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Sound level at two distances: intensity, decibels, and the inverse-square law

SPH3U Grade 11 Physics · Waves and Sound

The horn speaker at an outdoor community rink hangs from a pole above the boards, calling out skate times and music across the ice. As part of a noise survey, a student walks onto the pad with a sound-level meter and a measuring tape run out to the base of the pole. Held 8.00 m from the horn, the meter reads a sound intensity of 2.00 mW/m². The far end of the pad, where parents lean on the boards, lies 40.0 m from the speaker, and the survey sheet has a blank row waiting for the numbers there. Treating the speaker as a point source radiating equally in all directions, find its total acoustic power output, the sound level in decibels at the meter, and then the intensity and the sound level at the far end of the pad.

93.0 dB1.61 W horn2.00 mW/m² at 8.00 m79.0 dB at 40.0 m8.00 m

Every number in this problem is editable — change any value below and the whole chain recalculates.

Given
  • I = 2 mW/m² — Intensity read at the meter
  • r₁ = 8 m — Meter's distance from the horn
  • r₂ = 40 m — Far end of the pad
Determine
  1. (a)the speaker's total acoustic power output
  2. (b)the sound level at the meter
  3. (c)the intensity at the far end of the pad
  4. (d)the sound level there
Step 1 of 4(a) · solve for Acoustic power

A meter measures intensity at one spot, never power. Multiplying by the area of the sphere it sits on, 4πr², recovers the one quantity that belongs to the speaker rather than to your position.

PrI
Rearranged for P
P=4πr2IP = 4\pi r^{2} I
Your values, in your units
P=4π (8 m)2 (2 mW/m2)P = 4\pi \, \left( 8\ \text{m} \right)^{2} \, \left( 2\ \text{mW/m}^{2} \right)
Converted to base units
P=4π (8 m)2 (0.002 W/m2)P = 4\pi \, \left( 8\ \text{m} \right)^{2} \, \left( 0.002\ \text{W/m}^{2} \right)
Answer
P=1.6085 WP = 1.6085\ \text{W}

Carried onward at full precision, not this rounded figure.

Open the Inverse-Square Law for Sound solver →

Step 2 of 4(b) · solve for Sound level

The same reading on the logarithmic scale the ear actually works on. 2.00 mW/m² is two billion times the threshold of hearing, which the decibel scale compresses to 93.0.

Rearranged for β
β=10log⁡10 ⁣(II0)\beta = 10 \log_{10}\!\left(\tfrac{I}{I_0}\right)
Your values, in your units
β=10log⁡10 ⁣((2 mW/m2)10−12)\beta = 10 \log_{10}\!\left(\tfrac{\left( 2\ \text{mW/m}^{2} \right)}{10^{-12}}\right)
Converted to base units
β=10log⁡10 ⁣((0.002 W/m2)10−12)\beta = 10 \log_{10}\!\left(\tfrac{\left( 0.002\ \text{W/m}^{2} \right)}{10^{-12}}\right)
Answer
β=93.01 dB\beta = 93.01\ \text{dB}

Carried onward at full precision, not this rounded figure.

Open the Decibel Sound Level solver →

Step 3 of 4(c) · solve for Sound intensity

The speaker's power from step 1 is unchanged — the same joules per second are simply spread over a sphere 25 times larger in area, because 40.0 m is five times 8.00 m.

PrI
Rearranged for I
I=P4πr2I = \tfrac{P}{4\pi r^{2}}
1.6085 Wcarried from step 1
Your values, in your units
I=(1.6085 W)4π (40 m)2I = \tfrac{\left( 1.6085\ \text{W} \right)}{4\pi \, \left( 40\ \text{m} \right)^{2}}
Answer
I=80 μW/m2I = 80\ \mu\text{W/m}^{2}

Carried onward at full precision, not this rounded figure.

Open the Inverse-Square Law for Sound solver →

Step 4 of 4(d) · solve for Sound level

Convert the far-field intensity to decibels. The drop from step 2 is 14.0 dB, not the 25-fold collapse the intensity actually suffered.

Rearranged for β
β=10log⁡10 ⁣(II0)\beta = 10 \log_{10}\!\left(\tfrac{I}{I_0}\right)
80 μW/m²carried from step 3
Your values, in your units
β=10log⁡10 ⁣((0.00008 W/m2)10−12)\beta = 10 \log_{10}\!\left(\tfrac{\left( 0.00008\ \text{W/m}^{2} \right)}{10^{-12}}\right)
Answer
β=79.031 dB\beta = 79.031\ \text{dB}

Carried onward at full precision, not this rounded figure.

Open the Decibel Sound Level solver →

Answer

Therefore the horn radiates 1.61 W of acoustic power, the meter at 8.00 m reads 93.0 dB, and by the far boards the intensity has collapsed 25-fold to 8.00 × 10⁻⁵ W/m² — yet the level only eases to 79.0 dB, a drop of 14.0 dB.

Why this order

Step 1 is the move students skip, and skipping it is why they cannot answer part (c). Intensity is a property of a place, not of a source; it is meaningless without a distance attached. Acoustic power belongs to the speaker and is the same number everywhere, so converting the meter reading into a power is what lets the chain travel down the rink. The rest follows: at 40.0 m the same 1.61 W is smeared over a sphere of 25 times the area, so the intensity falls to a twenty-fifth, 8.00 × 10⁻⁵ W/m².

The decibel steps exist to make the logarithm concrete. Intensity fell by a factor of 25, but the level fell by only 10 log₁₀ 25 ≈ 14.0 dB, from 93.0 to 79.0 — because the scale is compressing twelve orders of magnitude of human hearing into about 130 numbers. That compression produces arithmetic that feels wrong until you trust it: two identical speakers side by side add 3 dB, not 93 dB, and it takes ten of them to sound roughly twice as loud. It also explains why moving back does less than you hope at a loud event, and why hearing-protection rules are written in 3 dB steps — every 3 dB doubles the energy arriving at the eardrum and halves the time it is safe to stand there.

Carried values move at full precision, not the rounded figure shown — chaining rounded numbers compounds error.