Exercises — Acoustic loads — SPL, octave band analysis

HUM KYA KARTE HAIN: numbers seedha SPL definition mein daalo. $\text{SPL}=20{\log }_{10}\left(\frac{p_{\text{rms}}}{p_{\text{ref}}}\right)$ KYUN: SPL by definition hearing threshold ke pressure ratio ke log ka 20 guna hai. Ratio measure karta hai "yeh kitne hearing-thresholds loud hai." $\frac{p_{\text{rms}}}{p_{\text{ref}}}=\frac{2\times 10^{-3}}{20\times 10^{-6}}=\frac{2\times 10^{-3}}{2\times 10^{-5}}=100$ $\text{SPL}=20{\log }_{10}(100)=20\times 2=40\mathbf{dB}$ Sanity (map chalo): $100\times$ threshold pressure figure pe do bade steps right hai, aur har step $+20$ dB hai, toh $+40$ dB. ✓

KYA: ek octave band $f_c/\sqrt{2}$ se $f_c\sqrt{2}$ tak span karta hai. KYUN $\sqrt{2}$: taaki band logarithmically centred ho — top/bottom ka ratio exactly $2$ ho (ek octave), aur $f_c$ geometric middle $\sqrt{f_{\text{low}}{f}_{\text{up}}}$ pe baithe. $f_{\text{low}}=\frac{500}{\sqrt{2}}\approx 353.6\text{Hz},f_{\text{up}}=500\sqrt{2}\approx 707.1\text{Hz}$ Check: $707.1/353.6=2.00$ ✓ (ek octave).

KYA: same definition, lekin ab pressure reference se neeche hai. $\frac{p_{\text{rms}}}{p_{\text{ref}}}=\frac{2\times 10^{-6}}{20\times 10^{-6}}=0.1$ $\text{SPL}=20{\log }_{10}(0.1)=20\times (-1)=-20\mathbf{dB}$ KYUN negative: 1 se kam ratio ka negative logarithm hota hai. Kuch toot nahin gaya — minus sign bas padhta hai "hearing ke threshold se 20 dB neeche." Figure mein yeh blue dot ke left wala shaded region hai. Decibels ka koi floor nahin hai: jitna quiet sound, utna zyada negative ($-\infty$ tak perfect silence par).

(a) Log invert karo. Agar $\text{SPL}=20{\log }_{10}(p_{\text{rms}}/p_{\text{ref}})$ hai, toh 20 se divide karke aur $10^{(\cdot )}$ apply karke log undo hota hai: $p_{\text{rms}}=p_{\text{ref}}\times 10^{\text{SPL}/20}=20\times 10^{-6}\times 10^{144/20}$ $=20\times 10^{-6}\times 10^{7.2}$ KYUN $10^{7.2}=10^7\times 10^{0.2}$ split karte hain? Hamare log map pe, exponent $7.2$ pressure axis pe ek position hai: pura number $7$ kehta hai "saat bade $\times 10$ steps right," aur bacha hua $0.2$ aathwe decade ke andar ek fractional step hai. Powers ke exponents add hote hain ($10^{a+b}=10^{a}10^b$), isliye hum easy whole-number part ($10^7$) alag kar lete hain aur sirf chhota fractional piece $10^{0.2}=1.585$ table se chahiye. Woh fractional part "decade mein kitna aage" ka fine-tuning hai. $10^{7.2}=10^7\times 1.585=1.585\times 10^7$ $p_{\text{rms}}=20\times 10^{-6}\times 1.585\times 10^7=20\times 1.585\times 10^1=316.9\mathbf{Pa}(\mathbf{rms})$

(b) Pressure se peak force — do sub-steps. KYUN "force = pressure × area" bilkul bhi: Pressure force hai jo area pe spread hai — yahi uski definition hai: $p=F/A$ (ek pascal matlab ek newton jo ek square metre pe push kar raha ho). Ek uniform pressure $p$ jo ek flat area $A$ pe press kar rahi hai isliye rearrange karke total force milti hai: $F=pA$. Area se multiply karna simply dish ke har square metre ki push ko ek net force mein "collect" karta hai.

Sub-step 1 — rms → peak pressure. Part (a) ne rms (effective) pressure di. Question peak oscillating force maang raha hai, toh pehle hume peak pressure chahiye. Sinusoid ke liye $p_{\text{rms}}=p_{\text{peak}}/\sqrt{2}$ (ek cycle mein ${\sin }^2$ ka mean $\frac{1}{2}$ hota hai), toh: $p_{\text{peak}}=\sqrt{2}\times p_{\text{rms}}=1.414\times 316.9=448.1\text{Pa (peak)}$ Sub-step 2 — peak force. $F_{\text{peak}}=p_{\text{peak}}\times A=448.1\times 1.5=672.2\mathbf{N}$ Picture: ~69 kg ki peak force dish ko aage-peeche ~hundreds of times per second hila rahi hai. Isliye dishes ko stiffening ribs chahiye. (Agar hum $\sqrt{2}$ bhool jaate aur $p_{\text{rms}}$ seedha use karte, toh hum rms force $475.4$ N report karte — "worst-case" push nahin, "typical" push.)

