Random vibration — PSD, RMS acceleration: Question bank

3.6.11 · D5 · Physics › Spacecraft Structures & Systems Engineering › Random vibration — PSD, RMS acceleration

Neeche ek quick vocabulary anchor hai taaki har reveal cleanly padha ja sake. Notation note: symbol clashes se bachne ke liye, is page mein record length ko $T_{rec}$ (recording ke seconds) likha gaya hai aur $T (f)$ ko transmissibility ke liye reserve kiya gaya hai; relative displacement ko $z_{rel}$ likha gaya hai taaki ise damping ratio $ζ$ se kabhi confuse na kiya ja sake.

Recall Symbol refresher (agar koi notation unfamiliar lage toh open karo)

$x (t)$ ::: time mein raw acceleration signal — ek wiggly, unpredictable curve.
$X (f, T_{rec})$ ::: $x (t)$ ka Fourier transform jo finite length $T_{rec}$ (seconds mein) ki recording ke upar compute kiya gaya hai; ye batata hai ki us record mein frequency $f$ kitni hai. $T_{rec}$ sirf record length hai — kabhi transmissibility nahi.
$G_{xx} (f)$ ::: Power Spectral Density — frequency ke har 1 Hz slice mein kitna mean-square "stuff" ( $g^{2}$ ) hai, isliye iske units hain $g^{2} / Hz$ . Yahan one-sided assume kiya gaya hai (sirf $f \geq 0$ ke liye defined).
$a_{RMS}$ ::: one-sided PSD curve ke neeche ke area ka square root; ek single number $g$ mein jo overall intensity describe karta hai.
$E [\cdot]$ ::: ensemble average — usi random environment ki bahut saari independent recordings ka average, na ki ek recording ka time average.
$r = f / f_{n}$ ::: dimensionless frequency ratio; $r = 1$ matlab "resonance par." $(f / f_{n})^{2}$ jaise terms simply $r^{2}$ hain.
$f_{n}$ , $ζ$ , $Q$ ::: natural frequency, damping ratio, aur quality factor $Q = 1/ (2 ζ)$ — resonance ki peakiness.
$z_{rel} = y - x$ , $H_{z}$ ::: relative displacement (mass motion $y$ minus base motion $x$ ) aur iska frequency-domain filter $∣ H_{z} (f) ∣^{2}$ ; yahi mounts ko bend karta hai aur stress drive karta hai. ( $z_{rel}$ ek displacement hai, $ζ$ ek damping ratio hai — inhe confuse mat karo.)
$T (f)$ ::: transmissibility — output acceleration divided by input base acceleration, frequency by frequency. Ye ek complex quantity hai jisme magnitude $∣ T ∣$ aur phase hoti hai; random-vibration RMS mein sirf $∣ T ∣^{2}$ use hota hai.

Reveals se pehle, teen pictures poore topic ko anchor karti hain. Har item argue karte waqt inhe refer karo.

Figure 1 — ek PSD ek area hai, height nahi. Parent note se flat aur piecewise PSDs; RMS hai $shaded area$ .

Figure — Random vibration — PSD, RMS acceleration

Figure 2 — transmissibility magnitude, teeno regimes. Low frequency 1 par ride karta hai, resonance $\approx Q$ tak spike karta hai, aur high frequency roll off karta hai — damping term ke saath jo ise kabhi zero nahi hone deta.

Figure 3 — base-excited mass. Base $x$ move karta hai, mass $y$ move karti hai; mounts relative displacement $z_{rel} = y - x$ se stretch hote hain, jo actually stress cause karta hai.

True or false — justify karo

PSD units g per Hz hain, isliye RMS sirf PSD times bandwidth hai

False — PSD units

g^{2} / Hz

hain (mean-square), isliye iska integral

g^{2}

deta hai;

g

paane ke liye end mein square root lena zaroori hai.

Flat PSD level ko double karne se RMS acceleration double ho jaata hai

False — RMS, PSD ke square root ke saath scale karta hai, isliye level double karne se RMS

2 \approx 1.41

se multiply hota hai, 2 se nahi.

Ek flat PSD ka matlab hai ki structure ko har frequency par equal acceleration feel hoti hai

Ye is sense mein True hai ki har bin mein equal

g^{2} / Hz

per bin hai, lekin total energy wide high-frequency span se dominate hoti hai kyunki wahan zyaada Hz hote hain (Figure 1) — ek flat PSD flat energy contribution nahi hai.

Agar flat PSD ka frequency band aadha kar do, toh RMS aadha ho jaata hai

False — area (mean-square) aadha hota hai, isliye RMS sirf

2

se girta hai, apni original value ka lagbhag

71%

tak.

Ek stationary random process ke liye, mean aur variance time mein constant hote hain

True — statistics ka yahi constant rehna stationary ki definition hai; individual samples phir bhi unpredictably jump karte rehte hain.

