Sound Intensity and the Inverse-Square Law
1 · Predict
A speaker radiates sound equally in all directions. If a listener moves twice as far away, how much does the sound intensity drop?
- Intensity drops to 1/4 — it falls off with the square of distance.
- Intensity drops to 1/2 — it falls off linearly with distance.
- Intensity stays the same regardless of distance.
2 · Set Up
- Open the sound-intensity preset and press Reset. The source radiates 0.05 W in all directions (spherical spreading).
- Enable the intensity readout at the observer.
- Set the observer's distance from the source for each trial and record the intensity.
3 · Collect Data
| Distance r (m) | Intensity I (nW/m²) |
|---|---|
| 150 | |
| 200 | |
| 300 |
Plot intensity I (y-axis) against 1/r² (x-axis) for your three trials. Is the line straight through the origin?
4 · Analyze
- For one trial, compute I = P/(4πr²) using P = 0.05 W, expressing your answer in nW/m² (1 nW/m² = 10⁻⁹ W/m²). Compare to the table.
- Explain, using the surface area of a sphere 4πr², why sound power spreading outward makes intensity fall off as 1/r² rather than 1/r.
5 · Extend
- Doubling distance quarters the intensity, but loudness (in decibels) only seems to drop by a fixed amount each time distance doubles. Look at the sound-decibel experiment — what kind of function turns a ×4 intensity ratio into a fixed dB difference?
- This experiment assumes the source radiates equally in every direction. A megaphone instead focuses sound into a narrow cone. How would you expect the intensity 1/r² law to change for a focused source at the same distance?
The Physics Behind This Experiment
Inverse-Square Law for Sound Intensity
Power radiated equally in all directions spreads over a sphere of area 4πr². Intensity — power per unit area — therefore falls off as 1/r² with distance from the source.