Environmental testing — thermal vacuum (TVAC), vibration, acoustic, EMC - EMI: Visual walkthrough

3.6.33 · D2 · Physics › Spacecraft Structures & Systems Engineering › Environmental testing — thermal vacuum (TVAC), vibration, ac

Yeh parent topic ke vibration section ka "movie" version hai. Hum yahan slower chalte hain aur sab kuch draw karte hain.

Step 1 — Satellite ka ek part basically ek springy stalk par rakha weight hai

KYA HAI. Koi bhi box dekho jo spacecraft panel par bolt ki hui ho: ek circuit board standoffs par, ek camera bracket par. Zoom out karo toh woh ek lump of stuff (box) hai jo kisi bendy cheez (bracket, bolts, panel) se tiki hui hai. Hum ise simplest possible cartoon se replace karte hain: ek block jo slide kar sakta hai, ek spring se wall se juda hua.

KYUN. Poora satellite haath se solve karna impossible hai — laakhon chhote pieces hain. Lekin ek single block-on-spring mein resonance ka poora raaz chhupa hai. Agar hum is cartoon ko poori tarah samajh lein, real cheez bas "bahut saare cartoons ek saath" hai. Yeh single-degree-of-freedom (SDOF) model hai — "one degree of freedom" matlab block sirf ek direction mein (left–right) move kar sakta hai, toh ek number $x$ poori tarah batata hai woh kahan hai.

PICTURE. Neeche: mass $m$ ka block, stiffness $k$ ki spring, aur ek chhota dashpot (shock-absorber) strength $c$ ka. Neeche ka floor woh hai jise shaker table pakad ke hilata hai.

Figure — Environmental testing — thermal vacuum (TVAC), vibration, acoustic, EMC - EMI

Step 2 — Block ke liye Newton ka law likho

KYA HAI. Ab hum table ko block ko wiggling force se push karne dete hain aur poochte hain: block kaise move karta hai? Newton kehta hai (mass)(acceleration) = (sum of forces). Block par teen forces act karti hain: spring, damper, aur driving push.

KYUN. Resonance dhundhne se pehle hume ek equation chahiye. Har force ko "block ki motion par kaise depend karta hai" mein translate kiya jaata hai, aur ek sentence algebra mein badal jaata hai jise hum actually solve kar saken.

PICTURE. Figure mein har arrow ek force hai; dekho woh kaunsi taraf point karta hai jab block right move karta hai.

Maano $x$ woh distance hai jitna block rest se slide hua (metres). Tab:

$\overset{x}{˙}$ (padho "x-dot") block ki velocity hai — $x$ kitni fast change ho rahi hai har second. Dot matlab "rate of change".
$\overset{x}{¨}$ ("x-double-dot") acceleration hai — velocity kitni fast change ho rahi hai.

Ab har force ko translate karo:

$mass \times accel m \overset{x}{¨} = spring pulls back - k x damper drags - c \overset{x}{˙} + table’s push F_{0} sin (ω t)$

Term by term padhte hain:

$- k x$ : spring force. Minus sign isliye kyunki spring hamesha block ko ghar ki taraf wapas kheenchti hai. Right slide kiya → left pull karta hai.
$- c \overset{x}{˙}$ : damper force. Minus sign isliye kyunki yeh hamesha motion ki direction ko oppose karta hai.
$F_{0} sin (ω t)$ : shaker ki push. $F_{0}$ kitna hard (peak newtons), $ω$ kitna fast wiggle karta hai (angular frequency, radians per second), $t$ time hai. $sin$ ise smoothly aage-peeche swing karata hai.

Standard form mein likhne ke liye sab kuch left side le jao:

$m \overset{x}{¨} + c \overset{x}{˙} + k x = F_{0} sin (ω t)$

Step 3 — Table band karo: part ki APNI favourite rhythm hoti hai

KYA HAI. Push hatao ( $F_{0} = 0$ ) aur damping ko bhi ignore karo ek second ke liye ( $c = 0$ ). Block ko ek baar pluck karo aur chodh do. Woh ek special frequency par hamesha ke liye oscillate karta rahega.

