Shock response spectrum (SRS): Worked examples — Hinglish deep dive

3.6.13 · D3 · Physics › Spacecraft Structures & Systems Engineering › Shock response spectrum (SRS)

Shuru karne se pehle, ek symbol jo hum har jagah use karte hain. Shock ke liye sabse important dimensionless number hai $f_{n} τ$ — natural frequency times the pulse duration. Yeh literally count karta hai ki chota oscillator pulse ke dauran kitne oscillation cycles complete karta hai jab tak pulse usse push kar raha hai.

Recall Symbols jo hum reuse karte hain (parent note se)

Is page pe har symbol parent mein build hua hai; yahan sab ek jagah hain taaki kuch "cold" na lage:

$f_{n}$ = chote oscillator ki natural frequency, Hz mein (cycles per second). $ω_{n} = 2 π f_{n}$ wahi cheez hai radians per second mein.
$T_{n} = 1/ f_{n}$ = oscillator ka period, ek poore swing ka time.
$τ$ = duration of the shock pulse (base kitni der push karta hai), seconds mein.
$A_{0}$ = base acceleration ki peak amplitude, $g$ mein quoted (gravity ke multiples, $1 g = 9.81 m/s^{2}$ ).
$ζ$ (Greek "zeta") = damping ratio, ek pure number: $ζ = 0$ matlab koi energy bleed nahi hoti (hamesha ring karta hai), $ζ = 1$ matlab oscillate hi nahi karta. Spacecraft components typically $ζ = 0.05$ (5%) use karte hain. Dekho Quality factor Q and damping.
$ω_{d} = ω_{n} 1 - ζ^{2}$ = damped natural frequency (rad/s): damping ke baad actual ringing rate thodi slow hoti hai. Chhote $ζ$ ke liye yeh lagbhag $ω_{n}$ hi hoti hai.
$z (t)$ = relative displacement (component minus base); $\overset{z}{˙}$ uski velocity hai, $\overset{z}{¨}$ uska acceleration.
$a_{abs} (t)$ = inertial frame mein chhote mass ka absolute acceleration — woh cheez jo actually component ko todti hai. Parent ke equation of motion se yeh equals $a_{abs} (t) = - 2 ζ ω_{n} \overset{z}{˙} (t) - ω_{n}^{2} z (t)$ , aur SRS hai $max_{t} ∣ a_{abs} (t) ∣$ .

Scenario matrix

Yeh hai har case-class jo ek SRS problem tumpe throw kar sakti hai. Har worked example neeche tagged hai us cell se jisme woh aata hai.

Cell	Case class	Kya special hai	Example
A	Short pulse, $f_{n} τ ≪ 1$	rising branch, $\propto f_{n}^{2}$	Ex 1
B	Pulse at the knee, $f_{n} τ \approx 1/ π$	maximum ring-up	Ex 2
C	Long pulse, $f_{n} τ ≫ 1$	flat branch, oscillator tracks base	Ex 3
D	Zero-damping limit, $ζ = 0$	undamped shock factor, no decay	Ex 4
E	Degenerate input, $a (t) = 0$	SRS har jagah zero hona chahiye	Ex 5
F	Residual vs primary peak	peak pulse khatam hone ke baad aati hai	Ex 6
G	Real-world qualification	capability vs requirement compare karo, same $ζ$	Ex 7
H	Exam twist: damping / velocity change	reasoning, plugging nahi	Ex 8

Backbone formulas jo hum reuse karte hain (sab parent se):

Figure s01 — SRS map. Horizontal axis natural frequency $f_{n}$ hai (Hz, log scale); vertical axis woh SRS acceleration hai jo us frequency ka component feel karega (in $g$ , log scale). Cyan curve hamari recurring half-sine pulse ( $A_{0} = 300 g$ , $τ = 0.2$ ms) ka poora SRS envelope hai. Isko left-to-right follow karo: yeh $f_{n}^{2}$ ki tarah steeply climb karta hai (rising branch), dashed amber knee line pe bend karta hai ( $f_{n} τ = 1/ π$ , lagbhag 1592 Hz), phir input peak ki taraf flatten ho jaata hai (flat branch). Teen dots neeche ke examples ke pins hain: Ex 1 steep ramp pe, Ex 2 amber knee pe, Ex 3 flat plateau pe kaafi door — same pulse, teen bilkul alag answers, sirf is baat se decide hote hain ki $f_{n}$ is curve pe kahan land karta hai. Hum is figure ko explicitly Ex 1, Ex 2 aur Ex 3 ke andar invoke karte hain jab hum har pin curve se read karte hain.

