Shock response spectrum (SRS): Question bank — Hinglish deep dive

3.6.13 · D5 · Physics › Spacecraft Structures & Systems Engineering › Shock response spectrum (SRS)

Prerequisites pehle review karne laayak hain: Single-degree-of-freedom systems, Duhamel's integral, Quality factor Q and damping, Pyroshock environments.

Is page par use hone waale symbols (pehle define kiye, phir milenge)

Definition Symbols ki cast

Ek bhi trap answer karne se pehle, har symbol ko pin down karo, kyunki zyaadatar traps asalon mein symbol confusions hi hote hain.

$a (t)$ — base acceleration time history: ek component ke support ko kitni force se shake kiya ja raha hai, time $t$ ke function ke roop mein. Yahi shock "input" hai.
$z (t)$ — relative displacement: component ke mass ki position minus uske base ki position. Yeh measure karta hai ki mass shaking base se kitna peechhe ya aage hai, room mein uski position nahi.
$\overset{z}{˙}$ , $\overset{z}{¨}$ — $z$ ke pehle aur doosre time derivatives: relative velocity aur relative acceleration.
$ω_{n}$ — natural angular frequency radians per second mein, $ω_{n} = k / m$ stiffness $k$ aur mass $m$ ke liye: oscillator agar uthake chhod diya jaaye toh kitni tez ring karega.
$f_{n}$ — natural frequency in hertz, $f_{n} = ω_{n} / (2 π)$ : wahi ringing rate lekin cycles per second mein count ki gayi.
$ζ$ (zeta) — damping ratio, ek dimensionless number, $ζ = c / (2 k m)$ damping coefficient $c$ ke liye: ringing kitni jaldi khatam hoti hai. $ζ < 1$ under-damped (ring karta hai), $ζ = 1$ critically damped (ring nahi), $ζ > 1$ over-damped (dhheere wapas aata hai).
$Q$ — quality factor, $Q = 1/ (2 ζ)$ : resonance par peak steady-state amplification. $ζ = 0.05$ matlab $Q = 10$ . Dekho Quality factor Q and damping.
$τ$ (tau) — pulse duration: shock input $a (t)$ kitni der tak rehta hai (jaise width $τ$ ka half-sine).

Figures jo traps ko anchor karte hain

Forcing term ka sign aur SRS ki shape — yahi do cheezein hain jo log sabse zyaada galat karte hain. Dono pictures se settle ho jaate hain.

Upar diye base-excitation diagram ko dekho: mass $m$ , spring $k$ aur damper $c$ par ek aisi base se lata hai jo khud $a (t)$ se uchhaali ja rahi hai. Kyunki hum mass ko us accelerating base ke relative track karte hain, ek inertial (pseudo-)force $- m a (t)$ mass par appear hoti hai — base acceleration ke opposite direction mein. Yahi $m \overset{z}{¨} + c \overset{z}{˙} + k z = - m a (t)$ mein minus sign ka origin hai.

Doosra figure width $τ$ ka ek half-sine pulse dikhata hai, aur uske neeche ek schematic SRS: ek steep rising branch ( $\propto f_{n}^{2}$ , $+ 40$ dB/decade), ek knee $f_{n} \approx 1/ (π τ)$ ke paas, aur ek flat high-frequency branch jo input peak par level off hoti hai. Jab bhi koi trap "branch", "knee", ya "asymptote" mention kare, ise refer karo.

True ya false — justify karo

SRS sirf shock pulse ki property hai, use compute karne waale oscillators par dependent nahi.

False. SRS ek response curve hai — yeh assumed damping ratio

ζ

par depend karta hai; wahi pulse high damping par lower, smoother SRS deta hai aur low damping par peakier.

Ek hi pulse ka SRS aur Fourier transform (FFT) same information carry karte hain.

False. FFT input ki frequency content describe karta hai; SRS describe karta hai ki resonators ki ek family kaise respond karti hai. Ek short pulse ka broadband FFT hoga lekin SRS mein ek distinct knee hoga.

Power spectral density (PSD) sirf FFT ka doosra naam hai.

False. Ek PSD, FFT ke squared magnitude ko bandwidth se divide karta hai (units

g^{2} / Hz

), isliye phase discard hota hai aur power per unit frequency express hoti hai; raw FFT amplitude aur phase dono rakhta hai. Dono mein se koi bhi SRS nahi hai.

Do bilkul alag dikhne waale time histories identical SRS curves rakh sakte hain.

True. Same net velocity change

Δ v = \int a d t

aur similar duration waale kai pulse shapes almost same SRS par collapse ho jaate hain — SRS phase aur bahut saara waveform detail discard karta hai, isliye SRS specs akele test signal uniquely define nahi karte.

Bahut high natural frequency par ek short pulse ka SRS input acceleration

a (t)

ke peak ke paas pahunch jaata hai.

