Random vibration — PSD, RMS acceleration: Worked examples

3.6.11 · D3 · Physics › Spacecraft Structures & Systems Engineering › Random vibration — PSD, RMS acceleration

Examples se pehle, teen chhote building blocks jo hum har jagah reuse karenge.

Neeche figure tumhari master key hai: yeh teen shapes dikhata hai jo ek PSD segment le sakta hai, aur har ek ke liye area recipe. Har example se pehle isko dekho — har worked problem inhi shapes mein se ek hai.

Figure — Random vibration — PSD, RMS acceleration

Teen coloured regions ko apni aankh se trace karo. Magenta rectangle (flat PSD) ka area = height $\times$ width hai — yeh Examples 1, 2, 4, 7-region-3, 8 hain. Violet ramp (log–log paper par ek straight line) ek curved area mein bulge karta hai jiske liye power-law integral chahiye — yeh Examples 3 aur 7-region-2 hain. Orange spike (resonance) ki almost saari area ek narrow band mein packed hoti hai — yeh Examples 5 aur 6 hain. Teen shapes → teen area recipes → poora topic.

The scenario matrix

Cell	Case class	Kya mushkil banata hai	Example
A	Flat PSD, single band	baseline area = $G Δ f$	Ex 1
B	Piecewise-flat (bands stacked)	sum of rectangles	Ex 2
C	Sloping segment (ramp, dB/oct)	log–log ramp ko special integral chahiye	Ex 3
D	Degenerate: zero-width band ( $f_{1} = f_{2}$ )	area zero ho jaati hai	Ex 4
E	Notched / dip spectrum	add mat karo, subtract karo	Ex 4
F	Resonance amplification (transmissibility)	flat input $\to$ peaked output	Ex 5
G	Limiting case: $ζ \to 0$ (undamped)	RMS $\to \infty$ , isliye testing mein damping zaroori hai	Ex 6
H	Real-world word problem (launch qual)	$f_{1}, f_{2}$ chuno, $g$ mein convert karo	Ex 7
I	Exam twist: work backwards (given RMS, find level)	area formula ko invert karo	Ex 8

Nau cells, aath examples (kuch do cells hit karte hain). Har ek labelled hai.

Example 1 — Cell A: Flat PSD, single band

Forecast: Aage padhne se pehle answer guess karo. Rectangle area, phir square-root. Kya yeh $2 g$ ke paas hai ya $20 g$ ke?

Area ko rectangle ki tarah likho. $\overline{a^{2}} = G \times Δ f = G \times (f_{2} - f_{1}) = 0.04 \times (2000 - 20)$ Yeh step kyun? Ek flat PSD constant height hoti hai, toh "curve ke neeche area" sirf height $\times$ width hai, jahan width bandwidth $Δ f = f_{2} - f_{1}$ hai — koi integral nahi chahiye.
Area compute karo. $\overline{a^{2}} = 0.04 \times 1980 = 79.2 g^{2}$ Yeh step kyun? Yeh mean-square hai (units $g^{2}$ mein). Yeh abhi answer nahi hai — units "typical acceleration" ke liye galat hain.
RMS pane ke liye square-root lo. $a_{RMS} = 79.2 \approx 8.90 g$ Yeh step kyun? RMS units ko $g$ par restore karta hai, ek aisa number deta hai jise tum ek steady acceleration se compare kar sako.

Verify: Units: $(g^{2} / Hz) \times Hz = g^{2}$ ; $g^{2} = g$ . ✓ Sanity: bahut wide band par $8.9 g$ real launch bracketry ke typical hai.

Example 2 — Cell B: Piecewise-flat spectrum

Forecast: Teen rectangles. Kaun zyada area hold karega — wide low bands, ya tall middle band?

Har band ke liye ek rectangle. $A_{1} = 0.01 (100 - 20) = 0.8, A_{2} = 0.04 (500 - 100) = 16.0, A_{3} = 0.01 (2000 - 500) = 15.0$ (sab $g^{2}$ mein) — Yeh step kyun? Areas simply add ho jaate hain — mean-square disjoint frequency bands mein additive hota hai kyunki alag bands mein energy interfere nahi karti.
Areas ka sum karo. $\overline{a^{2}} = 0.8 + 16.0 + 15.0 = 31.8 g^{2}$ Yeh step kyun? Total mean-square = pure (stepped) curve ke neeche total area.
Square-root lo. $a_{RMS} = 31.8 \approx 5.64 g$ Yeh step kyun? RMS mean-square ka square root hai; yeh units ko $g$ par restore karta hai taaki answer ek typical acceleration magnitude ki tarah read ho, na ki power ki tarah.

