Example 6 Sound Intensities

In Figure 16.21, 12×10^–5 W of sound power passes perpendicularly through the surfaces labeled 1 and 2. These surfaces have areas of A₁=4.0 m² and A₂=12 m². Determine the sound intensity at each surface and discuss why listener 2 hears a quieter sound than listener 1.

Reasoning The sound intensity I is the sound power P passing perpendicularly through a surface divided by the area A of that surface. Since the same sound power passes through both surfaces and surface 2 has the greater area, the sound intensity is less at surface 2.

Problem solving insight
Sound intensity I and sound power P are different concepts. They are related, however, since intensity equals power per unit area.

Solution The sound intensity at each surface follows from Equation 16.8:

The sound intensity is less at the more distant surface, where the same power passes through a threefold greater area. The ear of a listener, with its fixed area, intercepts less power where the intensity, or power per unit area, is smaller. Thus, listener 2 intercepts less of the sound power than listener 1. With less power striking the ear, the sound is quieter.