True. Bahut stiff/fast oscillator ki period pulse se kaafi choti hoti hai, isliye woh base ko rigidly track karta hai — uski absolute acceleration base acceleration ke barabar hoti hai, jiska maximum input peak hai.

Damping ratio

ζ

badhane se SRS har frequency par hamesha kam hoti hai.

Zyaadatar resonance peaks ke paas true hai, jahan damping amplification ko kaata hai; lekin low-frequency rising branch aur high-frequency flat branch par SRS

ζ

ke liye almost insensitive hai, isliye "hamesha" bahut strong claim hai.

"500 Hz par 1000g SRS" ka number damping ratio bataye bina meaningless hai.

True. SRS value frequency par

ζ

ke saath change hoti hai; component ki capability ko environment se compare karne ke liye dono ko ek hi same

ζ

par quote karna zaroori hai (commonly 5%, yaani Q

= 10

Primary (positive) SRS aur residual (negative) SRS alag ho sakte hain.

True. Pulse ke dauran response (primary) aur khatam hone ke baad (residual, free decay) alag values par peak kar sakta hai; standards exactly isi wajah se maximax, primary, aur residual SRS mein distinguish karte hain.

Galati dhundo

f_{n}

par SRS woh acceleration hai jo base

f_{n}

par vibrate karte time reach karta hai."

Base "

f_{n}

par vibrate" nahi karta — ek single transient pulse mein saari frequencies ek saath hoti hain.

f_{n}

par SRS $f_{n}$ par tuned oscillator ka peak response hai, base ki us frequency par koi property nahi.

"Kyunki

\overset{z}{¨}

woh acceleration hai jo hum compute karte hain, SRS ko

max ∣ \overset{z}{¨} ∣

plot karna chahiye."

\overset{z}{¨}

relative acceleration hai jo moving base se measure ki gayi hai. Hardware jo cheez todta hai woh hai absolute acceleration

a_{abs} = - 2 ζ ω_{n} \overset{z}{˙} - ω_{n}^{2} z

;

max ∣ \overset{z}{¨} ∣

plot karne se galat (base-frame) damage measure milega.

"Equation of motion hai

m \overset{z}{¨} + c \overset{z}{˙} + k z = + ma (t)

Sign galat hai. Base-relative frame mein inertial forcing term

- ma (t)

hai — base acceleration ek aisi force ki tarah act karta hai jo base motion ke opposite hoti hai (accelerating car mein "peeche dhakele jaane" ka effect; upar base-excitation figure dekho).

"Shorter shock pulse hamesha kam severe hota hai kyunki woh less energy deliver karta hai."

Shorter pulse knee ko higher frequency par le jaata hai (

f_{knee} \approx 1/ (π τ)

), isliye woh high-frequency components (relays, PCBs) ke liye zyaada damaging ho sakta hai chahe total energy kam ho — pyroshocks exactly yahi hain: brief lekin high-frequency-rich.

"SRS find karne ke liye main

a (t)

ka FFT leta hoon aur peaks read karta hoon."

Koi FFT involve nahi hota. Tum har

f_{n}

ke liye SDOF equation solve karte ho (Duhamel convolution ya numerical integration ke zariye), phir

∣ a_{abs} (t) ∣

ka time-domain peak lete ho.

ω_{d} = ω_{n} 1 + ζ^{2}

Root ke andar sign galat hai: damped natural frequency hai

ω_{d} = ω_{n} 1 - ζ^{2}

, jo undamped se slower hai — damping ringing frequency ko neeche kheenchti hai, upar nahi.

"Resonance par SDOF ki peak amplification hamesha lagbhag

2 \times

hoti hai."

Steady-state sinusoidal drive ke liye peak amplification

\approx Q = 1/ (2 ζ)

hoti hai (jaise 5% damping par

10

). Ek short transient sirf partially ring up karta hai, isliye woh

Q

se kaafi kam show kar sakta hai — amplification depend karta hai ki pulse kitne cycles excite karta hai.

Why questions

Hum absolute acceleration pane ke liye

a (t)

\overset{z}{¨}

mein kyun add karte hain, aur

\overset{z}{¨}

term kahan jaati hai?

Definition se

a_{abs} = a (t) + \overset{z}{¨}

. EOM

\overset{z}{¨} + 2 ζ ω_{n} \overset{z}{˙} + ω_{n}^{2} z = - a (t)

\overset{z}{¨}

solve karo:

\overset{z}{¨} = - a (t) - 2 ζ ω_{n} \overset{z}{˙} - ω_{n}^{2} z

, aur substitute karo: dono

a (t)

terms exactly cancel ho jaate hain, baki bachta hai

a_{abs} = - 2 ζ ω_{n} \overset{z}{˙} - ω_{n}^{2} z

— relative-acceleration term replace ho jaata hai spring aur damper terms se.

