Giancoli 7th Edition textbook cover
Giancoli's Physics: Principles with Applications, 7th Edition

12-1: Characteristics of Sound
12-2: Intensity of Sound; Decibels
12-3: Loudness
12-4: Sources of Sound: Strings and Air Columns
12-5: Quality of Sound, Superposition
12-6: Interference; Beats
12-7: Doppler Effect
12-8: Shock Waves; Sonic Booms

Question by Giancoli, Douglas C., Physics: Principles with Applications, 7th Ed., ©2014, Reprinted by permission of Pearson Education Inc., New York.
Problem 20
Q
  1. Estimate the power output of sound from a person speaking in normal conversation. Use Table 12–2. Assume the sound spreads roughly uniformly over a sphere centered on the mouth.
  2. How many people would it take to produce a total sound output of 60 W of ordinary conversation? [Hint: Add intensities, not dBs.]
A
  1. 9×106 W9 \times 10^{-6} \textrm{ W}
  2. 6 million people6 \textrm{ million people}
Giancoli 7th Edition, Chapter 12, Problem 20 solution video poster
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VIDEO TRANSCRIPT

This is Giancoli Answers with Mr. Dychko. Intensity is power divided by area. And so if we multiply both sides of this equation by A we'll solve for p. So, power is intensity times area. And the area over which this sound is distributed is going to be the surface area of a sphere centered at the person's mouth. And so that's going to be the formula for the surface area sphere is 4π times the distance from the sound source squared. So, that's 3 times 10 to the minus 6 watts per square meter in table 12-2. And that's the intensity at a 0.50 centimeters from the person. So, we have 4π times 50 times 10 to the minus 2 meters squared. And this gives a power of 9 times 10 to the minus 6 watts for 1 person. And if you want a total of 60 watts for, you know, some number of people, multiply 60 watts by 1 person generating this many watts. And the watts cancel leaving us with number of people. And this works out to about 6 million people will be needed talking at a regular conversation volume to produce a total of 60 watts power output.

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