Exercises — Acoustic loads — SPL, octave band analysis

WHAT we do: put the numbers straight into the SPL definition. $\text{SPL}=20{\log }_{10}\left(\frac{p_{\text{rms}}}{p_{\text{ref}}}\right)$ WHY: SPL is by definition 20 times the log of the pressure ratio to the hearing threshold. The ratio measures "how many hearing-thresholds loud" this is. $\frac{p_{\text{rms}}}{p_{\text{ref}}}=\frac{2\times 10^{-3}}{20\times 10^{-6}}=\frac{2\times 10^{-3}}{2\times 10^{-5}}=100$ $\text{SPL}=20{\log }_{10}(100)=20\times 2=40\mathbf{dB}$ Sanity (walk the map): $100\times$ threshold pressure is two big steps right on the figure, and each step is $+20$ dB, so $+40$ dB. ✓

WHAT: an octave band spans $f_c/\sqrt{2}$ to $f_c\sqrt{2}$. WHY $\sqrt{2}$: so the band is logarithmically centred — the ratio top/bottom is exactly $2$ (one octave), and $f_c$ sits at the geometric middle $\sqrt{f_{\text{low}}{f}_{\text{up}}}$. $f_{\text{low}}=\frac{500}{\sqrt{2}}\approx 353.6\text{Hz},f_{\text{up}}=500\sqrt{2}\approx 707.1\text{Hz}$ Check: $707.1/353.6=2.00$ ✓ (one octave).

WHAT: same definition, but now the pressure is below the reference. $\frac{p_{\text{rms}}}{p_{\text{ref}}}=\frac{2\times 10^{-6}}{20\times 10^{-6}}=0.1$ $\text{SPL}=20{\log }_{10}(0.1)=20\times (-1)=-20\mathbf{dB}$ WHY negative: a ratio less than 1 has a negative logarithm. Nothing broke — the minus sign simply reads "20 dB below the threshold of hearing." On the figure this is the shaded region left of the blue dot. Decibels have no floor: the quieter the sound, the more negative (all the way to $-\infty$ at perfect silence).

(a) Invert the log. If $\text{SPL}=20{\log }_{10}(p_{\text{rms}}/p_{\text{ref}})$, then dividing by 20 and applying $10^{(\cdot )}$ undoes the log: $p_{\text{rms}}=p_{\text{ref}}\times 10^{\text{SPL}/20}=20\times 10^{-6}\times 10^{144/20}$ $=20\times 10^{-6}\times 10^{7.2}$ WHY split $10^{7.2}=10^7\times 10^{0.2}$? On our log map, the exponent $7.2$ is a position on the pressure axis: the whole number $7$ says "seven big $\times 10$ steps to the right," and the leftover $0.2$ is a fractional step inside the eighth decade. Powers add exponents ($10^{a+b}=10^{a}10^b$), so we peel off the easy whole-number part ($10^7$) and only need the small fractional piece $10^{0.2}=1.585$ from a table. That fractional part is the "how far along the decade" fine-tuning. $10^{7.2}=10^7\times 1.585=1.585\times 10^7$ $p_{\text{rms}}=20\times 10^{-6}\times 1.585\times 10^7=20\times 1.585\times 10^1=316.9\mathbf{Pa}(\mathbf{rms})$

(b) From pressure to peak force — two sub-steps. WHY "force = pressure × area" at all? Pressure is force spread over area — that is its very definition: $p=F/A$ (a pascal is one newton pushing on one square metre). A uniform pressure $p$ pressing on a flat area $A$ therefore delivers a total force found by rearranging: $F=pA$. Multiplying by the area simply "collects" the push from every square metre of the dish into one net force.

