 1$ "no subsonic cone" confirm karta hai ✓. $sin 4 8^{\circ} = 0.7431$ aur $1/1.346 = 0.7431$ ✓, $M = 1.346 > 1$ ke saath (supersonic, jaisa chahiye) ✓. Yeh parent ka coherent pile-up rule nail karta hai: shock wall tabhi exist karti hai jab source apne ripples se aage nikal jaaye.

Recall Kaun sa cell kaun sa tha?

Forward substitution ::: Ex 1 Speed ke liye sine invert karna ::: Ex 2 $M = 1$ flat cone aur $M < 1$ no-cone ::: Ex 3 Hypersonic $θ \to 0$ limit ::: Ex 4 Boom-delay word problem ::: Ex 5 Same speed, changed sound speed ::: Ex 6 Back-out-the-height exam twist ::: Ex 7 Impossible-claim spot karna trap ::: Ex 8

Connections

Speed of sound in a medium — har $M$ ke peeche ek local $v_{s}$ chupi hoti hai (Ex 6 mein clearly dikhaya).
Doppler effect — Ex 5/7 mein trailing-cone lag, wavefront bunching ka $M \geq 1$ wala roop hai.
Superposition & constructive interference — cone banta kyun hai (Ex 8).
Wave drag and aerodynamic heating — Ex 4 mein hypersonic sliver ka consequence.
De Laval nozzle — design se supersonic Mach numbers.
Compressible flow / Bernoulli limits — Ex 3 ka $M \approx 1$ boundary.