Ek pulse SRS ki low-frequency branch

f_{n}^{2}

(+40 dB/decade) ki tarah kyun rise karti hai?

Ek soft, slow oscillator brief pulse ke dauran barely move karta hai — woh ek aisa static spring behave karta hai jo net impulse se displace hota hai. Uski peak absolute acceleration

ω_{n}^{2} z

ω_{n}^{2} \propto f_{n}^{2}

ke scale par hoti hai, isliye steep rise aati hai.

Hum real components ko SDOF oscillators ke roop mein model hi kyun karte hain?

Apne fundamental resonance ke paas almost har structure ek mass on one spring with one damper ki tarah behave karta hai (dekho Modal analysis); SRS har possible fundamental frequency ke liye worst-case response precompute karta hai taaki hum har part ko individually test na karein.

Qualification ke liye SRS raw shock time history se zyaada useful kyun hai?

Time history ek measurement point aur ek waveform ke liye specific hoti hai. SRS use ek frequency-indexed damage measure mein convert karta hai jo tumhe directly ek environment ko component ki rated capability se matching

ζ

par compare karne deta hai.

Half-sine of duration

τ

ke liye knee frequency

1/ (π τ)

ki tarah kyun scale karti hai?

Knee wahan mark hota hai jahan oscillator ki period roughly pulse duration se match karti hai — lagbhag ek cycle

τ

ke andar fit hota hai. Yeh

f_{n} τ \sim 1/ π

ke paas hota hai, isliye shorter pulses knee ko higher le jaate hain.

Damping, knee se bahut neeche aur bahut upar SRS ko barely kyun change karti hai?

Dono branches par oscillator ya toh impulse ko quasi-statically follow karta hai (low

f

) ya base peak ko rigidly track karta hai (high

f

); resonant ring-up — jo ek cheez hai damping strongly suppress karti hai — mainly knee ke paas hi hoti hai.

Edge cases

SRS kaisi dikhti hai jab

f_{n} \to 0

(infinitely soft mount)?

Woh zero ki taraf jaati hai: bahut soft oscillator impulse se displace hota hai lekin

ω_{n}^{2}

vanishing hoti hai, isliye uski absolute acceleration

\to 0

— yahi perfect shock isolation ka ideal hai.

Exactly

ζ = 1

(critical damping) par response formula ka kya hota hai?

Underdamped Duhamel form break down karta hai kyunki

ω_{d} = ω_{n} 1 - ζ^{2} = 0

; tumhe sine ki jagah

t e^{- ω_{n} t}

term waala critically damped solution use karna padta hai. SRS khud finite aur smooth rehta hai.

Over-damped regime

ζ > 1

mein — kya SRS abhi bhi exist karta hai?

Haan. Ab

1 - ζ^{2} < 0

hai, isliye

ω_{d}

imaginary ho jaati hai aur sine real decaying exponentials

e^{(- ζ ω_{n} \pm ω_{n} ζ^{2} - 1) t}

ke sum mein convert ho jaata hai — bilkul oscillation nahi, bas ek sluggish return. SRS abhi bhi

max_{t} ∣ a_{abs} ∣

hai aur critically damped case se bhi zyaada smooth aur lower hai, kyunki heavy damping ring-up ko completely suppress kar deti hai; yeh input-peak envelope ki taraf smoothly approach karta hai bina kisi resonant hump ke.

Kya SRS value kabhi input peak acceleration se neeche ja sakti hai?

Haan — low-frequency branch par ek soft oscillator base peak se kam respond karta hai (de-amplification). Input peak se upar amplification sirf knee ke paas aur uske upar appear hoti hai.

Kya ek supplier ki 1500g SRS at

ζ = 0.03

ko flight spec of 1800g at

ζ = 0.05

se compare karna valid hai?

Nahi — alag damping matlab alag curves. Tumhe dono ko ek common

ζ

par re-derive karna hoga; lower-damping rating generally zyaada conservative hoti hai (same pulse ke liye higher), isliye inhe mix karna margin ko chupa ya fake kar sakta hai.

Bahut lamba pulse (

f_{n} τ ≫ 1

) oscillator ke relative mein kya response deta hai?

Oscillator ke paas dhheere badal rahi base ko quasi-steady follow karne ka time hota hai, isliye uski absolute acceleration input ko closely track karti hai — SRS input peak ki taraf flatten hoti hai, yaani high-frequency asymptote.

Agar do shock events time mein overlap karein — kya SRS simply add ho jaate hain?

Nahi. SRS ek nonlinear "max of a response" operation hai, linear transform nahi; tumhe pehle time histories superpose karni padhti hain (SDOF response linear hai) aur phir combined signal ka ek single SRS compute karna padta hai.