Verify: Middle band akela ( $16 g^{2}$ ) total energy ka aadha hai bawajood narrow hone ke — kyunki iska height doosron se 4× hai. Isliye 100–500 Hz "bump" Fatigue Analysis ke liye important hai. ✓

Example 3 — Cell C: Sloping segment (dB/octave ramp)

Forecast: Log–log paper par ek ramp linear paper par ek curve hai. Kya answer plain trapezoid $\frac{1}{2} (G_{1} + G_{2}) Δ f$ se bada hoga ya chhota?

Neeche figure mein violet region exactly yahi shape hai — ek log–log straight line jo ordinary linear axes par dekhi jaaye toh naive straight chord (dashed magenta) se upar bulge karti hai. Hum sahi bulged area compute karenge.

Power law ka exponent $n$ dhundo. Log–log axes par ek straight line matlab $G (f) = G_{1} (f / f_{1})^{n}$ . Do endpoints padhte hue, exponent hai $n = \frac{l n ( G _{2} / G _{1} )}{l n ( f _{2} / f _{1} )} = \frac{l n ( 0.08/0.01 )}{l n ( 400/100 )} = \frac{l n 8}{l n 4} = \frac{2.0794}{1.3863} = 1.5$ Yeh step kyun? Log–log paper par slope $\frac{Δ ( lo g G )}{Δ ( lo g f )}$ hai, jo exactly exponent $n$ hai. Hume $n$ chahiye sahi area integral choose karne ke liye. (Yeh $lo g_{2} 8 = 3$ nahi hai; woh exponent tabhi hoga jab frequency ratio bhi single doubling ho, jo yahan nahi hai — frequency 2 octaves span karti hai.)
Log–log ramp area formula use karo. $A = \int_{f_{1}}^{f_{2}} G_{1} (\frac{f}{f _{1}})^{n} df = \frac{G _{1} f _{1}}{n + 1} [(\frac{f _{2}}{f _{1}})^{n + 1} - 1]$ Yeh step kyun? Ek power law $f^{n}$ power rule se integrate hota hai (valid kyunki $n = 1.5 \neq = - 1$ ); $G_{1} f_{1} / (n + 1)$ factor out karne se tidy rehta hai.
$n = 1.5$ , $G_{1} = 0.01$ , $f_{1} = 100$ , $f_{2} / f_{1} = 4$ daalo. $A = \frac{0.01 \times 100}{2.5} [4^{2.5} - 1] = \frac{1}{2.5} \times (32 - 1) = \frac{31}{2.5} = 12.4 g^{2}$ Yeh step kyun? $4^{2.5} = 4^{2} \cdot 4^{0.5} = 16 \times 2 = 32$ ; minus 1 se 31 milta hai; prefactor $G_{1} f_{1} / (n + 1) = 1/2.5 = 0.4$ hai.
(Optional) Sirf is segment ka RMS: $12.4 \approx 3.52 g$ .

Verify: Naive linear trapezoid deta hai $\frac{1}{2} (0.01 + 0.08) (300) = 13.5 g^{2}$ . Sahi power-law area $12.4 g^{2}$ close but smaller hai yahan (gentle $n = 1.5$ ramp ke liye curve chord ke neeche thoda hi bow karta hai). ✓ Units: $(g^{2} / Hz) \cdot Hz = g^{2}$ . ✓

Example 4 — Cells D & E: Zero-width band aur ek notch

Forecast: (a) ek trap hai — zero width wale rectangle ki area kya hai? (b) Kya notch zyada fark karta hai?