Example 1 — Short pulse (Cell A: rising branch)

Worked example Circuit board, short shock

Ek half-sine base shock ki amplitude $A_{0} = 300 g$ aur duration $τ = 0.2 ms = 0.0002 s$ hai. Ek circuit board ka $f_{n} = 500$ Hz hai, damping $ζ = 0.05$ . Uska SRS acceleration estimate karo.

Forecast: Andaza lagao — board ka peak input $300 g$ se upar hoga ya neeche? Roughly kitna?

$f_{n} τ$ compute karo. $f_{n} τ = 500 \times 0.0002 = 0.1$ . Yeh step kyun? Yeh ek number humein batata hai ki hum kis branch pe hain. $0.1 ≪ 1$ → rising branch pe deep hain, toh hum input se neeche peak expect karte hain. Figure s01 pe yeh white "Ex 1" pin hai jo steep left ramp pe neeche hai.
Low-frequency asymptote apply karo. $SRS \approx A_{0} \cdot \frac{π ^{2} f _{n}^{2} τ ^{2}}{2} = 300 \cdot \frac{π ^{2} ( 500 ) ^{2} ( 0.0002 ) ^{2}}{2}$ Yeh step kyun? Rising branch pe response base velocity kick se dominated hota hai aur $f_{n}^{2}$ ki tarah badhta hai; asymptote formula exactly wahi capture karta hai.
Calculate karo. $f_{n}^{2} τ^{2} = (f_{n} τ)^{2} = 0.01$ , toh $SRS \approx 300 \cdot \frac{π ^{2} \cdot 0.01}{2} = 300 \cdot 0.04935 \approx 14.8 g .$ Yeh step kyun? Dhyaan do ki sab kuch $(f_{n} τ)^{2}$ pe collapse ho jaata hai — rising branch sirf us ek number pe depend karta hai.

Verify: $14.8 g ≪ 300 g$ , toh board ne shock barely feel kiya — exactly wahi jo "rising branch pe deep" promise karta hai. Units: $g \cdot (dimensionless)^{2} = g$ . ✓ Hamara forecast hona chahiye tha "input se kaafi neeche."

Example 2 — Pulse at the knee (Cell B: maximum ring-up)

Worked example Component tuned to the knee

Same half-sine: $A_{0} = 300 g$ , $τ = 0.0002$ s. Ab component exactly knee frequency pe baitha hai. Knee $f_{n}$ dhundho aur uska SRS estimate karo.

Forecast: Knee pe oscillator "ring up" karta hai. Amplification factor andaza lagao — 1× ke kareeb ya 10× ke?

Knee dhundho. Half-sine ke liye knee $f_{n} τ \approx 1/ π$ hai, toh $f_{knee} = \frac{1}{π τ} = \frac{1}{π \cdot 0.0002} \approx 1592 Hz .$ Yeh step kyun? Knee woh jagah hai jahan rising branch flat branch mein bend hoti hai — woh frequency jahan oscillator pulse ke dauran lagbhag half cycle complete karta hai aur sabse zyada ring karta hai. Figure s01 pe yeh amber "Ex 2" pin exactly dashed amber knee line pe baitha hai.
Exact undamped half-sine result se amplification estimate karo. Ek undamped SDOF ko half-sine se drive karne ka classic closed-form maximax amplification factor $\approx 1.77$ pe peak karta hai (Duhamel-integral solution mein derived; Ex 4 bhi dekho jahan hum yeh ceiling quote aur check karte hain). Light damping $ζ = 0.05$ ise thoda trim karta hai, practical near-knee factor ko $\approx 1.6$ – $1.8$ band mein le aata hai. Yeh step kyun? Hum 1.6–1.8 air se assert nahi kar rahe — yeh proven undamped ceiling $1.77$ ka damped neighbourhood hai jo Ex 4 establish karta hai. Ek shock full sustained-resonance $Q = 1/ (2 ζ) = 10$ build nahi kar sakta, toh factor 1.7 ke kareeb rehta hai.
SRS estimate karo. $SRS \approx 1.7 \times 300 g = 510 g$ . Yeh step kyun? Figure s01 se amber pin read karne pe wahan curve ki height thodi input se kam double hai — hamare $1.7 \times$ estimate se consistent hai.

