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Chapter 17: Problem 95

The sound intensity is \(0.0080 \mathrm{~W} / \mathrm{m}^{2}\) at a distance of \(10 \mathrm{~m}\) from an isotropic point source of sound. (a) What is the power of the source? (b) What is the sound intensity \(5.0 \mathrm{~m}\) from the source? (c) What is the sound level \(10 \mathrm{~m}\) from the source?

Short Answer

Expert verified
(a) Power is 10.05 W. (b) Intensity at 5 m is 0.032 W/m虏. (c) Sound level at 10 m is 89 dB.

Step by step solution

01

Define Intensity-Power Relationship

The sound intensity \( I \) at distance \( r \) from a point source is given by the formula \( I = \frac{P}{4 \pi r^2} \), where \( P \) is the power of the source. Our goal is to find \( P \).
02

Solve for Power of the Source

Given \( I = 0.0080 \mathrm{~W} / \mathrm{m}^2 \) and \( r = 10 \mathrm{~m} \), we substitute these values into the intensity formula: \( 0.0080 = \frac{P}{4 \pi (10)^2} \). Rearranging gives \( P = 0.0080 \times 4 \pi \times 100 = 10.05 \mathrm{~W} \).
03

Find Sound Intensity at 5.0 m

To find the intensity at \( 5.0 \mathrm{~m} \), use \( I = \frac{P}{4 \pi r^2} \) with \( r = 5.0 \mathrm{~m} \) and \( P = 10.05 \mathrm{~W} \). Thus, \( I = \frac{10.05}{4 \pi (5.0)^2} = 0.032 \mathrm{~W} / \mathrm{m}^2 \).
04

Calculate Sound Level at 10 m

Sound level in decibels (dB) is calculated using \( L = 10 \cdot \log_{10}(\frac{I}{I_0}) \), where \( I_0 = 10^{-12} \mathrm{~W}/\mathrm{m}^2 \) is the reference intensity. At \( 10 \mathrm{~m} \), \( L = 10 \cdot \log_{10}(\frac{0.0080}{10^{-12}}) = 89 \mathrm{~dB} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Intensity-Power Relationship
Understanding the intensity-power relationship is crucial when dealing with sound waves. Sound intensity, represented by \( I \), is a measure of how much sound power \( P \) passes through a specific area. Think of it as the concentration of sound energy hitting a surface. This relationship is mathematically described by the equation \( I = \frac{P}{4 \pi r^2} \), where \( r \) represents the distance from the sound source.

The formula highlights that intensity decreases as you move away from the source because the same amount of energy spreads over a larger area. Imagine the sound spreading in a spherical pattern around the source, the further you are, the thinner the layer of sound energy.

In practical terms:
  • If you know the intensity at a point, you can calculate the power of the source, and vice versa.
  • For example, if a sound has an intensity of \( 0.0080 \, \mathrm{W/m^2} \) at \( 10 \mathrm{~m} \), and you want to determine the power \( P \), you rearrange the formula to get \( P = I \times 4 \pi r^2 \), resulting in \( P = 10.05 \, \mathrm{W} \).
Isotropic Point Source
An isotropic point source is a special type of sound source that emits sound energy uniformly in all directions. Think of it as a perfect tiny loudspeaker placed in open space. The sound doesn't favor any particular direction and spreads out evenly to form spheres of increasing size as it moves away from the source.

This concept helps simplify the calculations of sound intensity. Since the sound spreads out spherically, the intensity at any point on a sphere around the source is the same. This symmetry removes the need to consider directional biases, making it easier to predict and calculate sound-related quantities at different points.

In practice:
  • For problems involving isotropic sources, you can easily predict how intensity changes with distance. This allows straightforward application of the formula \( I = \frac{P}{4 \pi r^2} \).
  • As distance doubles, the intensity drops to one-quarter, due to the spherical spreading of sound.
Sound Level in Decibels
Sound level in decibels (dB) is a way to express the loudness of a sound, relative to a reference intensity. It uses a logarithmic scale, because human ears perceive sound intensity logarithmically. The equation is \( L = 10 \cdot \log_{10}\left(\frac{I}{I_0}\right) \), where \( I_0 = 10^{-12} \, \mathrm{W/m^2} \) is the threshold of hearing.

This scale helps compress a vast range of intensities into a manageable number. Unlike sound intensity, which can range enormously, sound levels range from 0 dB (threshold of hearing) to about 120-130 dB (threshold of pain).

Key points to remember:
  • The dB scale is not linear; an increase of 10 dB means the sound is perceived as twice as loud.
  • For our problem, calculating the sound level at \( 10 \mathrm{~m} \) involves substituting \( I = 0.0080 \, \mathrm{W/m^2} \) into the formula, yielding a sound level of \( 89 \, \mathrm{dB} \).

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Most popular questions from this chapter

Two sounds differ in sound level by \(1.00 \mathrm{~dB}\). What is the ratio of the greater intensity to the smaller intensity?

A stationary detector measures the frequency of a sound source that first moves at constant velocity directly toward the detector and then (after passing the detector) directly away from it. The emitted frequency is \(f\). During the approach the detected frequency is \(f_{\text {app }}^{\prime}\) and during the recession it is \(f_{\text {rec }}^{\prime}\). If \(\left(f_{\text {app }}^{\prime}-f_{\text {rec }}^{\prime}\right) / f=0.500,\) what is the ratio \(v_{s} / v\) of the speed of the source to the speed of sound?

Pipe \(A\) has only one open end; pipe \(B\) is four times as long and has two open ends. Of the lowest 10 harmonic numbers \(n_{B}\) of pipe \(B,\) what are the (a) smallest, (b) second smallest, and (c) third smallest values at which a harmonic frequency of \(B\) matches one of the harmonic frequencies of \(A ?\)

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