KYUN. Hume ek reference frequency chahiye taaki table ki shaking se compare kar sakein. Yeh natural rhythm poori kahaani ki star hai — resonance tab hoti hai jab table ki rhythm is se match karti hai. Toh pehle hume yeh dhundhna hoga.

PICTURE. Stiff spring fast snap back karti hai (high pitch); heavy block sluggish hota hai (low pitch). Figure dono extremes dikhata hai.

$c = 0, F_{0} = 0$ ke saath equation hai $m \overset{x}{¨} + k x = 0$ . Ise solve karne wali motion ek pure sine wave hai, aur ise plug in karne par frequency yeh honi chahiye:

$ω_{n} = \frac{k}{m} (radians per second)$

$ω_{n}$ : natural angular frequency. Subscript $n$ = "natural".
$k$ upar: stiffer spring → zyada hard snap back → faster wiggle. Sense banta hai.
$m$ neeche: heavier block → zyada sluggish → slower wiggle. Yeh bhi sense banta hai.

Engineers cycles-per-second (hertz) prefer karte hain, toh ek full circle ke radians, $2 π$ , se divide karo:

$f_{n} = \frac{ω _{n}}{2 π} = \frac{1}{2 π} \frac{k}{m} (Hz)$

Step 4 — Pehle friction number $ζ$ ka naam rakho, phir frequency $ω$ par push karo

KYA HAI. Koi bhi amplification formula aane se pehle, damping $c$ ko ek clean, unitless number $ζ$ mein bottle karte hain. Phir table ko wapas on karte hain, frequency $ω$ par push karte hue, aur jo steady wiggle aata hai use dekhte hain.

KYUN. Step 5 mein friction $c$ ek square root ke andar $2 ζ r$ ke roop mein aayega. Agar pehle $ζ$ define nahi kiya, toh woh symbol reader ko achanak surprise kar dega. Toh hum ise yahan define karte hain, apne aap, use karne se pehle.

PICTURE. Input arrow (table) chhota aur steady; output arrow (block) bada hota ja raha hai jaise $ω$ $ω_{n}$ ki taraf creep karta hai.

Friction number. Raw damping $c$ ko "critical" amount $2 k m$ se compare karo (woh exact amount jo saari oscillation khatam kar deta). Unka ratio hai damping ratio:

$ζ = \frac{c}{2 k m}$

$ζ$ (Greek "zeta") batata hai part mein kitna friction hai, ek clean unitless fraction ke roop mein.
$ζ = 0$ = bilkul bhi friction nahi; $ζ = 1$ = itna friction ki block bina overshoot kiye slowly ghar aa jaaye.
Spacecraft structure lightly damped hota hai: $ζ \approx 0.02$ – $0.05$ .

Do dials. Ab doosra clean dial define karo, frequency ratio:

$r = \frac{ω}{ω _{n}}$

$r$ batata hai "push frequency part ki favourite frequency se kaise compare hoti hai?"
$r < 1$ : natural se slower push. $r = 1$ : bilkul natural par push. $r > 1$ : faster push.

Hum kya dhundh rahe hain. Thodi der baad block same frequency $ω$ par ek steady sine wiggle mein settle ho jaata hai, lekin amplitude $X$ ke saath. Us output wiggle ka input wiggle se ratio transmissibility $T$ hai. Agar $T = 1$ , part table jitna hi hard hilata hai. Agar $T = 20$ , ek gentle $1$ g input part par brutal $20$ g ban jaata hai. $ζ$ aur $r$ dono ab table par hain — hum $T$ banane ke liye ready hain.

Step 5 — Rotating-arrow (phasor) picture se transmissibility banao

KYA HAI. Hum $T (r)$ ko heavy algebra ke bina derive karte hain har force ko ek rotating arrow ("phasor") ki tarah draw karke aur arrows ko tip-to-tail add karke.