Verify: $510 g > 300 g$ (amplified, jaisa ek resonant knee hona chahiye) aur $510 g$ bahut neeche hai $10 \times 300 = 3000 g$ se (koi sustained resonance nahi) — dono sanity bounds pass hain. ✓ Forecast: ~1.7× amplification.

Example 3 — Long pulse (Cell C: flat branch)

Worked example Fast relay under a slow pulse

Half-sine $A_{0} = 300 g$ , lekin ab ek long pulse $τ = 10 ms = 0.01$ s. Ek stiff relay ka $f_{n} = 5000$ Hz hai. Uska SRS estimate karo.

Forecast: Ek bahut fast oscillator ek slow push dekh raha hai — kya woh amplify karega, ya bas saath chalega?

$f_{n} τ$ compute karo. $f_{n} τ = 5000 \times 0.01 = 50$ . Yeh step kyun? $50 ≫ 1$ → oscillator 50 baar swing karta hai jabki pulse abhi bhi rise aur fall kar raha hai. Woh khud se aage nahi ja sakta; woh bas base ke saath chalta hai. Figure s01 pe yeh pin flat cyan plateau ke kaafi right pe hai.
High-frequency asymptote apply karo. $SRS \to A_{0} = 300 g$ . Yeh step kyun? Jab base oscillator ke comparison mein slowly move karta hai, toh relative displacement $z \to 0$ aur $a_{abs} \to a (t)$ ; $a (t)$ ka peak $A_{0}$ hai.
Result batao. $SRS (5000 Hz) \approx 300 g$ . Yeh step kyun? Hum number explicitly spell out karte hain taaki ise Figure s01 ke flat plateau pe seedha drop kiya ja sake aur doosre pins se compare kiya ja sake — flat-branch SRS hai input peak, aur ise name dena "same pulse, teen answers" contrast ko concrete banata hai.

Verify: Flat branch pe answer input peak $300 g$ ke equal hai — ek fast oscillator ka defining behavior. Ex 1 se contrast karo (same input, tiny $14.8 g$ ) aur Ex 2 se (knee, $510 g$ ): teen cells, same pulse, teen answers. ✓

Example 4 — Zero damping limit (Cell D: $ζ = 0$ )

Worked example Undamped shock amplification factor

Ek long half-sine lo ( $f_{n} τ ≫ 1$ ) lekin damping zero kar do, $ζ = 0$ . Half-sine ke liye maximum possible amplification kya hai, aur $A_{0} = 300 g$ kya SRS deta hai?

Forecast: Koi damping nahi toh oscillator kabhi energy bleed nahi karta. Kya amplification infinity tak blow up hoga, ya ek finite ceiling pe rukta hai?

Exact undamped half-sine result recall karo. Ek undamped SDOF ke liye classic maximax shock amplification factor of a half-sine ek finite value $\approx 1.77$ pe top out karta hai (yeh diverge nahi karta, kyunki ek single pulse oscillator ko sirf finitely pump karta hai). Yeh step kyun? Shock ek finite-energy event hai; undamped bhi, half-sine ki peak amplification bounded hai — yeh theoretical ceiling hai jisko hamare saare "1.6–1.8" estimates approach karte hain.
Apply karo. $SRS_{m a x} \approx 1.77 \times 300 g = 531 g$ . Yeh step kyun? Zero damping upper bound SRS deta hai; real $ζ = 0.05$ thoda neeche rehta hai.
Steady resonance se contrast karo. Ek sustained sine $Q = 1/ (2 ζ) \to \infty$ deta as $ζ \to 0$ . Shock nahi deta, kyunki drive ruk jaata hai. Yeh step kyun? Yahi poora point hai zero-damping edge case ka — yeh "transient shock" ko "steady resonance" se alag karta hai.

