Review Conceptual Example 8 as background for this problem. A loudspeaker is generating sound in a room. At a certain point, the sound waves coming directly from the speaker (without reflecting from the walls) create an intensity level of 75.0 dB. The waves reflected from the walls create, by themselves, an intensity level of 72.0 dB at the same point. What is the total intensity level? The answer is not 147.0 dB.

Conceptual Example 8

Suppose that the person singing in the shower in Figure 16.24 produces a sound power P. Sound reflects from the surrounding shower stall. At a distance r in front of the person, does the expression I = P/(4????r²) (Equation 16.9) (a) overestimate, (b) underestimate, or (c) give the correct total sound intensity?

Reasoning In arriving at Equation 16.9, it was assumed that the sound spreads out uniformly from the source and passes only once through the imaginary surface that surrounds it (see Figure 16.22). In Figure 16.24, only part of this imaginary surface (colored blue) is shown, but nonetheless, if Equation 16.9 is to apply, the same assumption must hold.

Answers (a) and (c) are incorrect. Equation 16.9 cannot overestimate the sound intensity, because it assumes that the sound passes through the imaginary surface only once and, hence, does not take into account the reflected sound within the shower stall. For the same reason, neither can Equation 16.9 give the correct sound intensity.

Answer (b) is correct. Figure 16.24 illustrates three paths by which the sound passes through the imaginary surface. The “direct” sound travels along a path from its source directly to the surface. It is the intensity of this sound that is given by I = P/(4????r²). The remaining paths are two of the many that characterize the sound reflected from the shower stall. The total sound power that passes through the surface is the sum of the direct and reflected powers. Thus the total sound intensity at a distance r from the source is greater than the intensity of the direct sound alone, so Equation 16.9 underestimates the sound intensity from the singing. People like to sing in the shower because their voices sound so much louder due to the enhanced intensity caused by the reflected sound.

Transcribed Image Text:

Reflected sound Direct sound Reflected sound