Sound frequency intensity and safe exposure

Sound is a longitudinal pressure wave in a medium. Frequency $f$ (in hertz) sets the pitch, while intensity $I$ is the energy flow per unit area, $[I] = \text{W/m}^2$. Human hearing spans roughly $20$ Hz to $20$ kHz: below that lies infrasound, above it lies ultrasound. Because audible intensities differ by about $10^{12}$, it is convenient to use a logarithmic level $L = 10\log_{10}\frac{I}{I_0}, \qquad I_0 = 10^{-12}\ \text{W/m}^2.$ Hearing risk depends on level and duration: damage is governed by total acoustic energy, so every $+3$ dB halves the permitted exposure time. Practical reference points: $85$ dB(A) is permitted for $8$ h, $100$ dB for about $15$ min, and $120$ dB is the pain threshold and an immediate hazard. Risk can be reduced in three ways: lower the level (quieter source, greater distance), shorten the time, or use hearing protection.

A machine produces a sound intensity of $I = 10^{-4}\ \text{W/m}^2$ at a workstation. Find the sound intensity level $L$, taking $I_0 = 10^{-12}\ \text{W/m}^2$.

Two identical loudspeakers each create $70$ dB at the listening position. (a) What is the level when both play together? (b) How many such loudspeakers would be needed to reach $85$ dB?

A rehearsal room measures $91$ dB. Using the equal-energy rule (3 dB doubling) and the reference of $85$ dB for $8$ h: (a) find the maximum permitted exposure time; (b) find the daily noise dose for a musician who plays for $3$ h.

Safe self-study investigation. Do not generate loud sounds: treat anything above about $80$ dB by calculation, not by listening.

1. Install a sound-level meter app on a phone and measure the level in a quiet room, during normal conversation, and with quiet music (a clearly comfortable volume, phone at arm's length). Record the three values of $L$ in dB.
2. Convert each value to intensity with the inverse formula $I = I_0 \cdot 10^{L/10}, \qquad I_0 = 10^{-12}\ \text{W/m}^2,$ and compare the intensity ratios of the quiet room and the conversation.
3. In a spreadsheet, build a "safe-time calculator" for levels from $85$ to $100$ dB: time $= 8\ \text{h} \cdot 2^{-(L-85)/3}$. Ask "how many minutes may one listen at $97$ dB?" and check the answer against the graph.
4. State a conclusion: how does permitted time change when the level rises by $10$ dB, and why is controlling time as important as controlling loudness.

Show that the permitted exposure time falls steeply as the sound level rises, and indicate where the audible band lies on the frequency axis.

Graph plus schematic. Upper part — a semi-logarithmic graph: horizontal axis is sound level in dB from $80$ to $120$, vertical axis (logarithmic scale) is permitted time in hours. A straight line passes through the marked points $(85;\ 8\ \text{h})$, $(88;\ 4\ \text{h})$, $(91;\ 2\ \text{h})$, $(94;\ 1\ \text{h})$; the region to the right of $100$ dB is shaded as the hazardous-exposure zone. Lower part — a horizontal frequency bar from $10$ Hz to $100$ kHz with boundaries at $20$ Hz and $20$ kHz, divided into the zones "infrasound", "audible sound", and "ultrasound".

On a bus a teenager listens to music through in-ear earbuds with the volume at maximum. A microphone that mimics the ear canal reads about $100\ \text{dB}$ SPL. Sound level is linked to intensity by $L = 10\log_{10}(I/I_0)$ with $I_0 = 10^{-12}\ \text{W/m}^2$, so $100\ \text{dB}$ corresponds to $I = 10^{-2}\ \text{W/m}^2$. The safe limit is $85\ \text{dB}$ for $8\ \text{h}$; every $+3\ \text{dB}$ halves the allowed time: $88\ \text{dB}$ — $4\ \text{h}$, $94\ \text{dB}$ — $1\ \text{h}$, $100\ \text{dB}$ — about $15\ \text{min}$. The earbud sits $1\ \text{cm}$ from the eardrum: if the same power were radiated spherically, moving from $1\ \text{cm}$ to $1\ \text{m}$ (100 times farther) would cut intensity by a factor of $10^4$, that is by $40\ \text{dB}$.

The example exercises the logarithmic sound-level scale with its reference intensity $I_0$, the equal-energy rule for safe exposure (dose proportional to $I\cdot t$), and the inverse-square law when the distance changes.

A referee blows a pea whistle: the fundamental is near $4\ \text{kHz}$ and the level at $1\ \text{m}$ from the whistle is $110\ \text{dB}$. That matches an intensity $I = 10^{-1}\ \text{W/m}^2$; spread over a sphere of area $4\pi r^2 = 12.6\ \text{m}^2$ this is an acoustic power $P = I\cdot A \approx 1.3\ \text{W}$. A player $10\ \text{m}$ away receives $I = P/(4\pi r^2) \approx 1.0\times 10^{-3}\ \text{W/m}^2$, that is $90\ \text{dB}$ — $20\ \text{dB}$ lower. Young hearing spans roughly $20\ \text{Hz}$ to $20\ \text{kHz}$, and sensitivity peaks at $2$–$5\ \text{kHz}$ because of the ear-canal resonance, so the $4\ \text{kHz}$ tone sounds especially loud.

This checks the conversion from intensity to level via $L = 10\log_{10}(I/I_0)$, the inverse-square law for a point source, and the frequency dependence of ear sensitivity that A-weighting accounts for.

A motorcyclist rides on a highway at $90\ \text{km/h}$; a microphone at the ear reads $90\ \text{dB}$ — wind noise inside the helmet plus the engine. By the equal-energy rule such riding is allowed for no more than $2.5\ \text{h}$ per day, whereas $85\ \text{dB}$ is safe for $8\ \text{h}$: a level of $90\ \text{dB}$ corresponds to $I = 10^{-3}\ \text{W/m}^2$ and $85\ \text{dB}$ to $I = 3.16\times 10^{-4}\ \text{W/m}^2$, which is $3.2$ times less. The spectrum is uneven: engine rumble lies below $500\ \text{Hz}$, while wind noise is broadband with a rise around $500$–$2000\ \text{Hz}$, where a helmet attenuates sound least.

The example applies the equal-energy rule (each $3\ \text{dB}$ rise requires halving the time) and the spectral content of the noise, which is why limits are written in dBA — with a correction for the ear's frequency sensitivity.