Random vibration aur sine vibration ek hi failure mechanism test karte hain

False — sine ek time mein ek frequency excite karta hai resonances dhundne ke liye, jabki random saari frequencies ek saath excite karta hai, real launch ke zyaada kareeb aur cumulative fatigue ke liye better.

Resonance par, response PSD input PSD ke equal hota hai

False — response PSD

∣ T (f) ∣^{2}

times input hota hai, aur

f_{n}

ke paas yeh factor roughly

Q^{2}

hota hai, isliye ek flat input ek sharp spike ban jaata hai (Figure 2).

Damping

ζ

badhane se response RMS hamesha kam hota hai

Resonance ke paas mostly true hai (peak ko flat karta hai), lekin ye resonance ke upar transmissibility ko badhata hai kyunki damper force transmit karta hai, isliye net effect depend karta hai ki input energy kahan hai.

Ek one-sided PSD aur ek two-sided PSD sahi se use karne par same RMS dete hain

True — two-sided PSD same energy ko

\pm f

par half height ke saath spread karta hai, isliye saare

f

par iska integral one-sided integral ke barabar hota hai

f \geq 0

par; conventions mix karna (one-sided height, two-sided limits) hi ek spurious factor of 2 introduce karta hai.

Error dhundho

a_{RMS} = \int G_{xx} (f) df

" — kya missing hai?

Square root — woh integral

g^{2}

mein mean-square hai;

a_{RMS} = \int G_{xx} df

"Maine ek one-sided PSD ko

- f_{2}

f_{2}

tak integrate kiya."

Wrong limits — ek one-sided PSD sirf

f \geq 0

ke liye defined hai, isliye

0

f_{2}

tak integrate karo; negatives tak extend karna double-count karta hai aur RMS ko

2

se inflate karta hai.

"Resonance factor

1/ [(1 - r^{2})^{2} + (2 ζ r)^{2}]

transmissibility hai."

Yeh sirf denominator (dynamic amplification) shape hai; true base-acceleration transmissibility ise numerator

1 + (2 ζ r)^{2}

se multiply karta hai, jo low frequency par

∣ T ∣ \to 1

force karta hai (jahan

r \to 0

"Ek piecewise PSD ke liye, har band ka RMS sum karo:

a = 0.8 + 16 + 15

Galat — pehle mean-squares (areas) add karo, phir ek square root lo:

0.8 + 16 + 15

. RMS values linearly add nahi hote.

"Peak transmissibility magnitude

∣ T ∣^{2} = Q

hai."

Squared magnitude

\approx Q^{2}

hai (yahan 100); transmissibility magnitude khud

\approx Q = 10

hai. "

∣ T ∣ \approx Q

" approximation ko "

∣ T ∣^{2} \approx Q^{2} + 1

" statement se confuse mat karo.

"Kyunki

Q = 1/ (2 ζ)

aur

ζ = 0.05

, isliye

Q = 0.1

Sign/reciprocal slip —

Q = 1/ (2 \times 0.05) = 10

, na ki

0.1

; chhoti damping matlab bada

Q

T (f)

ek real number hai, isliye main response contributions arithmetically add kar sakta hoon."

Galat —

T (f)

complex hai, jisme ek phase hai jo resonance par

- 9 0^{\circ}

se swing karta hai; sirf iska squared magnitude

∣ T ∣^{2}

PSD relation mein enter karta hai, isliye tum PSDs multiply karte ho, signed transmissibilities kabhi add nahi karte.

"Relative-displacement filter

∣ H_{z} ∣^{2}

aur transmissibility

∣ T ∣^{2}

same value par peak karte hain."

Ye lagbhag same frequency par peak karte hain lekin same value par nahi;

∣ H_{z} ∣^{2}

displacement ke liye

1/ ω_{n}^{4}

scaling carry karta hai, jabki

∣ T ∣^{2}

dimensionless hai — bilkul alag quantities hain.

"Structure mein stress absolute acceleration

\overset{y}{¨}

se set hota hai."

Stress relative displacement

z_{rel} = y - x

se drive hota hai (mass apne mounts ko kitna bend karta hai, Figure 3), jo

∣ H_{z} ∣^{2}

use karta hai, absolute-acceleration transmissibility nahi.

Why questions

PSD definition mein

∣ X (f, T_{rec}) ∣^{2}

ko record length

T_{rec}

se kyun divide karte hain?

Kyunki ek longer recording zyaada cycles accumulate karta hai, isliye

∣ X (f, T_{rec}) ∣^{2}

T_{rec}

ke saath badhta hai; divide karne se ek rate (power per unit time) milti hai jo ek steady value par converge karta hai.

Hum ek measurement ki jagah ensemble average

E [\cdot]

(bahut saari recordings ka average) kyun use karte hain?

Ek single random realization noisy aur non-repeatable hoti hai; ensemble par average karne se stable statistical content extract hota hai — aur ek ergodic process ke liye ek record ka lamba time average usi ensemble average par converge karta hai, isliye ek lamba test bahut saare tests ki jagah le sakta hai.