Sub-step 1 — rms → peak pressure. Part (a) gave the rms (effective) pressure. The question asks for the peak oscillating force, so we first need the peak pressure. For a sinusoid $p_{\text{rms}}=p_{\text{peak}}/\sqrt{2}$ (the mean of ${\sin }^2$ over a cycle is $\frac{1}{2}$), so: $p_{\text{peak}}=\sqrt{2}\times p_{\text{rms}}=1.414\times 316.9=448.1\text{Pa (peak)}$ 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 of peak force flapping the dish back and forth ~hundreds of times per second. That is why dishes need stiffening ribs. (Had we forgotten the $\sqrt{2}$ and used $p_{\text{rms}}$ directly, we would have reported the rms force $475.4$ N — the "typical," not the "worst-case" push.)

WHAT: convert to intensity, add, convert back — never add the dB numbers. WHY: intensity (energy flow) is physically additive for uncorrelated sources; the cross-terms $⟨p_1{p}_2⟩$ average to zero. Decibels are log-axis labels and cannot be summed. For $N$ equal, uncorrelated sources of level $L$: $L_{\text{total}}=L+10{\log }_{10}(N)$ $=142+10{\log }_{10}(3)=142+10(0.4771)=142+4.77=146.8\mathbf{dB}$ Sanity: doubling ($N=2$) gives $+3$ dB; tripling gives a bit under $+5$ dB. ✓

WHAT: each band contributes a relative intensity $10^{L_i/10}$; sum them, then take $10{\log }_{10}$ of the total. WHY: bands are non-overlapping frequency ranges, so their energies add cleanly. $10^{12.8}=6.310\times 10^{12}$ $10^{13.4}=2.512\times 10^{13}$ $10^{13.7}=5.012\times 10^{13}$ $10^{13.3}=1.995\times 10^{13}$ $10^{12.7}=5.012\times 10^{12}$ Sum (in units of $10^{12}$): $6.310+25.12+50.12+19.95+5.012=106.5\times 10^{12}=1.065\times 10^{14}$. $L_{\text{OASPL}}=10{\log }_{10}(1.065\times 10^{14})=10(14+0.0274)=140.3\mathbf{dB}$ Insight: the peak band is 137 dB; the overall is only $+3.3$ dB above it. The 500 Hz band alone carries ~47% of the energy — one band dominates.

Relative intensities: $10^{15.0}=1.000\times 10^{15}$ and $10^{13.8}=6.310\times 10^{13}$. $I_{\text{tot}}=1.000\times 10^{15}+0.0631\times 10^{15}=1.0631\times 10^{15}$ $L=10{\log }_{10}(1.0631\times 10^{15})=10(15+0.0266)=150.3\mathbf{dB}$ WHY negligible: a 12 dB gap means the quieter source carries only $10^{-1.2}=6.3\%$ of the louder one's energy — it adds a mere $+0.27$ dB. Rule of thumb: sources >10 dB apart, ignore the quieter one to within ~0.4 dB.

Step 1 — band pressure (rms). Convert 140 dB to pressure: $p_{\text{band}}=20\times 10^{-6}\times 10^{140/20}=20\times 10^{-6}\times 10^7=200\text{Pa (rms)}$ Step 2 — resonant amplification. The quality factor $Q$ answers: "how many times bigger is the response at resonance than the static push?" When the forcing frequency matches $f_n$, the natural frequency response is amplified by $Q$: $p_{\text{resp}}=Q\times p_{\text{band}}=12\times 200=2400\text{Pa (rms)}$ WHY multiply by $Q$: near resonance the panel's inertia and stiffness nearly cancel, so tiny pushes build up over many cycles — like timing shoves on a swing. Step 3 — rms → peak. Everything so far is rms (the effective steady value). The peak of a sine wave is $\sqrt{2}$ times its rms, because $p_{\text{rms}}=p_{\text{peak}}/\sqrt{2}$ for a sinusoid (the mean of ${\sin }^2$ over a cycle is $\frac{1}{2}$). The problem asks for the peak, so: $p_{\text{peak}}=\sqrt{2}\times p_{\text{resp}}=1.414\times 2400=3394\text{Pa (peak)}$ Step 4 — peak force. Using $F=pA$ (because pressure is force per unit area, $p=F/A$): $F_{\text{peak}}=p_{\text{peak}}\times A=3394\times 0.8=2715\mathbf{N}$ Picture: ~277 kg of peak force on a panel smaller than a tabletop, hammering at 250 Hz — a clear qualification concern. This is the acoustic sibling of Miles' equation, which does the same $Q$-amplification bookkeeping for random base vibration.