KYA: intensity mein convert karo, add karo, wapas convert karo — dB numbers kabhi add mat karo. KYUN: intensity (energy flow) uncorrelated sources ke liye physically additive hai; cross-terms $⟨p_1{p}_2⟩$ average hokar zero ho jaate hain. Decibels log-axis labels hain aur unhe sum nahin kiya ja sakta. Level $L$ ke $N$ equal, uncorrelated sources ke liye: $L_{\text{total}}=L+10{\log }_{10}(N)$ $=142+10{\log }_{10}(3)=142+10(0.4771)=142+4.77=146.8\mathbf{dB}$ Sanity: doubling ($N=2$) se $+3$ dB milta hai; tripling se thoda $+5$ dB se kam. ✓

KYA: har band ek relative intensity $10^{L_i/10}$ contribute karta hai; unhe sum karo, phir total ka $10{\log }_{10}$ lo. KYUN: bands non-overlapping frequency ranges hain, isliye unki energies cleanly add hoti hain. $10^{12.8}=6.310\times 10^{12}$ $10^{13.4}=2.512\times 10^{13}$ $10^{13.7}=5.012\times 10^{13}$ $10^{13.3}=1.995\times 10^{13}$ $10^{12.7}=5.012\times 10^{12}$ Sum ($10^{12}$ ki units mein): $6.310+25.12+50.12+19.95+5.012=106.5\times 10^{12}=1.065\times 10^{14}$. $L_{\text{OASPL}}=10{\log }_{10}(1.065\times 10^{14})=10(14+0.0274)=140.3\mathbf{dB}$ Insight: peak band 137 dB hai; overall sirf usse $+3.3$ dB upar hai. 500 Hz band akela ~47% energy carry karta hai — ek band dominate karta hai.

Relative intensities: $10^{15.0}=1.000\times 10^{15}$ aur $10^{13.8}=6.310\times 10^{13}$. $I_{\text{tot}}=1.000\times 10^{15}+0.0631\times 10^{15}=1.0631\times 10^{15}$ $L=10{\log }_{10}(1.0631\times 10^{15})=10(15+0.0266)=150.3\mathbf{dB}$ KYUN negligible: 12 dB ka gap matlab quieter source louder wale ki energy ka sirf $10^{-1.2}=6.3\%$ carry karta hai — yeh sirf $+0.27$ dB add karta hai. Rule of thumb: >10 dB apart sources, quieter ko ~0.4 dB ke andar ignore karo.

Step 1 — band pressure (rms). 140 dB ko pressure mein convert karo: $p_{\text{band}}=20\times 10^{-6}\times 10^{140/20}=20\times 10^{-6}\times 10^7=200\text{Pa (rms)}$ Step 2 — resonant amplification. Quality factor $Q$ jawaab deta hai: "resonance pe response static push se kitni baar bada hai?" Jab forcing frequency $f_n$ se match kare, toh natural frequency response $Q$ se amplify hoti hai: $p_{\text{resp}}=Q\times p_{\text{band}}=12\times 200=2400\text{Pa (rms)}$ KYUN $Q$ se multiply karte hain: resonance ke paas panel ki inertia aur stiffness almost cancel ho jaati hain, isliye tiny pushes kaafi cycles mein build up hoti hain — jaise swing pe pushes timing karna. Step 3 — rms → peak. Ab tak sab kuch rms hai (effective steady value). Ek sine wave ka peak uske rms ka $\sqrt{2}$ guna hota hai, kyunki sinusoid ke liye $p_{\text{rms}}=p_{\text{peak}}/\sqrt{2}$ (ek cycle mein ${\sin }^2$ ka mean $\frac{1}{2}$ hota hai). Question peak maang raha hai, toh: $p_{\text{peak}}=\sqrt{2}\times p_{\text{resp}}=1.414\times 2400=3394\text{Pa (peak)}$ Step 4 — peak force. $F=pA$ use karte hain (kyunki pressure force per unit area hai, $p=F/A$): $F_{\text{peak}}=p_{\text{peak}}\times A=3394\times 0.8=2715\mathbf{N}$ Picture: ek tabletop se chhote panel pe ~277 kg ki peak force, 250 Hz pe hammer kar rahi hai — yeh clearly ek qualification concern hai. Yeh Miles' equation ka acoustic sibling hai, jo random base vibration ke liye same $Q$-amplification bookkeeping karta hai.

