Visual walkthrough — Shock waves — Mach number, Mach cone — - CRITICAL for rockets -

1.6.22 · D2 · Physics › Oscillations & Waves › Shock waves — Mach number, Mach cone — - CRITICAL for rocke

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Step 1 — Ek source, ek ripple

KYA. Still air mein ek single point rakh do. Ek instant mein woh ek choti si push deta hai — ek pressure pulse. Woh pulse bahar ki taraf ek badhte hue sphere ki tarah phailti hai (agar hum flat cross-section draw karein toh circle).

YE STEP KYUN. Shocks ke baare mein sab kuch sirf bahut saari aise ripples ka sum hai. Agar ek ripple ki size samajh lo, toh poora cone samajh lo. Ripple hamare liye atom hai.

PICTURE. Dot hai source. Blue circle hai wavefront — un saare points ka set jahan tak pulse ab tak pahunchi hai. Uska radius badhta hai kyunki sound fixed speed se chalti hai.

Figure ko term by term padhte hain:

$\underset{\text{ripple kitni door pahunchi}}{\underbrace{r}}=\underset{\text{ripple edge speed}}{\underbrace{v_s}}\times \underset{\text{push ke baad ka time}}{\underbrace{t}}$

Yahan har symbol picture mein hai: $r$ hai blue radius, $v_s$ hai bahar ki taraf jaata arrow, $t$ hai "kitni der pehle." Jitna zyada wait karo, ripple utni badi — yahi $r=v_{s}t$ ka poora content hai.

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Step 2 — Source move karta hai aur emit karta rehta hai

KYA. Ab source ko speed $v_{\text{obj}}$ se daayein slide karne do, har instant mein ek fresh ripple drop karte hue, jaise breadcrumbs. Har purani ripple badi hai (use grow karne ka zyada time mila); har nayi chhoti hai.

KYUN. Ek chalti hui seeti sirf ek baar nahi bajti — woh lagaataar bajti hai. Un overlapping ripples ka jo pattern banta hai wahi physics hai. Hume dekhna hai ki woh kahan bheedti hain.

PICTURE. Char ripples, char past positions se emit ki gayi hain. Notice karo ki sabse purani (sabse baayi) sabse badi circle hai, nayi sirf ek dot hai. Source aage-aage chalta rehta hai.

Jab se ripple drop ki, source ne kitni door travel ki:

$\underset{\text{source ka apne track par head-start}}{\underbrace{d}}=\underset{\text{source speed}}{\underbrace{v_{\text{obj}}}}\times \underset{\text{us ripple ke baad ka time}}{\underbrace{t}}$

Toh kisi bhi moment par hum do lengths compare kar rahe hain jo dono $t$ ke saath badhti hain: ripple ka radius $r=v_{s}t$ aur source ka travel $d=v_{\text{obj}}t$. In dono ke beech ki race hi poori kahaani hai.

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Step 3 — Subsonic case: source apni ripples ke andar rehta hai

KYA. Maano source sound se slower hai, $v_{\text{obj}}<v_s$ (toh $M<1$). Toh same time $t$ mein, ripple ($v_{s}t$) source ($v_{\text{obj}}t$) se aage nikal jaati hai. Source hamesha apni banaayi har ripple ke andar rehta hai.

PEHLE YEH KYUN DIKHAYEIN. Shock ki appreciation ke liye pehle usse pehle ki shaanti dekhni zaruri hai. Subsonically ripples aage bunch up ho jaati hain (woh forward crowding Doppler effect hai) lekin kabhi ek sharp wall nahi banati — hamesha ek gap rehta hai, koi common edge nahi.

PICTURE. Moving dot apne saare circles ke andar baitha hai. Ripples aage denser hain, peechhe sparse, lekin kaheen bhi ek shared line ko touch nahi karti.

$\text{if }{v}_{\text{obj}}<v_s⟹\underset{\text{source travel}}{\underbrace{v_{\text{obj}}t}}<\underset{\text{ripple radius}}{\underbrace{v_{s}t}}⟹\text{source andar phansa}$

Yahan koi pile-up surface nahi hai. Is contrast ko pakad ke rakho — agla step ise palat deta hai.