Energy view: 500 Hz (137 dB) has $10^{(137-134)/10}=2.0\times$ the energy of the 250 Hz band — twice the raw acoustic input. Resonance view: but the panel amplifies only what lands near $f_n=250$ Hz, by $\sim Q=12$. The 500 Hz energy is off-resonance (an octave away) and gets almost no amplification (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: the on-resonance 250 Hz band dominates by ~70×, even though it is 3 dB quieter acoustically. Where the energy sits (relative to $f_n$) beats how much energy there is. This is why octave-band analysis exists: to line the spectrum up against the mode shapes.

(a) Required test level $=145+3=148$ dB. Achieved $=143.5$ dB. $\text{Shortfall}=148-143.5=4.5\mathbf{dB}$ (b) Intensity factor. dB in intensity: $\Delta L=10{\log }_{10}(I_2/I_1)$, so undo the log: $\frac{I_2}{I_1}=10^{\Delta L/10}=10^{4.5/10}=10^{0.45}=2.82\times$ WHY $10\log$ here: intensity is a power quantity, so it wears the $10\log$ form; inverting gives $10^{\Delta L/10}$. (c) Pressure factor. Since $I\propto p_{\text{rms}}^2$, level in pressure uses $\Delta L=20{\log }_{10}(p_2/p_1)$: $\frac{p_2}{p_1}=10^{\Delta L/20}=10^{4.5/20}=10^{0.225}=1.68\times$ Cross-check: pressure factor squared $=1.68^2=2.82=$ intensity factor. ✓ Consistent, because doubling pressure quadruples intensity. Takeaway: a modest-sounding $4.5$ dB shortfall demands almost tripling the acoustic energy in the chamber — decibels compress big physical changes into small-looking numbers, which is exactly why margins are policed so carefully.

(a) $\text{SPL}=20{\log }_{10}(1)=20\times 0=0\mathbf{dB}$. That is why $p_{\text{ref}}$ was chosen — it fixes the zero of the scale at the hearing threshold. 0 dB is not silence, it is "just barely audible." (b) For any $0<p_{\text{rms}}<p_{\text{ref}}$ the ratio is between 0 and 1, its log is negative, so SPL is negative (as in Problem 1.3). As $p_{\text{rms}}\to 0$, the ratio $\to 0$ and ${\log }_{10}(0)\to -\infty$, so $\text{SPL}\to -\infty \mathbf{dB}$. The scale has no floor — perfect silence is minus infinity decibels; negative dB simply means "quieter than the reference." (c) Correlated case breaks the rule. For in-phase sources the pressures add, not the intensities: $p_{\text{tot}}=2p_{\text{rms}}⟹\Delta L=20{\log }_{10}(2)=6\mathbf{dB}(\text{not }3).$ WHY the difference: the "$+3$ dB" rule assumed the cross-term $⟨p_1{p}_2⟩$ averaged to zero (uncorrelated). If the sources are locked in phase, that cross-term is maximal, and you get $⟨(2p{)}^2⟩=4⟨p^2⟩\Rightarrow +6$ dB. Real launch acoustics are diffuse and uncorrelated → the $+3$ dB rule is the right default.

The SPL formula uses a coefficient of 20 because the quantity inside the log is a pressure (a field amplitude), and $10\log (p^2)=20\log (p)$. When combining uncorrelated sources, intensities add, giving +3 dB per doubling. OASPL is always above (never the average of) the individual band levels. Resonant amplification multiplies the band pressure by the quality factor Q, and rms → peak multiplies by √2. When $p_{\text{rms}}<p_{\text{ref}}$ the SPL is negative (not an error).