Acoustic loads — SPL, octave band analysis: Worked examples

3.6.12 · D3 · Physics › Spacecraft Structures & Systems Engineering › Acoustic loads — SPL, octave band analysis

Ye page parent topic ka drill hall hai. Parent ne formulas build kiye; yahan hum har case ko cover karte hain jo wo formulas produce kar sakte hain — number ki har sign, har degenerate input, har limiting value, plus ek real-world word problem aur ek exam trap.

Shuru karne se pehle, har symbol ko pin down karte hain jo hum use karte hain, taaki kuch unexplained na rahe:

Recall Jo formulas hum use karte hain

SPL (forward): $L = 20 lo g_{10} (\frac{p _{rms}}{p _{ref}})$ , jahan $p_{ref} = 20 \times 1 0^{- 6}$ Pa. SPL (inverse): $p_{rms} = p_{ref} \times 1 0^{L /20}$ . Bands/sources combine karna (energy sum): $L_{tot} = 10 lo g_{10} (\sum_{i} 1 0^{L_{i} /10})$ , kyunki har term $1 0^{L_{i} /10} = I_{i} / I_{ref}$ us source ki ground-level intensity hai. Yahan $lo g_{10} (x)$ poochta hai "10 ki kaunsi power $x$ deti hai?"; $1 0^{a}$ usse undo karta hai. dB ek ratio on a log ruler hai, kabhi bhi absolute pressure nahi.

Scenario matrix

Is topic ka har problem in cells mein se kisi ek mein aata hai. Neeche ke examples mein us cell ka label diya gaya hai jo wo cover karte hain.

Cell	Kya tricky hai	Example
A. Forward dB → pressure → force	inverse formula lagao	Ex 1
B. Backward pressure → dB	log apply karo	Ex 2
C. Equal sources add	sum ka log, dono equal	Ex 3
D. Unequal sources add	ek term dominate karta hai	Ex 4
E. Degenerate: ek source ZERO / silence hai	$lo g_{10} (0) = - \infty$	Ex 5
F. Limiting: bahut saare equal bands	$+ 10 lo g_{10} N$ growth	Ex 6
G. Real-world word problem	sahi tool chunna	Ex 7
H. Exam twist: subtraction / negative dB	source "remove" karna, negative $Δ$	Ex 8

Figure — Acoustic loads — SPL, octave band analysis

Figure 1 — dB levels as heights on a log ladder (left) vs. unki intensities as ground-level bars jo sum ho sakti hain (right).

Example 1 — Cell A: dB → pressure → force

Worked example Fairing SPL se panel force

Diya hai: Ek fairing interior 148 dB measure karta hai. Area $A = 1.5 m^{2}$ ka ek flat antenna wave ka samna karta hai. Dhundho: RMS pressure, aur antenna ko feel hone wali RMS force.

Forecast: Andaza lagao — kya pressure kuch Pa hogi, kuch sau, ya hazaron? Apna guess likho.

Step 1. SPL formula ko invert karo: $p_{rms} = p_{ref} \times 1 0^{L /20}$ . Yeh step kyun? Humein ek dB number diya gaya hai aur physical pressure chahiye, isliye humein log undo karna hai — exactly yahi $1 0^{(\cdot)}$ karta hai. $p_{rms} = 20 \times 1 0^{- 6} \times 1 0^{148/20} = 20 \times 1 0^{- 6} \times 1 0^{7.4}$

Step 2. $1 0^{7.4} = 1 0^{7} \times 1 0^{0.4} = 1 0^{7} \times 2.512 = 2.512 \times 1 0^{7}$ evaluate karo. Yeh step kyun? Exponent ko ek whole part aur ek decimal part mein split karne se tum magnitude ( $1 0^{7}$ ) aur mantissa ( $1 0^{0.4} \approx 2.51$ ) alag-alag aankhon se padh sakte ho. $p_{rms} = 20 \times 1 0^{- 6} \times 2.512 \times 1 0^{7} = 502.4 Pa$

Step 3. Force $= p_{rms} \times A = 502.4 \times 1.5 = 753.6 N$ . Yeh step kyun? Pressure force per area hai; exposed area se multiply karne par actual push milti hai. Yahi oscillating force structure ko fatigue karti hai.

Verify: Units: $Pa \times m^{2} = \frac{N}{m ^{2}} \times m^{2} = N$ ✓. Sanity: 148 dB, parent note ke 150 dB / 632 Pa case se 2 dB neeche hai; 2 dB kam ≈ factor $1 0^{- 2/20} = 0.794$ , aur $632 \times 0.794 \approx 502$ Pa ✓.