Peak acceleration report karne ki jagah RMS kyun lete hain?

Ek random signal ki koi fixed peak nahi hoti — extreme values rarely aur unpredictably occur karte hain, isliye RMS design ke liye ek repeatable, physically meaningful "typical" magnitude deta hai.

Low frequency par absolute-acceleration transmissibility 1 kyun approach karta hai (0 nahi)?

f_{n}

se bahut neeche (jahan

r \to 0

) mass rigidly base ke saath ride karta hai, isliye output equals input (Figure 2 ka left edge); numerator mein

+ 1

exactly is rigid-body following ko encode karta hai.

Ek flat input PSD bhi peaked response PSD kyun produce karta hai?

Kyunki structure ek resonant filter ki tarah act karta hai — transmissibility

∣ T (f) ∣^{2}

f_{n}

ke paas ki band ko amplify karta hai (Figure 2), isliye uniform input energy wahan concentrated ho jaati hai.

Launch spectrum mein 100–500 Hz "bump" wo region kyun hai jiske baare mein engineers chintit rehte hain?

Isme sabse zyaada energy hoti hai (sabse bada area), aur structural resonances often wahan hoti hain, isliye amplification right wahan land karti hai jahan input already strong hai.

Fourier transform apne building blocks ke roop mein complex exponentials

e^{- i 2 π f t}

kyun use karta hai?

Ye pure single-frequency tones hain jo ek complete basis form karte hain, isliye kisi bhi signal

x (t)

ko inke weighted sum ke roop mein likha ja sakta hai — yahi hume "frequency

f

par energy" ki baat karne deta hai.

SDOF formulas mein hamesha akele

f

ki jagah ratio

r = f / f_{n}

kyun appear karta hai?

Kyunki physics sirf yahi care karta hai ki tum natural frequency se upar ya neeche drive kar rahe ho, na ki absolute Hz;

f_{n}

se normalize karne se response shape universal aur dimensionless ban jaata hai.

Edge cases

Ek PSD ka RMS kya hoga jo har jagah zero hai?

Zero — kisi bhi band mein koi energy nahi matlab integral zero hai aur

a_{RMS} = 0

; ek silent input silent response deta hai.

Damping

ζ \to 0

hone par peak transmissibility ka kya hota hai?

Q = 1/ (2 ζ) \to \infty

, isliye resonance par

∣ T ∣^{2} \approx 1 + 1/ (4 ζ^{2})

blow up ho jaata hai.

+ 1

sirf low-frequency (rigid-body) asymptote hai jo saath ride kar raha hai; resonant growth

1/ (4 ζ^{2}) \approx Q^{2}

term se aati hai, isliye chhote

ζ

ke liye exact peak

Q^{2} + 1

hai, literally

Q^{2}

nahi. Real damping ise hamesha finite rakhta hai.

Absolute-acceleration transmissibility ka high-frequency limit kya hai (

f ≫ f_{n}

, yaani

r \to \infty

∣ T ∣^{2} \to (2 ζ r)^{2} / r^{4} = (2 ζ)^{2} / r^{2} \to 0

, isliye magnitude mein

1/ r

ki tarah roll off karta hai (Figure 2 ki right tail). Critically roll-off damper term se set hoti hai, isliye zyaada damping ka matlab resonance ke upar slower fall-off hai — mass base se increasingly isolated hota hai, lekin kabhi perfectly nahi, kyunki damper force pass karta rehta hai.

Agar input PSD mein exactly

f_{n}

par zero content hai, kya phir bhi amplification hoti hai?

Jahan koi input energy amplify hone ke liye nahi hai wahan koi response energy appear nahi ho sakti; transmissibility input ko multiply karta hai, aur kuch bhi times zero us frequency par zero hai.

f = 0

par transmissibility kya hai (static offset,

r = 0

Exactly 1 — mass base ke saath rigid body ki tarah move karta hai, koi relative motion nahi aur koi dynamic amplification nahi (Figure 2 ka left edge).

Kya do bahut alag PSD shapes same

a_{RMS}

de sakte hain?

Haan — RMS sirf total area measure karta hai, isliye ek tall narrow PSD aur ek low wide PSD RMS mein match kar sakte hain jabki bahut alag fatigue cause karte hain kyunki damaging frequencies differ karti hain.

Agar frequency band

[f_{1}, f_{2}]

resonance

f_{n}

ko exclude kar de?

Tum amplified band ko poori tarah miss karte ho, isliye computed RMS true structural response ko understate karta hai — hamesha ek range par integrate karo jo resonances ko contain kare.

Agar pehle se bahut upar ek second band same PSD level ke saath add kiya jaye toh kya RMS bahut zyaada badalta hai?

Agar woh band wide hai toh substantial area add karta hai aur RMS dominate kar sakta hai, kyunki zyaada Hz par ek flat level zyaada mean-square contribute karta hai chahe level "chhota" lage.