Intensity of Sound

What this figure shows. A point sound source is drawn at the centre, surrounded by three concentric spherical surfaces at increasing distances, labelled Distance 1, Distance 2 (= 2r) and Distance 3 (= 3r), with their corresponding surface areas labelled Area 1, Area 2 and Area 3; alongside, formulas indicate the intensity falls from I at the first sphere to I/4 at the sphere of twice the distance, and to I/9 at the sphere of three times the distance. The figure visually derives the inverse-square law: since the SAME total sound power radiated by the source must spread out over an ever-larger spherical surface area ($4\pi r^2$, growing with the square of distance) as it travels outward, the intensity -- power divided by that area -- must fall off in exact proportion to $1/r^2$, so doubling the …

Worked out. A baby's cry is detected at a distance of 3.0 m with intensity $10^{-2}\ \text{W/m}^2$, and the task is to find the intensity of the same cry at a greater distance of 6.0 m. Using the inverse-square law, $I\propto1/r^2$, the ratio of intensities at the two distances is $I_1/I_2=r_2^2/r_1^2$, so $I_2=I_1\times(r_1/r_2)^2=10^{-2}\times(3.0/6.0)^2=10^{-2}\times0.25=0.25\times10^{-2}\ \text{W/m}^2$. Since doubling the distance from the fixed-power source (from 3.0 m to 6.0 m) always reduces the intensity by exactly a factor of $2^2=4$ regardless of the source's actual power output, this example is a direct, concrete demo …