Verify: $531 g /300 g = 1.77$ ✓ — finite, infinite nahi, exactly jaisa boundedness argument demand karta hai. Aur $1.77 > 1.7$ (hamara damped near-knee estimate), consistent hai "less damping → slightly more amplification" ke saath. ✓

Example 5 — Degenerate input (Cell E: $a (t) = 0$ )

Worked example No-shock sanity check

Base koi acceleration experience nahi karta: $a (t) = 0$ sabhi $t$ ke liye, rest se starting. $SRS (f_{n}, ζ)$ har frequency pe kya hai?

Forecast: Trivial — lekin kya tum jaante ho kyun math se zero hai, sirf "obviously" se nahi?

Duhamel mein substitute karo. Parent ka relative-displacement integral hai $z (t) = - \frac{1}{ω _{d}} \int_{0}^{t} a (τ) e^{- ζ ω_{n} (t - τ)} sin [ω_{d} (t - τ)] d τ .$ $a (τ) \equiv 0$ ke saath integrand $0$ hai, toh $z (t) = 0$ aur $\overset{z}{˙} (t) = 0$ . Yeh step kyun? SRS input se driven hota hai; input kill karo toh har response identically vanish ho jaata hai.
Absolute acceleration compute karo. Is page ke upar recall ki gayi definition use karke, $a_{abs} (t) = - 2 ζ ω_{n} \overset{z}{˙} - ω_{n}^{2} z = 0$ . Yeh step kyun? Confirm karta hai ki response frame bhi zero deta hai, sirf relative nahi — aur yeh symbol $a_{abs}$ jo humne pehle recall kiya tha usse dikhata hai jaisa hona chahiye.
Max lo. $SRS (f_{n}, ζ) = max_{t} ∣ a_{abs} (t) ∣ = 0$ sab $f_{n}$ ke liye aur sab $ζ$ ke liye. Yeh step kyun? SRS defined hai $∣ a_{abs} (t) ∣$ ke time pe maximum ke roop mein; max lena woh final, mandatory operation hai jo time-history ko ek SRS number mein convert karta hai — aur ek function ka maximum jo identically zero hai, woh zero hai, har $f_{n}$ aur har $ζ$ pe. Yeh anchor case hai jo certify karta hai ki algorithm mein koi spurious offsets nahi hain.

Verify: Ek flat-zero curve. Yeh degenerate anchor hai: koi bhi SRS algorithm jo zero input ke liye nonzero return karta hai usmein bug hai (usually ek spurious initial condition). ✓

Example 6 — Residual vs primary peak (Cell F)

Worked example Woh peak jo late aati hai

Ek bahut short pulse ( $f_{n} τ ≪ 1$ ) ek velocity change $Δ v$ deliver karta hai. $A_{0} = 300 g$ aur $τ = 0.0002$ s wale half-sine ke liye, $f_{n} = 500$ Hz pe, maximum response pulse ke dauran (primary) hai ya uske khatam hone ke baad (residual)? Residual SRS estimate karo.

Forecast: Ek quick kick ke baad oscillator swing karna shuru ho jaata hai aur swinging karta rehta hai. Sabse bada acceleration kahan hai — kick ke dauran, ya baad mein free ringing ke dauran?

Figure s02 — kyun late peak jeetti hai. Top panel: short amber base pulse $a (t)$ , $τ$ tak khatam. Bottom panel: oscillator ka absolute acceleration $a_{abs} (t)$ cyan mein. Pulse ke dauran (shaded band) mass ne barely move kiya hai, toh $a_{abs}$ chhota hai (primary region). Sirf pulse khatam hone ke baad free ringing apna full swing reach karti hai — band ke right mein pehla cyan crest residual peak hai, band ke andar kuch bhi se uncha. Yahi geometric reason hai ki short pulse ke liye late peak dominate karti hai.