KYUN. Formula ke teen terms — $k x$ , $c \overset{x}{˙}$ , $m \overset{x}{¨}$ — plain numbers ki tarah add nahi hote; woh alag alag directions mein point karte hain kyunki velocity aur acceleration position se "quarter-turn aage" hote hain. Teen arrows ki picture square root ko magical ki jagah obvious bana deti hai.

PICTURE. Neeche: driving push (yellow) teen response arrows ke tip-to-tail sum ke barabar hona chahiye. Right triangle padho jo woh banata hai.

Poora idea teen sentences mein:

Jab block $x = X sin (ω t)$ ki tarah wiggle karta hai, uski spring force $k x$ motion ke saath point karti hai, size $k X$ .
Uski velocity $\overset{x}{˙}$ quarter-turn ( $9 0^{\circ}$ ) aage hai, toh damper force $c \overset{x}{˙}$ sideways point karti hai, size $c ω X$ .
Uski acceleration $\overset{x}{¨}$ half-turn ( $18 0^{\circ}$ ) round hai, toh inertia force $m \overset{x}{¨}$ motion ke ulti direction mein point karti hai, size $m ω^{2} X$ .

Table ki push $F_{0}$ ko inke sum ko balance karna hoga. Motion ki line ke saath dono forces combine hoke $(k - m ω^{2}) X$ banate hain; sideways wala $c ω X$ hai. Woh right angles par hain, toh Pythagoras se push magnitude hypotenuse hai:

$F_{0} = X (k - m ω^{2})^{2} + (c ω)^{2}$

$(k - m ω^{2})$ : spring pull minus inertia — triangle ki "along-motion" leg. Yeh bilkul zero ho jaata hai jab $ω^{2} = k / m$ , yaani resonance par.
$(c ω)$ : friction leg, hamesha sideways, kabhi zero nahi jab tak block move kare.

Output amplitude $X$ ke liye solve karo, aur ise us amplitude se compare karo jo table ek rigid part ko deta, $X_{0} = F_{0} / k$ (push divided by stiffness). Unka ratio $T$ hai:

$T = \frac{X}{X _{0}} = \frac{k}{( k - m ω ^{2} ) ^{2} + ( c ω ) ^{2}}$

Ab top aur bottom ko $k$ se divide karo aur apne do dials use karo. Kyunki $m ω^{2} / k = (ω / ω_{n})^{2} = r^{2}$ aur $c ω / k = 2 ζ r$ (using $ζ = c / (2 k m)$ aur $ω_{n} = k / m$ ), sab kuch collapse ho jaata hai:

$T = \frac{1}{( 1 - r ^{2} ) ^{2} + ( 2 ζ r ) ^{2}}$

Square root ke andar term-by-term:

$(1 - r^{2})$ : along-motion leg, "resonance se kitna door". Jab $r = 1$ yeh zero hai — danger point.
$(2 ζ r)$ : friction leg. Yahi ek cheez hai jo $T$ ko blow up hone se rokti hai jab $(1 - r^{2}) = 0$ .

Step 6 — $r$ ke har case se guzro

KYA HAI. Formula mein $r$ ke teen regions daalo aur dekho kya hota hai.

KYUN. Contract yeh hai: reader koi aisa case nahi dekhega jise humne dikhaya nahi. Resonance se neeche, resonance par, resonance se upar — har ek bilkul alag behave karta hai, aur real testing mein sab matter karta hai.

PICTURE. Famous transmissibility curve, teeno zones coloured ke saath.

Case A — slow push, $r < 1$ (resonance se bahut neeche). Tab $r^{2}$ chhota hai, $(1 - r^{2}) \approx 1$ , friction term tiny, toh $T \approx \frac{1}{1} = 1.$ Block bas table ke saath ride karta hai. Koi amplification nahi, koi danger nahi. Part wahi feel karta hai jo use diya gaya hai.

