Decibels on the Factory Floor
A factory floor runs three machines. Measured one at a time, machine A registers 78 dB, machine B registers 75 dB and machine C registers 72 dB, where the sound level in decibels is L = 10 log₁₀(I/I₀) and I₀ = 1 × 10⁻¹² W m⁻². The legal exposure limit in the room is 85 dB. Production wants to install extra machines identical to A. The floor manager argues that since A is 78 dB and B is 75 dB, running both already gives 153 dB, so the limit is being smashed and no further machines can be contemplated. A safety officer disagrees and says extra machines are 'almost free' in decibel terms because doubling the number of identical machines adds only about 3 dB. The night shift, meanwhile, wants a practical answer: how many extra A-type machines can run while all three existing machines are switched on?
1. Show that running A and B together gives about 79.8 dB, and explain briefly why the manager's 153 dB is not how sound levels combine.
Read the worked answer
Convert dB to intensity ratios
I_A/I₀ = 10^(78/10) = 10^7.8 ≈ 6.3096 × 10⁷
I_B/I₀ = 10^(75/10) = 10^7.5 ≈ 3.1623 × 10⁷
Add the intensities, not the decibels
Combined ratio = (6.3096 + 3.1623) × 10⁷ = 9.4719 × 10⁷
L = 10 log₁₀(9.4719 × 10⁷) = 10 × 7.9764 ≈ 79.8 dB
Why the manager is wrong
Decibels are logarithms of intensity. Intensities from independent sources add, but the logarithm of a sum is not the sum of the logarithms: log(a + b) ≠ log a + log b. Adding 78 + 75 would correspond to multiplying the intensities, which no physical source arrangement does. The true combined level, 79.8 dB, is dominated by the louder machine and is only 1.8 dB above A alone.
2. Determine the greatest number of extra A-type machines that can be added while all three existing machines remain running, and still stay within 85 dB.
Read the worked answer
Set up the inequality in intensity units of I₀
Let k be the total number of A-type machines (the existing A plus the extras). Requirement:
k(10^7.8) + 10^7.5 + 10^7.2 ≤ 10^8.5
i.e. k(6.3096 × 10⁷) + 3.1623 × 10⁷ + 1.5849 × 10⁷ ≤ 3.1623 × 10⁸
Test integer values of k
k = 4: 2.5238 × 10⁸ + 0.4747 × 10⁸ = 2.9985 × 10⁸ → L = 10 log(2.9985 × 10⁸) = 84.8 dB ✓ (within limit)
k = 5: 3.1548 × 10⁸ + 0.4747 × 10⁸ = 3.6295 × 10⁸ → L = 85.6 dB ✗ (over limit)
Answer
Four A-type machines in total, i.e. three extra machines, giving about 84.8 dB. The fifth machine pushes the room over the limit. Because integer counts are involved, solving the inequality exactly and then rounding down is the correct judgement - rounding to the nearest whole number would illegally give five.
3. The night shift proposes switching machine C off so that one more A-type machine can be added. Test this proposal numerically and comment on the safety officer's '3 dB per doubling' rule.
Read the worked answer
Testing the proposal
With C off and five A-type machines plus B running:
5(6.3096 × 10⁷) + 3.1623 × 10⁷ = 3.1548 × 10⁸ + 0.3162 × 10⁸ = 3.4710 × 10⁸
L = 10 log(3.4710 × 10⁸) = 85.4 dB > 85 dB
The proposal fails: removing C (only 72 dB, the quietest machine) buys less than the extra A machine costs. In fact, even with both B and C switched off, five A machines give 10 log(3.1548 × 10⁸) = 85.0 dB - right on the limit, with no margin at all, so that is not a sensible operating state either.
Checking the officer's rule
Doubling identical sources: L = 10 log(2 × 10^7.8) = 78 + 10 log 2 = 78 + 3.01 = 81.0 dB. So the 3 dB-per-doubling rule is correct for identical machines (a doubling of intensity, not a doubling of decibels).
Judgement
The rule is right in principle but misleading here: 85 dB is reached at 84.99 dB for five A machines alone, so the headroom is about 3 dB in total - one doubling of A-type machines only. The room is close to the limit because A already sits high in the logarithmic scale, not because the added machines are 'free'.