Part (a): degenerate band. $A = G \times (f_{2} - f_{1}) = G \times 0 = 0 g^{2}$ Yeh step kyun? Ek single frequency ki bandwidth zero hoti hai ( $Δ f = 0$ ). Ek true pure tone ek Dirac spike hoga (infinite height, finite area) — woh Sine Vibration Testing se belong karta hai, random vibration se nahi. Random PSD ko hamesha energy carry karne ke liye finite width chahiye.
Part (b): agar notch na hota toh full area. $A_{full} = 0.05 \times (2000 - 20) = 0.05 \times 1980 = 99.0 g^{2}$ Yeh step kyun? Un-notched rectangle se shuru karo, phir dip ke liye correct karo.
Jo notch remove karta hai use subtract karo. Notch $0.05$ ko $0.005$ se replace karta hai $40$ Hz window par, yani $(0.05 - 0.005) \times 40$ remove karta hai. $Δ A = (0.05 - 0.005) \times (220 - 180) = 0.045 \times 40 = 1.8 g^{2}$ Yeh step kyun? Ek dip area subtract karta hai. Missing rectangle compute karo aur le lo — use re-add mat karo.
Net area aur RMS. $\overline{a^{2}} = 99.0 - 1.8 = 97.2 g^{2}, a_{RMS} = 97.2 \approx 9.86 g$ Yeh step kyun? Jab corrected total area (mean-square) pata ho, uska square root lene se RMS $g$ mein milta hai — woh single final operation jo $g^{2}$ power ko physical acceleration mein badalta hai.

Verify: Notch ke bina, $a_{RMS} = 99 = 9.95 g$ . Notch zyada help nahi karta (9.95 → 9.86 g) kyunki yeh sirf 40 Hz wide hai 1980 mein se — ek accha reminder ki narrow Transmissibility dip broadband RMS ke liye khaas nahi karta, chahe woh kisi specific resonant part ko bachata ho. ✓

Example 5 — Cell F: Resonance amplification (transmissibility)

Is example se pehle, notation ke do pieces earn karne hain.

Formula Transmissibility

$∣ T (f) ∣^{2}$ — poori frequency-dependent form Natural frequency $f_{n}$ aur damping ratio $ζ$ wale ek single-degree-of-freedom (SDOF) oscillator ke liye, jo apne base se driven hai, frequency ratio $r \equiv f / f_{n}$ likho. Frequency Response Function (FRF) jo input base-acceleration PSD ko output absolute-acceleration PSD par map karta hai woh transmissibility hai: $∣ T (f) ∣^{2} = \frac{1 + ( 2 ζ r ) ^{2}}{[ 1 - r ^{2} ] ^{2} + ( 2 ζ r ) ^{2}}, r = \frac{f}{f _{n}}$ Teen regimes se padhte hain:

Low frequency $r \to 0$ : numerator $\to 1$ , denominator $\to 1$ , toh $∣ T ∣^{2} \to 1$ — mass sirf base ke saath ride karta hai (koi amplification nahi).
Resonance par $r = 1$ : $[1 - r^{2}]^{2}$ term vanish ho jaata hai, chhod jaata hai $∣ T ∣^{2} = \frac{1 + ( 2 ζ ) ^{2}}{( 2 ζ ) ^{2}} = 1 + \frac{1}{4 ζ ^{2}}$ — ek tall peak jo $ζ$ se controlled hai.
High frequency $r \to \infty$ : numerator $\sim (2 ζ r)^{2}$ , denominator $\sim r^{4}$ , toh $∣ T ∣^{2} \to (2 ζ / r)^{2} \to 0$ — mass isolated hai, uski motion khatam ho jaati hai.

Response PSD tab, band by band, $G_{\overset{y}{¨} \overset{y}{¨}} (f) = ∣ T (f) ∣^{2} G_{\overset{x}{¨} \overset{x}{¨}} (f)$ hai.

Forecast: Input flat aur calm hai. $f = f_{n}$ par output spike kitni tall ho jaati hai?

Figure mein calm flat magenta input line aur violet response dikhta hai jo $f_{n}$ par 100-fold spike karta hai — orange dotted line natural frequency mark karti hai. Almost saari response energy us narrow violet peak ke neeche hai.

Resonance par transmissibility. $r = 1$ set karo upar wali full form mein; $[1 - r^{2}]^{2}$ term vanish ho jaata hai: $∣ T (f_{n}) ∣^{2} = \frac{1 + ( 2 ζ ) ^{2}}{( 2 ζ ) ^{2}} = 1 + \frac{1}{4 ζ ^{2}} = 1 + \frac{1}{4 ( 0.05 ) ^{2}} = 101$ Yeh step kyun? FRF denominator resonance par apna bada term kho deta hai, sirf chhota $(2 ζ)^{2}$ bachta hai — toh ratio blow up karta hai. Yeh resonance peak hai.
Peak response PSD. $G_{\overset{y}{¨} \overset{y}{¨}} (f_{n}) = ∣ T (f_{n}) ∣^{2} G_{\overset{x}{¨} \overset{x}{¨}} = 101 \times 0.02 = 2.02 g^{2} / Hz$ Yeh step kyun? Response PSD = transmissibility $^{2}$ × input PSD, band by band. Output $f_{n}$ par flat input se 101× spike karta hai.
Miles' equation se response RMS (peak ke neeche area). Ek mode se flat input ke liye classic Modal Analysis result hai $a_{RMS,resp} = \frac{π}{2} f_{n} Q G_{\overset{x}{¨} \overset{x}{¨}}, Q = \frac{1}{2 ζ} = 10$ $= \frac{π}{2} \times 200 \times 10 \times 0.02 = 62.83 \approx 7.93 g$ Yeh step kyun? Response area narrow resonant peak se dominate hoti hai; Miles' equation us peak ka closed-form area hai (bandwidth $\sim \frac{π}{2} f_{n} / Q$ times peak height). Yeh pure $∣ T ∣^{2}$ curve integrate karne se bachata hai.

