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Adding sound levels from several sources

Short answer

Combined SPL = 10 log(10⁸·⁵ + 10⁸·⁸ + 10⁸) = 90.2 dB. With only the 85 and 88 dB machines, level is 89.8 dB. Two equal sources add 3 dB (10 log 2) because doubling energy adds 3 dB, which is the maximum possible increase; adding lower-level sources gives a smaller rise.

Three machines in a workshop produce sound pressure levels of 85 dB, 88 dB and 80 dB at a worker's position when operating individually. Find the combined SPL when all work together. What is the level if the 80 dB machine is switched off? Also, why can two equal sources never raise the level by more than 3 dB?

Sound levels are logarithmic and cannot be added directly. Add the intensities (or mean-square pressures):

Ltotal=10log⁡10(∑i10Li/10)L_{total} = 10\log_{10}\left(\sum_i 10^{L_i/10}\right)

All three machines

108.5=3.162×108108.8=6.310×108108.0=1.000×108Sum=10.472×108\begin{aligned} 10^{8.5} &= 3.162\times10^{8} \\ 10^{8.8} &= 6.310\times10^{8} \\ 10^{8.0} &= 1.000\times10^{8} \\ \text{Sum} &= 10.472\times10^{8} \end{aligned} Ltotal=10log⁡10(10.472×108)=10×9.0200=90.2 dBL_{total} = 10\log_{10}(10.472\times10^{8}) = 10\times9.0200 = 90.2\ \text{dB}

Without the 80 dB machine

L=10log⁡10[(3.162+6.310)×108]=10log⁡10(9.472×108)=89.8 dBL = 10\log_{10}\left[(3.162+6.310)\times10^{8}\right] = 10\log_{10}(9.472\times10^{8}) = 89.8\ \text{dB}

Switching off the quietest source reduces the total only by 0.4 dB.

Two equal sources

Ltotal=10log⁡10(2×10L/10)=L+10log⁡102=L+3 dBL_{total} = 10\log_{10}(2\times10^{L/10}) = L + 10\log_{10}2 = L + 3\ \text{dB}

Doubling the energy adds 3 dB. If the two levels differ, the increase is smaller (for a difference of 10 dB it is only about 0.4 dB), so 3 dB is the maximum. Thus nn equal sources give L+10log⁡10nL + 10\log_{10}n.

Answer: Combined level = 90.2 dB; without the 80 dB machine = 89.8 dB.

✓ Answer checked by Claude Opus 5.5 (high effort)

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