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Step 4 — Supersonic case: source apni ripples se nikal jaata hai

KYA. Ab source ko sound se faster banao, $v_{\text{obj}}>v_s$ (toh $M>1$). Same $t$ mein, source zyada door jaata hai ($v_{\text{obj}}t$) compared to ripple ke radius ($v_{s}t$) se. Source apne khud ke wavefront ke bahar nikal jaata hai.

YEH TURNING POINT KYUN HAI. Jab source apni ripples se aage nikal jaata hai, saari ripples ek saath peechhe chhoot jaati hain — aur unke outer edges ek single straight tangent share kar sakte hain. Woh shared tangent woh jagah hai jahan har ripple ki crest ek saath land karti hai.

PICTURE. Dot har circle ke sabse baaye hisse se aage nikal gaya hai. Ek green straight line saari circles ko ek saath graze karti hai (tangent hai). Woh green line hai shock front.

$\text{if }{v}_{\text{obj}}>v_s⟹\underset{\text{source travel}}{\underbrace{v_{\text{obj}}t}}>\underset{\text{ripple radius}}{\underbrace{v_{s}t}}⟹\text{source nikal gaya}$

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Step 5 — Right triangle nikalo

KYA. Picture ko ek moment par freeze karo. Teen points mark karo:

• $A$ = jahan time $0$ par ek ripple emit ki gayi thi.
• $B$ = jahan source ab hai, time $t$ par (woh apne track par aage nikal gaya hai).
• $P$ = tangent point: woh single spot jahan shock line us ripple ko just touch (graze) karti hai. Ripple ka radius uske centre $A$ se seedha $P$ tak khiincho — kyunki radius hamesha tangent se right angle par milta hai, yeh radius $AP$ shock line se $90^{\circ }$ par milta hai.

Yeh teen points $A$, $B$, $P$ ek right triangle banate hain, right angle $P$ par.

TRIANGLE KYUN, AUR YEH WALA KYUN. Hume cone ke angle ka ek number chahiye. Angle ko trap karne ka sabse clean tarika ek right triangle mein hai, kyunki tab angle do sides ke ratio mein lock ho jaata hai — koi calculus nahi, sirf shape. Yahi woh side-ratio question ka type hai jiske liye sine function banaya gaya tha.

PICTURE. Triangle $ABP$: $B$ par apex angle $\theta$, $P$ par right angle, do known sides label kiye hue.

Is triangle ke teen parts naam dete hain:

• Hypotenuse $AB=v_{\text{obj}}t$ — lamba slanted side, source kitna gaya.
• Opposite side $AP=v_{s}t$ — ripple radius (centre $A$ se tangent point $P$ tak), angle $\theta$ ke samne khada hua.
• Angle $\theta$ — Mach angle, source apex $B$ par baitha; yeh cone ki half-opening hai.

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Step 6 — Angle padho; dekho time cancel hota hai

KYA. Do side lengths ko sine ratio mein daalo.

KYUN. Yeh picture ko ek aisi formula mein convert karta hai jisse hum compute kar saken.

$\sin \theta =\frac{\text{opposite}}{\text{hypotenuse}}=\frac{\stackrel{\text{ripple radius}}{\overbrace{v_{s}t}}}{\underset{\text{source travel}}{\underbrace{v_{\text{obj}}t}}}=\frac{v_s}{v_{\text{obj}}}$

Upar aur neeche $t$ cancel ho jaata hai — yeh ek quiet miracle hai. Dono sides time ke saath same rate se badhti hain, isliye unka ratio kabhi nahi badlta.

PICTURE. Do alag moments ke do triangles (chhhota aur bada) ek doosre par rakhe hue: alag sizes, identical angle $\theta$. Cone ki opening hamesha ke liye fixed hai.