Pulse ka velocity change. $Δ v = \frac{2 A _{0} τ}{π}$ . $A_{0} = 300 g = 300 \times 9.81 = 2943 m/s^{2}$ ke saath: $Δ v = \frac{2 \cdot 2943 \cdot 0.0002}{π} = \frac{1.1772}{π} \approx 0.3747 m/s .$ Yeh step kyun? Ek short pulse ke liye oscillator sirf net velocity kick "remember" karta hai, pulse ki shape nahi — exactly isliye Figure s02 ka top panel ek single number $Δ v$ tak collapse ho sakta hai.
Residual (post-pulse) peak acceleration. Velocity $Δ v$ se start hua ek free undamped oscillator $z = (Δ v / ω_{n}) sin ω_{n} t$ ki tarah oscillate karta hai, toh uska peak absolute acceleration $ω_{n} Δ v$ hai. $ω_{n} = 2 π \cdot 500 = 3141.6$ rad/s ke saath: $a_{res} = ω_{n} Δ v = 3141.6 \times 0.3747 \approx 1177 m/s^{2} .$ Yeh step kyun? Yeh residual SRS hai — peak free ringing ke dauran hoti hai, pulse khatam hone ke baad, yani Figure s02 mein shaded band ke right mein woh uncha cyan crest.
$g$ mein convert karo. $a_{res} = 1177/9.81 \approx 120 g$ ; equivalently, clean algebra mein $a_{res} = ω_{n} Δ v = 2 π f_{n} \cdot \frac{2 A _{0} τ}{π} = 4 f_{n} A_{0} τ = 4 \cdot 500 \cdot 300 g \cdot 0.0002 = 120 g$ . Yeh step kyun? Clean algebra $a_{res} = 4 A_{0} (f_{n} τ)$ dikhata hai ki residual sirf $f_{n}^{1}$ ki tarah badhta hai (velocity limit), Ex 1 ke primary $f_{n}^{2}$ branch se gentler.

Verify: $a_{res} = 4 \cdot 300 \cdot 0.1 = 120 g$ . Ex 1 ke primary estimate $14.8 g$ se compare karo: residual dominate karta hai yahan ( $120 g > 14.8 g$ ), toh is short pulse ke liye peak actually pulse khatam hone ke baad aati hai — exactly wahi jo Figure s02 dikhata hai. Isliye SRS software hamesha simulation ko pulse ke past continue karta hai. ✓

Example 7 — Real-world qualification (Cell G)

Worked example Kya relay qualified hai?

Ek relay datasheet capability $1500 g$ list karta hai $2000$ Hz pe, $ζ = 0.05$ . Tumhara flight pyroshock spec hai $1800 g$ at $2000$ Hz, bhi $ζ = 0.05$ . Phir koi propose karta hai ek soft mount jo relay ki effective $f_{n}$ ko $700$ Hz tak lower kar de, jahan spec SRS sirf $900 g$ hai. Mount ke baad qualified?

Forecast: As-is: pass ya fail? Shock-isolate karne ke baad: kya frequency shift karna actually help karta hai?

Same- $ζ$ check. Dono numbers $ζ = 0.05$ use karte hain. Yeh step kyun? SRS values sirf identical damping pe comparable hoti hain — alag $ζ$ poori curve ko shift kar deta hai. Yahan match karte hain. ✓
As-is comparison. Requirement $1800 g$ vs capability $1500 g$ : $1800 > 1500$ → not qualified. Yeh step kyun? Flight relay ko uske rated survival level se aage drive karega.
Soft mount ke baad. Relay ke $f_{n}$ pe new environment: $700$ Hz pe spec $900 g$ hai. Capability $1500 g$ se compare karo: $900 < 1500$ → qualified, margin factor $1500/900 = 1.667$ ke saath. Yeh step kyun? Shock isolation component ko SRS curve pe lower-response frequency par le jaata hai — parent note se classic fix.