Verify: Peak PSD $2.02 g^{2} / Hz$ vs input $0.02$ → exactly 101× ✓. Miles' $g$ : band par input RMS tiny hoga, lekin resonance use $\approx 7.9 g$ tak pump karta hai. Structural Damping $ζ$ peak (101) aur RMS ( $\propto Q$ ) dono control karta hai. ✓

Example 6 — Cell G: Limiting case $ζ \to 0$ (undamped)

Forecast: Koi friction nahi, toh kya answer finite rehta hai ya infinity par chala jaata hai?

Peak transmissibility limit. $∣ T (f_{n}) ∣^{2} = 1 + \frac{1}{4 ζ ^{2}} ζ \to 0 \infty$ Yeh step kyun? Zero damping matlab resonance par kuch bhi energy remove nahi karta, toh amplification $ζ \to 0$ hone par bina bound ke badhti hai.
Response RMS limit. $a_{RMS,resp} = \frac{π}{2} f_{n} Q G_{\overset{x}{¨} \overset{x}{¨}}, Q = \frac{1}{2 ζ} ζ \to 0 \infty$ Yeh step kyun? Kyunki quality factor $Q = 1/ (2 ζ)$ Miles square root ke andar appear karta hai, aur $Q \to \infty$ jab $ζ \to 0$ , RMS $\propto 1/ ζ$ bhi diverge karta hai. Physically iska matlab hai: koi energy loss nahi, toh har resonant cycle pichle par add hota hai aur response bina limit ke build hota hai — undamped model predict karta hai infinite RMS, jo koi real structure kabhi reach nahi kar sakta.
Physical resolution. Yeh step kyun? Real structures mein hamesha kuch Structural Damping hota hai ( $ζ$ typically $0.01$ – $0.05$ ), aur nonlinear effects (joint friction, material hysteresis, yielding) response ko cap karte hain before it runs away. Isliye divergent limit ek warning hai, prediction nahi: yeh batata hai damping exactly wahi hai jo random-vibration response ko finite rakhti hai, isliye har qualification test realistic, non-zero $ζ$ ke saath run hota hai aur kabhi undamped mode assume nahi karta.

Verify: Ek chhota $ζ = 0.01$ daalo: $∣ T ∣^{2} = 1 + 1/ (4 \cdot 1 0^{- 4}) = 2501$ ; $Q = 50$ ; RMS $= \frac{π}{2} \cdot 200 \cdot 50 \cdot 0.02 = 314.2 \approx 17.7 g$ . Jab $ζ$ 5× shrink hota hai (0.05→0.01), RMS $5 \approx 2.24$ × badhta hai (7.93→17.7). ✓ Confirm karta hai $\propto 1/ ζ$ aur trend infinity ki taraf ja raha hai.

Example 7 — Cell H: Real-world launch qualification

Forecast: Teen shapes — rectangle, ramp, rectangle. Predict karo kaun zyada contribute karta hai.

