Inverse-Square Law for Sound

I=P4πr2I = \frac{P}{4\pi r^{2}}

Worked example: 100 W at 10 m → I = 1/(4*pi) W/m^2 — press Try an example to run it live, then adjust anything.

Enter your known values, leave one input blank, and solves for the missing one. Tap a variable’s symbol to see what it means, with a typical value. Try different units for next level excitement!

Here the solver did the work — could you?

How loud is loud →

Grade 11Grade 11 Physics

Test your skills in the Exam Room: new numbers every attempt — free lessons for students, no sign-up, just pure learning.

See your Report Card
Compete with your friends
share your results
Learning zone

Inverse-Square Law for Sound explained

PrI

Sound from a small source spreads out as an ever-growing sphere, and a sphere's surface area grows as r², so the intensity thins out as 1/r². Double your distance from a firecracker and the intensity drops to a quarter; step back tenfold and it falls a hundredfold — a 20 dB drop. A 0.5 W siren heard from 10 m delivers 0.5/(4π × 100) ≈ 4 × 10⁻⁴ W/m², about 86 dB; from 100 m it is down to 66 dB, no louder than animated conversation.

The law assumes the sound spreads freely in all directions, which is why real environments bend it. A megaphone or a stage horn concentrates power into a cone instead of a sphere, staying loud farther out; a highway heard across open fields behaves more like a line source, fading as 1/r rather than 1/r². Concert engineers exploit the pure form in reverse: measure the intensity at a known distance and the formula reveals the source's total acoustic power.

Inverse-Square Law for Sound formula

I=P4πr2I = \frac{P}{4\pi r^{2}}
Where
  • II= Sound intensity (W/m²)
  • PP= Acoustic power (W)
  • rr= Distance from source (m)

Missing one of these? Work it out first, then come back