Ab Step 2 mein define kiya hua Mach number use karo, $M=v_{\text{obj}}/v_s$. Fraction $v_s/v_{\text{obj}}$ exactly $M$ ulta hai, yaani $1/M$:

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Step 7 — Har case, broken wale bhi

KYA. Formula $\sin \theta =1/M$ ko boundaries par test karo, kyunki jis formula par trust karo usse extremes par survive karna chahiye.

KYUN. Koi bhi physicist result accept nahi karta bina use $M=1$, $M\to \infty$, aur $M<1$ par push kiye. Formula ya toh kuch sensible dega ya batayega ki physics impossible hai.

PICTURE. Teen cones side by side: low supersonic $M$ par wide, high $M$ par thin, aur $M=1$ par flat plane.

• $M=1$ exactly (source sound speed par): $\sin \theta =1/1=1$, toh $\theta =90^{\circ }$. Cone poori tarah ek flat wall mein khul jaata hai motion ke perpendicular — "sound barrier." Ripples saari ki saari nose par ek plane pe pile ho jaati hain.
• $M\to \infty$ (hypersonic): $\sin \theta =1/M\to 0$, toh $\theta \to 0^{\circ }$ — ek infinitely thin, sharply swept cone flight path se chipka hua. Faster ⇒ thinner. (Yeh Wave drag and aerodynamic heating ka regime hai re-entry par.)
• $M<1$ (subsonic): $\sin \theta =1/M>1$. Lekin sine kabhi $1$ se zyada nahi ho sakta — koi real angle iska solution nahi hai. Math khud refuse karta hai, jo Step 3 se match karta hai: source apni ripples ke andar rehta hai, isliye koi cone exist nahi karta. Equation toota nahi; woh sach bol raha hai.

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Ek picture mein poora summary

Sab kuch ek canvas par: kai growing ripples, source aage, tangent cone, right triangle $v_{s}t$ aur $v_{\text{obj}}t$ ke saath, aur boxed result. Agar tum yeh memory se redraw kar sako, derivation tumhari hai.

Recall Poore walkthrough ki Feynman retelling

Ek pathar girao: ek ripple ring bahar ki taraf sound speed par baadhti hai (Step 1). Ab paani tap karte hue chalo — tum rings ki ek trail chhodte ho, badi purani peechhe, chhoti nayi paon ke neeche (Step 2), aur hum $M$ se score rakhte hain, yaani tumhari speed ripples se kitni compare karti hai. Dheere chalo ($M<1$) aur rings tumhe gheer leti hain; tum hamesha unke andar ho (Step 3). Ripples se faster sprint karo ($M>1$) aur tum aage nikal jaate ho — ab saare ring edges ek straight slanted wall ke saath line up ho jaate hain jo tumhare peechhe drag hoti hai V mein (Step 4). Us V ke angle ko ek right triangle mein pin karo: ek side hai kitni door ripple grow hui ($v_{s}t$), slanted side hai kitni door tum daude ($v_{\text{obj}}t$), aur woh tangent point $P$ par milti hain (Step 5). Unka ratio V ke half-angle ka sine hai, aur times cancel ho jaate hain isliye V kabhi shape nahi badlti: $\sin \theta =v_s/v_{\text{obj}}=1/M$ (Step 6). Push karo: exactly sound speed par V ek wall mein flat ho jaati hai; isse kaafi upar woh razor-thin ho jaati hai; sound se slower formula ek impossible sine maangta hai, jo nature ka tarika hai yeh kehne ka ki "yahaan neeche koi shock nahi" (Step 7). Squished air ka woh patla V sonic boom hai jo tum sunte ho jab woh tumhare paas se guzarta hai.

Connections

• Speed of sound in a medium — $v_s$ provide karta hai, Step 1 mein ripple-edge speed.
• Doppler effect — subsonic Step 3 mein ripples ka forward crowding.
• Superposition & constructive interference — Step 4 mein tangent line shock kyun banti hai.
• Wave drag and aerodynamic heating — Step 7 ka thin hypersonic cone.
• De Laval nozzle — jahan supersonic $M$ deliberately engineer kiya jaata hai.
• Compressible flow / Bernoulli limits — kyun $M\approx 1$ simple flow assumptions ko todta hai.