Region 1 (flat, 20–80 Hz). $A_{1} = 0.026 \times (80 - 20) = 0.026 \times 60 = 1.56 g^{2}$ Yeh step kyun? Flat band ke liye rectangle rule.
Region 2 (log–log ramp, 80–350 Hz). Exactly Example 3 ki tarah do endpoints se exponent lo: $n = \frac{l n ( 0.11375/0.026 )}{l n ( 350/80 )} = \frac{l n 4.375}{l n 4.375} = 1$ toh $G (f) = 0.026 (f /80)^{1}$ . Ramp area formula apply karo $n = 1$ ke saath: $A_{2} = \frac{G _{1} f _{1}}{n + 1} [(f_{2} / f_{1})^{n + 1} - 1] = \frac{0.026 \times 80}{2} [(4.375)^{2} - 1]$ $= 1.04 \times (19.1406 - 1) = 1.04 \times 18.1406 = 18.866 g^{2}$ Yeh step kyun? Ramp ek power law $f^{n}$ hai; ise $(n + 1)$ rule se integrate karo, trapezoid se nahi. Yahan numbers clean $n = 1$ dete hain (level frequency ke same ratio mein rise karta hai).
Region 3 (flat, 350–2000 Hz). $A_{3} = 0.16 \times (2000 - 350) = 0.16 \times 1650 = 264.0 g^{2}$ Yeh step kyun? Rectangle rule phir; note karo yeh tall, wide band dominate karta hai.
Total aur RMS. $\overline{a^{2}} = 1.56 + 18.866 + 264.0 = 284.43 g^{2}, a_{RMS} = 284.43 \approx 16.9 g$ Yeh step kyun? Teen sub-areas ko total mean-square mein sum karo, phir ek baar square-root lo overall RMS $g$ mein paane ke liye.

Verify: Region 3 mein $264/284 \approx 93%$ energy hai — samajh mein aata hai, yeh sabse high level hai sabse wide span par. Ek realistic small-sat qual RMS $\sim 17 g$ hai. ✓ Root se pehle sab units $g^{2}$ hain. ✓ Yeh directly Fatigue Analysis aur Acoustic Loading margins mein jaata hai.

Example 8 — Cell I: Exam twist (ulta kaam)

Forecast: Hume answer (RMS) pata hai aur input (level) recover karna hai. Area formula invert karo.

Forward relation likho. $a_{RMS}^{2} = G Δ f = G (f_{2} - f_{1}) \Rightarrow 1 0^{2} = G \times (2000 - 20)$ Yeh step kyun? Flat spectrum ke liye, mean-square = level × bandwidth. Hum sirf unknown $G$ solve karte hain.
$G$ solve karo. $G = \frac{100}{1980} = 0.05051 g^{2} / Hz$ Yeh step kyun? Required mean-square ( $100 g^{2}$ ) ko bandwidth ( $1980$ Hz) se divide karne par per-Hz level milta hai.
Test spec ke liye round karo. $G \approx 0.0505 g^{2} / Hz$ Yeh step kyun? Test consoles ko PSD level chahiye, RMS nahi; yeh woh number hai jo tum type karte ho.

Verify: Forward-check: $0.05051 \times 1980 = 100.0 = 10.0 g$ ✓. Yeh inversion exactly wahi hai jo ek Sine Vibration Testing engineer karta hai jab "10 g RMS" requirement ko shaker command mein translate karta hai. ✓

Recall Self-test (guess karne ke baad reveal karo)

$Δ f$ kya hai? ::: Bandwidth $f_{2} - f_{1}$ , Hz mein band ki width jis par tum integrate karte ho. $\overline{a^{2}}$ mein overline ka matlab kya hai? ::: Squared acceleration ka average (ensemble / time mean) — mean-square, $g^{2}$ mein. Log–log par ramp area kaun sa exponent use karta hai? ::: $n = ln (G_{2} / G_{1}) / ln (f_{2} / f_{1})$ , phir $(n + 1)$ power-law area formula — linear trapezoid NAHI. Zero-width band kitna mean-square contribute karta hai? ::: Zero — random PSD ko energy carry karne ke liye finite bandwidth chahiye. Ek notch (dip) area ke saath kya karta hai? ::: Removed rectangle subtract karta hai; use kabhi re-add mat karo. Sirf positive frequencies par integrate kyun karte hain? ::: One-sided PSD symmetric negative-frequency half ko $f \geq 0$ par fold karta hai; negative $f$ test spec mein meaningless hai. Resonance $r = 1$ par full transmissibility form kya hai? ::: $∣ T ∣^{2} = 1 + 1/ (4 ζ^{2})$ ; low- $f$ par yeh $\to 1$ , high- $f$ par yeh $\to 0$ . Response RMS damping ke saath kaise scale karta hai? ::: $\propto 1/ ζ$ (via $Q = 1/2 ζ$ ), $ζ \to 0$ par diverge karta hai. Ek flat band par target RMS diya ho toh level kaise dhundho? ::: $G = a_{RMS}^{2} /Δ f$ .