gen-define-sound-intensity-and-the-inverse-square-law-and-give-o-80df

Define sound intensity and the inverse-square law and give one short worked example.

# Sound Intensity and the Inverse-Square Law ## Sound Intensity **Sound intensity** (I) is the sound power per unit area, measured in watts per square meter (W/m²). It represents the rate at which sound energy flows through a unit area perpendicular to the direction of propagation. $$I = \frac{P}{A}$$ where P = power (W) and A = area (m²) ## Inverse-Square Law For a point source radiating sound uniformly in all directions, the intensity decreases with the square of the distance from the source: $$I = \frac{P}{4\pi r^2}$$ or equivalently: $$\frac{I_1}{I_2} = \frac{r_2^2}{r_1^2}$$ This means **doubling the distance reduces the intensity to one-quarter** of its original value. ## Worked Example **Problem:** A sound has an intensity of 1.0 × 10⁻⁴ W/m² at 2 m from a point source. What is the intensity at 8 m from the source? **Solution:** Using the inverse-square relationship: $$\frac{I_1}{I_2} = \frac{r_2^2}{r_1^2}$$ $$I_2 = I_1 \times \frac{r_1^2}{r_2^2} = 1.0 \times 10^{-4} \times \frac{(2)^2}{(8)^2}$$ $$I_2 = 1.0 \times 10^{-4} \times \frac{4}{64} = 6.25 \times 10^{-6} \text{ W/m}^2$$ The distance increased by a factor of 4, so the intensity decreased by a factor of 16.