Energy view: 500 Hz (137 dB) mein 250 Hz band se $10^{(137-134)/10}=2.0\times$ energy hai — do guna raw acoustic input. Resonance view: lekin panel sirf wahi amplify karta hai jo $f_n=250$ Hz ke paas land ho, $\sim Q=12$ se. 500 Hz energy off-resonance hai (ek octave door) aur almost koi amplification nahin milti (transmissibility $\approx 1$). Net effective response $\propto$ (band energy) × (amplification${}^2$):

• 250 Hz: $1.0\times 12^2=144$ (relative units)
• 500 Hz: $2.0\times 1^2=2.0$ Conclusion: on-resonance 250 Hz band ~70× dominate karta hai, bhaawajood iske ki acoustically 3 dB quieter hai. Kahan energy hai ($f_n$ ke relative) yeh beat karta hai kitni energy hai. Isliye octave-band analysis exist karta hai: spectrum ko mode shapes ke saath line up karne ke liye.

(a) Required test level $=145+3=148$ dB. Achieved $=143.5$ dB. $\text{Shortfall}=148-143.5=4.5\mathbf{dB}$ (b) Intensity factor. Intensity mein dB: $\Delta L=10{\log }_{10}(I_2/I_1)$, toh log undo karo: $\frac{I_2}{I_1}=10^{\Delta L/10}=10^{4.5/10}=10^{0.45}=2.82\times$ KYUN $10\log$ yahan: intensity ek power quantity hai, isliye yeh $10\log$ form pehanta hai; invert karne se $10^{\Delta L/10}$ milta hai. (c) Pressure factor. Kyunki $I\propto p_{\text{rms}}^2$ hai, pressure mein level $\Delta L=20{\log }_{10}(p_2/p_1)$ use karta hai: $\frac{p_2}{p_1}=10^{\Delta L/20}=10^{4.5/20}=10^{0.225}=1.68\times$ Cross-check: pressure factor squared $=1.68^2=2.82=$ intensity factor. ✓ Consistent hai, kyunki pressure double karna intensity chaar guni karta hai. Takeaway: ek modest-sounding $4.5$ dB shortfall almost acoustic energy ko teen guna chamber mein karne ki demand karta hai — decibels bade physical changes ko chhote-lagte numbers mein compress karte hain, yahi exactly reason hai ki margins itni carefully police ki jaati hain.

(a) $\text{SPL}=20{\log }_{10}(1)=20\times 0=0\mathbf{dB}$. Yahi reason hai ki $p_{\text{ref}}$ choose kiya gaya — yeh scale ka zero hearing threshold pe fix karta hai. 0 dB silence nahin hai, yeh hai "just barely audible." (b) Kisi bhi $0<p_{\text{rms}}<p_{\text{ref}}$ ke liye ratio 0 aur 1 ke beech hai, uska log negative hai, isliye SPL negative hota hai (jaise Problem 1.3 mein). Jab $p_{\text{rms}}\to 0$, ratio $\to 0$ aur ${\log }_{10}(0)\to -\infty$, toh $\text{SPL}\to -\infty \mathbf{dB}$. Scale ka koi floor nahin hai — perfect silence minus infinity decibels hai; negative dB simply matlab hai "reference se quieter." (c) Correlated case rule todhta hai. In-phase sources ke liye pressures add hoti hain, intensities nahin: $p_{\text{tot}}=2p_{\text{rms}}⟹\Delta L=20{\log }_{10}(2)=6\mathbf{dB}(\text{not }3).$ KYUN difference: "$+3$ dB" rule assume karta tha ki cross-term $⟨p_1{p}_2⟩$ average hokar zero ho gaya (uncorrelated). Agar sources phase mein lock hain, toh woh cross-term maximal hai, aur tumhe milta hai $⟨(2p{)}^2⟩=4⟨p^2⟩\Rightarrow +6$ dB. Real launch acoustics diffuse aur uncorrelated hain → $+3$ dB rule sahi default hai.

SPL formula 20 coefficient use karta hai kyunki log ke andar quantity ek pressure hai (ek field amplitude), aur $10\log (p^2)=20\log (p)$. Uncorrelated sources combine karte waqt, intensities add hoti hain, jo doubling pe +3 dB deta hai. OASPL hamesha individual band levels ke above hota hai (kabhi unka average nahin). Resonant amplification band pressure ko quality factor Q se multiply karta hai, aur rms → peak √2 se multiply karta hai. Jab $p_{\text{rms}}<p_{\text{ref}}$ ho toh SPL negative hota hai (koi error nahin).