Case B — matched push, $r = 1$ (resonance PAR). $r = 1$ set karo: $(1 - r^{2})^{2}$ term bilkul vanish ho jaata hai. Sirf friction bachti hai: $T = \frac{1}{0 + ( 2 ζ \cdot 1 ) ^{2}} = \frac{1}{2 ζ} .$ Is peak value ka apna naam hai, quality factor: $Q = \frac{1}{2 ζ} .$ $ζ = 0.02$ ke saath, $Q = 25$ . 1 g input 25 g monster ban jaata hai. Yahan hardware marti hai. Dhyan do: friction ke bina ( $ζ = 0$ ) formula deta hai $T = 1/0 = \infty$ — block khud ko shake karke toot jaata. Friction hi ek cheez hai jo use bachati hai.

Case C — fast push, $r > 2$ (resonance se upar). Ab $r^{2} > 2$ , toh $(1 - r^{2})^{2}$ bada ho jaata hai, denominator bada hai, aur $T < 1.$ Part table se kam hilta hai — mass itna sluggish hai ki keep up nahi kar sakta. Yahi vibration isolation ka principle hai: ek delicate part ko soft spring par mount karo taaki uska $ω_{n}$ shaking se bahut neeche rahe, tum safe zone mein chale jao.

Step 7 — Degenerate aur heavy-damping cases

KYA HAI. Formula ko uske limits tak push karo: koi friction nahi, rigid support, huge mass — aur important opposite extreme, itna zyada friction ki peak bilkul gayab ho jaaye.

KYUN. Edge cases woh jagah hain jahan intuition toot jaata hai aur bugs chhup jaate hain. Spacecraft brackets lightly damped hote hain, lekin rubber isolators, foams aur fluid dampers jo aur jagah use hote hain woh heavily damped ho sakte hain — reader ko woh bhi pata hona chahiye kaisa dikhta hai.

PICTURE. Left: teen quick thought-experiments. Right: peak flatten hota ja raha hai jaise $ζ$ $1/ 2$ se bada hota jaata hai.

Zero damping, $ζ \to 0$ : $r = 1$ par, $T = 1/ (2 ζ) \to \infty$ . Infinite amplitude — block bina ruke accelerate karta jaata hai. Real parts kabhi true zero damping nahi hit karte, lekin low damping exactly isliye hai kyun light structures itni dangerous hoti hain.
Bahut stiff support, $k \to \infty$ : $ω_{n} = k / m \to \infty$ , toh uski natural frequency kisi bhi launch shaking se bahut upar hai. Tab $r ≪ 1$ hamesha → $T \approx 1$ , koi amplification nahi. Ek rock-solid rigid mount test band mein kabhi resonate nahi karta. (Yahi design goal hai: $f_{n}$ ko 50 Hz se upar push karo taaki random-vibration band use ring nahi kar sake.)
Bahut heavy block, $m \to \infty$ : $ω_{n} \to 0$ , toh part hamesha $r > 1$ isolation zone mein hota hai — lekin woh kisi cheez par barely respond bhi karta hai. Massive, sluggish, safe lekin structure ke roop mein useless.
Heavy damping, $ζ \geq 1/ 2 \approx 0.707$ : peak gayab ho jaata hai. Peak hone ke liye, $T$ ko kaheen $1$ se upar uthna hoga; $T$ ka maximum actually $r_{p e ak} = 1 - 2 ζ^{2}$ par hota hai, jo sirf exist karta hai (real, non-zero number hota hai) jab $2 ζ^{2} < 1$ , yaani $ζ < 1/ 2$ . Ek baar $ζ \geq 1/ 2$ ho jaaye, $T$ bas $1$ se seedha neeche slide karta hai bina kisi bump ke — system itna gooey hai ki resonate nahi kar sakta. Isliye ek achha rubber isolator apni khud ki natural frequency se guzarte waqt barely wobble karta hai.

Recall Real life mein "at resonance" amplitude infinite kyun nahi ho sakta?