Example 2 — Cell B: pressure → dB

Example 3 — Cell C: do EQUAL uncorrelated sources

Worked example Twin engines, equal level

Diya hai: Do uncorrelated engines dono 144 dB produce karte hain. Dhundho: Combined SPL.

Forecast: Kya answer 288, 147, ya 144 dB hai? Padhne se pehle guess karo.

Step 1. Uncorrelated ⇒ intensities $I$ add hoti hain, dB nahi. $N$ equal sources of level $L$ ke liye: $L_{tot} = L + 10 lo g_{10} N$ . Yeh step kyun? Random phases cross-term $⟨ 2 p_{1} p_{2} ⟩ = 0$ ko khatam kar dete hain, isliye $⟨(p_{1} + p_{2})^{2} ⟩ = ⟨ p_{1}^{2} ⟩ + ⟨ p_{2}^{2} ⟩$ — energies simply stack ho jaati hain.

Step 2. $N = 2$ ke saath: $10 lo g_{10} 2 = 3.01$ dB. $L_{tot} = 144 + 3.01 = 147.0 dB$ Jump itna chhota kyun? Energy double karna sirf +3 dB hai kyunki log ladder sab kuch compress kar deta hai.

Verify: Ground pe sum karo: $2 \times 1 0^{14.4} = 2 \times 2.512 \times 1 0^{14} = 5.024 \times 1 0^{14}$ ; $10 lo g_{10} (5.024 \times 1 0^{14}) = 10 (14.701) = 147.0$ dB ✓.

Example 4 — Cell D: do UNEQUAL sources

Worked example Loud source plus ek quiet wala

Diya hai: Source 1 = 150 dB, Source 2 = 138 dB, uncorrelated. Dhundho: Combined SPL.

Forecast: Kya 138 dB wala source bahut, thoda, ya kuch nahi add karega? Guess karo.

Step 1. Energy sum: $L_{tot} = 10 lo g_{10} (1 0^{150/10} + 1 0^{138/10})$ . Yeh step kyun? Sirf energies (ground values $I / I_{ref}$ ) add ho sakti hain — neeche utro, add karo, upar chado.

Step 2. $1 0^{15.0} = 1.000 \times 1 0^{15}$ , $1 0^{13.8} = 6.310 \times 1 0^{13}$ . Sum $= 1.0631 \times 1 0^{15}$ . Dono kyun rakhen? Dominance dekhne ke liye: quiet source, loud wale ki energy ka sirf $6.3%$ hai.

Step 3. $L_{tot} = 10 lo g_{10} (1.0631 \times 1 0^{15}) = 10 (15.0266) = 150.27 dB$ . Yeh step kyun? Ground pe energies sum karne ke baad, hum $10 lo g_{10}$ se log ladder pe wapas chadhte hain taaki answer dB mein report kar sakein — down–add–up cycle ki final leg.

Verify: Difference 12 dB hai. Rule of thumb: 10 dB down wala source ≈ 0.4 dB add karta hai; 12 dB down thoda aur kam add karna chahiye. Humein $+ 0.27$ dB mila ✓.

Figure 2 exactly is rule ko plot karta hai: horizontal axis do sources ke beech gap hai, curve us dB ko dikhata hai jo louder wale ke upar add hota hai. Orange dot ko 12 dB gap pe trace karo $+ 0.27$ dB tak (yeh example) aur green dot ko 0 dB gap pe $+ 3.0$ dB tak (Ex 3). Curve ka steady fall isliye hai ki ek door, quieter source practically koi farak nahi dalta.

Figure 2 — louder source ke upar add hone wala dB, do uncorrelated sources ke beech level gap ka function hai.

Example 5 — Cell E: degenerate input, ek source SILENCE hai

Worked example Dead engine add karna

Diya hai: Engine A = 145 dB. Engine B zero pressure fire karta hai ( $p_{rms} = 0$ , yaani silence). Dhundho: Combined SPL.

Forecast: Kya silence level ko lower, raise, ya unchanged chhod deti hai? Guess karo.

Step 1. Silence ke liye, $p_{rms} = 0$ , isliye uski intensity $I$ bhi $0$ hai, aur uska "level" $20 lo g_{10} (0) = - \infty$ dB hai. Yeh step kyun? $lo g_{10} (0)$ hota hai $- \infty$ — isliye silence $- \infty$ dB hoti hai, kabhi 0 dB nahi. (0 dB ka matlab hai "hearing threshold ke barabar," jo silence NAHI hai.)