Verify: Margin $= 1500/900 = 1.667 \approx 1.67$ ( $> 1$ , toh qualified). As-is deficit $1800 - 1500 = 300 g$ short tha; mount ek $- 300 g$ deficit ko $+ 600 g$ margin mein convert karta hai. ✓

Example 8 — Exam twist: reason karo, plug mat karo (Cell H)

Worked example Do boards, ek shock

Board X: $f_{n} = 500$ Hz, $ζ = 0.05$ . Board Y: identical $f_{n} = 500$ Hz lekin higher damping $ζ = 0.20$ (ek lower-Q potted board). Same short half-sine ( $A_{0} = 300 g$ , $τ = 0.0002$ s). Kaun zyada residual acceleration dekhta hai, aur roughly kya hai Board X ka residual pehle free swing ke baad?

Forecast: Higher damping — kya yeh pehli peak zyada lower karta hai, ya mainly baad ki rings ko?

Same $f_{n}$ , same $Δ v$ . Dono boards ko identical velocity kick $Δ v = 2 A_{0} τ / π$ milta hai, $ζ$ se independent. Yeh step kyun? Damping impulse delivered nahi badalta — woh sirf change karta hai ki ringing baad mein kitni jaldi die out karti hai.
First residual peak factor. Light-to-moderate damping ke liye pehli peak decay factor $e^{- ζ ω_{n} t_{1}}$ se reduce hoti hai, jahan $t_{1} = \frac{1}{4} T_{n}$ pehle velocity reversal tak ka quarter-period hai. Board X ke liye, $ζ ω_{n} t_{1} = 0.05 \cdot 3141.6 \cdot \frac{1}{4 \cdot 500} = 0.0785$ , toh factor hai $e^{- 0.0785} \approx 0.9245$ . Yeh step kyun? Yeh dikhata hai ki pehli peak barely dented hai — damping mainly baad ke cycles khaata hai, pehle wale ko nahi.
Dono boards ke numbers. Board X first-peak residual $\approx 120 g \times 0.9245 \approx 111 g$ . Board Y ka factor hai $e^{- 0.20 \cdot 3141.6 \cdot \frac{1}{2000}} = e^{- 0.3142} \approx 0.7304$ , jo deta hai $\approx 120 g \times 0.7304 \approx 87.6 g$ . Yeh step kyun? Ab hum dono boards ko directly rank kar sakte hain: Board Y (more damping) lower first peak dekhta hai, $87.6 g < 111 g$ — toh potted, lower-Q board jeet jaata hai.

Verify: Ordering $87.6 g < 111 g < 120 g$ (undamped baseline) damping mein monotone hai — zyada damping, lower peak, exactly jaisa physics demand karta hai. Dono Ex 6 ke undamped $120 g$ ke karib hain, confirming ki damping trim karta hai lekin pehli residual peak ko dominate nahi karta. ✓ Forecast answered: higher damping pehli peak ko sirf modestly lower karta hai; uska asli kaam hai baad ki rings ko khana.

Recall Self-test

Kaun sa dimensionless number SRS branch decide karta hai? ::: $f_{n} τ$ — pulse ke dauran complete hue cycles. Pulse duration ke terms mein half-sine knee frequency? ::: $f_{knee} = 1/ (π τ)$ . Rising branch $f_{n}$ ke kiski power ki tarah scale karta hai? ::: $f_{n}^{2}$ (via $SRS \approx A_{0} π^{2} f_{n}^{2} τ^{2} /2$ ). Short pulse ka residual (velocity-limited) peak $f_{n}$ ke kiski power ki tarah scale karta hai? ::: $f_{n}^{1}$ , kyunki $a_{res} = 4 A_{0} f_{n} τ$ . $a (t) = 0$ ke liye SRS? ::: har frequency aur damping pe exactly zero. Kya undamped half-sine amplification diverge karta hai? ::: Nahi — yeh $1.77 \times$ ke kareeb cap hota hai.