Kyunki real parts mein hamesha kuch friction hoti hai ( $ζ > 0$ ), denominator $2 ζ r$ kabhi truly zero nahi hota, toh $T$ $Q = 1/ (2 ζ)$ par finite rehta hai. ::: Friction (damping) peak ko cap karti hai.

Recall Kitni damping par resonant peak bilkul gayab ho jaata hai?

Jab $ζ \geq 1/ 2 \approx 0.707$ , kyunki peak location $r_{p e ak} = 1 - 2 ζ^{2}$ ek real number banana band ho jaata hai. ::: $ζ \geq 1/ 2$ .

Step 8 — Ek number par: parent note ka cubesat

KYA HAI. $Q = 1/ (2 ζ)$ use karo parent ke spacecraft damping ke saath.

KYUN. Taaki abstract " $T = 20$ " ek real force ban jaaye jo tum feel kar sako.

Ek-picture summary

Ab apni nazar final board par left-to-right chalao; har jagah exactly yeh dekho.

Intuition Summary figure kaise padhen

Bilkul left — cartoon. Block+spring poora model hai. Baaki sab kuch isi se derived hai. $m$ , $k$ , $c$ par point karo aur yaad karo: mass, stiffness, friction.
Arrow $ω_{n} = k / m$ ki taraf. Yeh Step 3 hai — part ki apni rhythm, sirf $k$ aur $m$ se janam leti hai. Yeh horizontal axis par "resonance" ki jagah fix karta hai.
Curve $T (r)$ . Yeh Steps 5–6 hai. Ise left se trace karo ( $T \approx 1$ , safe), spike tak (danger), $r = 2$ (dashed crossover) se neeche, isolation tail mein ( $T < 1$ ).
Peak par label $Q = 1/ (2 ζ)$ . Yeh Step 6B hai — spike kitna tall hai, bilkul friction se decide hota hai. Low $ζ$ → tall, deadly spike; high $ζ$ → koi spike nahi (Step 7).
Takeaway jo zaur bol ke padho: "Rhythm match karo aur force $Q$ se multiply ho jaati hai; friction hi ek cheez hai jo ise cap karti hai."

Recall Feynman retelling — plain words mein bolo

Ek bachche ko swing par socho. Agar tum random moments par dhaka do (wrong frequency, $r \neq = 1$ ) swing barely hilta hai — yahi hai $T \approx 1$ . Lekin agar tum exactly swing ke natural aage-peeche ke saath time karke dhaka do ( $r = 1$ ), har chhota push add hota jaata hai aur bachcha dangerously zyada upar jaata hai — yahi resonance hai, $T = Q$ . Swing ko top se neeche girne se rokne wali ek cheez hai air drag aur rusty chain — yahi hai damping $ζ$ , aur yeh maximum height set karta hai $Q = 1/ (2 ζ)$ . Agar chain thick mud mein coat ho jaaye ( $ζ \geq 1/ 2$ ) swing barely hilega — koi resonance peak nahi. Ek satellite part ek rocket par bolted swing hai. Rocket launch ke dauran ise har frequency par hilata hai. Agar kisi part ki natural swing-rhythm shaking band ke andar aati hai, woh part 10–25× pump up ho jaata hai aur toot jaata hai. Toh ground par hum deliberately shaker ko saari frequencies par sweep karte hain taaki har part ki rhythm dhundh sakein, phir ya toh use damp karein, stiffen karein ( $f_{n}$ ko band se upar uthao), ya soft mount par isolate karein ( $r > 2$ zone mein laao). Yahi poora game hai.

Connections

Parent: full environmental-testing note
Launch Vehicle Dynamics — shaking kahan se aati hai
Structural Mechanics — $k$ aur $ω_{n}$ kahan se aate hain
Reliability Engineering & Quality Assurance in Aerospace — hum ground par failures dhundhne ke liye test kyun karte hain
Related heat-side story: Spacecraft Thermal Control Systems