Step 2. Energy sum: $I_{A} + 0 = I_{A}$ , isliye $L_{tot} = 145 + 10 lo g_{10} (1) = 145 + 0 = 145 dB$ . Unchanged kyun? Zero energy add karne se ground pe kuch nahi add hota, isliye level untouched rehta hai.

Verify: $10 lo g_{10} (1 0^{14.5} + 0) = 10 \times 14.5 = 145$ dB ✓. Degenerate case confirmed: ek silent source total ko kabhi change nahi karta.

Example 6 — Cell F: limiting behaviour, bahut saare equal bands

Worked example Das equal octave bands

Diya hai: Ek spectrum mein 10 octave bands hain, har ek 130 dB pe. Dhundho: OASPL (Overall SPL), aur general rule.

Forecast: Kya yeh 130 ke paas hai, 140 ke paas, ya 200 dB se kaafi upar? Guess karo.

Step 1. $N$ equal bands ⇒ $L_{OASPL} = L + 10 lo g_{10} N = 130 + 10 lo g_{10} 10$ . Yeh step kyun? Non-overlapping bands independent energy $I$ hold karte hain, isliye unke ground-values sirf sum ho jaate hain — equal sources jaisi bilkul same maths.

Step 2. $10 lo g_{10} 10 = 10$ dB exactly. $L_{OASPL} = 130 + 10 = 140 dB$

Step 3 — the limit. Generally $N$ equal bands $10 lo g_{10} N$ dB add karte hain. Yeh logarithmically badhta hai: $N = 2 \to + 3$ , $N = 10 \to + 10$ , $N = 100 \to + 20$ . $N$ double karna hamesha exactly $+ 3$ dB add karta hai, hamesha ke liye. Yeh fast blow up kyun nahi karta? Kyunki log ladder bade se bade $N$ ko bhi compress kar deta hai — bands stack karke runaway dB total nahi mil sakta.

Verify: $10 lo g_{10} (10 \times 1 0^{13}) = 10 lo g_{10} (1 0^{14}) = 140$ dB ✓.

Figure 3 is ceiling ko draw karta hai: curve $10 lo g_{10} N$ hai $N$ ke against log axis pe, dots ke saath $N = 2, 10, 100$ pe jo $+ 3, + 10, + 20$ dB pe land karte hain. Notice karo ki bands ki sankhya mein har tenfold jump sirf aur $+ 10$ dB khareedta hai — woh visual flattening hi poora safety argument hai.

Figure 3 — added level $10 lo g_{10} N$ sirf logarithmically badhta hai equal bands $N$ ki sankhya ke saath.

Example 7 — Cell G: real-world word problem

Worked example Kya solar array survive karega?

Diya hai: Ek solar array (area $A = 3 m^{2}$ ) ek fairing mein baitha hai jiske octave-band SPLs hain: 125 Hz → 134, 250 Hz → 139, 500 Hz → 141, 1000 Hz → 137 dB. Array ki structural qualification limit ek peak oscillating force of 3000 N hai. (Peak $\approx 2 p_{rms}$ lo, aur effective broadband pressure ke liye OASPL use karo.) Dhundho: Kya array pass karta hai?

Forecast: Margin ke saath pass, just pass, ya fail? Guess karo.

Step 1. OASPL $= 10 lo g_{10} (1 0^{13.4} + 1 0^{13.9} + 1 0^{14.1} + 1 0^{13.7})$ . Yeh step kyun? Pressure mein convert karne se pehle humein bands ko ek single broadband level mein combine karna hoga. Terms: $2.512 \times 1 0^{13}, 7.943 \times 1 0^{13}, 1.259 \times 1 0^{14}, 5.012 \times 1 0^{13}$ . Sum $= 3.0165 \times 1 0^{14}$ . $OASPL = 10 lo g_{10} (3.0165 \times 1 0^{14}) = 144.80 dB$

Step 2. $p_{rms}$ mein convert karo: $p_{rms} = 20 \times 1 0^{- 6} \times 1 0^{144.80/20} = 20 \times 1 0^{- 6} \times 1 0^{7.2398}$ . Yeh step kyun? Force physical world mein rehti hai, isliye dB ko peeche chhodo. $1 0^{7.2398} = 1.738 \times 1 0^{7}$ , giving $p_{rms} = 347.5 Pa$ .

Step 3. Peak pressure $\approx 2 \times 347.5 = 491.4 Pa$ . Peak force $= 491.4 \times 3 = 1474.3 N$ . $2$ kyun? Ek sinusoid ka peak uske rms ka $2$ times hota hai — hum average ke khilaf nahi, peak ke khilaf design karte hain.

Step 4. Compare karo: $1474 N < 3000 N$ ⇒ PASS, margin ke saath $3000/1474 = 2.03 \times$ . Yeh step kyun? Qualification ek pass/fail gate hai: predicted peak load tested limit ke neeche hona chahiye, aur dono ka ratio woh safety margin hai jo engineers hardware sign off karne ke liye report karte hain.

Verify: Units $Pa \cdot m^{2} = N$ ✓. Dominant band ko cross-check karo: 500 Hz band akela 141 dB hai; OASPL of 144.8 dB usse sirf $+ 3.8$ dB upar hai — consistent hai kyunki ek band ~42% energy carry karta hai ✓. Yeh wahi energy-summing logic hai jaisi random vibration PSD integration mein hai.

Example 8 — Cell H: exam twist, source subtraction (negative dB)

Worked example Background noise hatana

Diya hai: Ek test rig shaker chalu hone par 147 dB measure karta hai. Shaker band hone par (sirf background), mic 141 dB read karta hai. Ek examiner poochhta hai: shaker-only SPL kya hai? Dhundho: Sirf shaker ka contribution.

Forecast: Kya shaker-only level 6 dB, 141 dB, ya ~146 dB hai? Guess karo.

Step 1. Energies ground pe subtract hoti hain: $I_{shaker} = I_{tot} - I_{bg}$ . Yeh step kyun? Total energy $I$ = shaker + background (uncorrelated), isliye shaker ko isolate karne ka matlab background energy ko hatana hai — ek subtraction, Ex 4 ka mirror image.

Step 2. $L_{shaker} = 10 lo g_{10} (1 0^{147/10} - 1 0^{141/10})$ . $1 0^{14.7} = 5.012 \times 1 0^{14}$ , $1 0^{14.1} = 1.259 \times 1 0^{14}$ . Difference $= 3.753 \times 1 0^{14}$ . Yeh kyun ho sakta hai? Sirf isliye kyunki $I_{tot} > I_{bg}$ hai; agar background kabhi total se zyada ho, toh measurement impossible hai (tum ek negative ka log loge — exam ki hidden degeneracy check). $L_{shaker} = 10 lo g_{10} (3.753 \times 1 0^{14}) = 145.74 dB$

Step 3. "Negative dB" idea ko interpret karo: correction hai $145.74 - 147 = - 1.26$ dB. Kyunki background sirf 6 dB down hai, yeh reading se non-trivial $1.26$ dB chheen leta hai. Care kyun karo? Agar background total se <10 dB neeche hai, to tum zaroor use subtract karo warna shaker ko overstate karoge.

Verify: Re-add karo: $10 lo g_{10} (1 0^{14.574} + 1 0^{14.1}) = 10 lo g_{10} (3.753 \times 1 0^{14} + 1.259 \times 1 0^{14}) = 10 lo g_{10} (5.012 \times 1 0^{14}) = 147.0$ dB ✓. Humne original total recover kar liya.

Recall check

Recall dB arithmetically add kyun nahi hote?

Kyunki dB ek log ladder pe height hai; sirf ground-level energies ( $1 0^{L /10} = I / I_{ref}$ ) sum ho sakti hain. Ground pe add karo, phir $10 lo g_{10}$ se wapas upar chado aur dB mein return karo. ::: Log-scale ⇒ intensity mein convert karo, intensities sum karo, wapas convert karo.

Equal sources double karne se kitne dB add hote hain?

Exactly

10 lo g_{10} 2 \approx 3.01

dB.

Das equal bands level ko kitna raise karte hain?

10 lo g_{10} 10 = 10

dB.

True silence ka SPL kya hai?

- \infty

dB (0 dB nahi).

Measurement se known background remove karne ke liye tum…

intensities subtract karo, phir wapas dB mein convert karo.

Yeh bhi dekho: 3.6.10 Structural Natural Frequencies and Mode Shapes · 3.6.13 Shock Loads and SRS · 3.6.14 Combined Environmental Testing · 2.5.8 Acoustic Impedance and Transmission.