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Chapter 16: Problem 67

A listener doubles his distance from a source that emits sound uniformly in all directions. There are no reflections. By how many decibels does the sound intensity level change?

Short Answer

Expert verified
The sound intensity level decreases by 6 decibels.

Step by step solution

01

Understanding Sound Intensity and Distance

The intensity of sound depends inversely on the square of the distance from the source. This means if the distance is doubled, the intensity decreases to a quarter of its original value. Mathematically, if the original intensity is \( I_0 \) at distance \( d \), at distance \( 2d \), the new intensity \( I \) is \( I = \frac{I_0}{4} \).
02

Sound Intensity Level in Decibels

Sound intensity level \( eta \) in decibels is given by \( eta = 10 \log_{10} \left( \frac{I}{I_0} \right) \), where \( I \) is the intensity at the new distance, and \( I_0 \) is the reference intensity. We need the change in decibels when the listener is at a new distance, i.e., \( eta - eta_0 \).
03

Calculate the Intensity Level Change

Substitute \( I = \frac{I_0}{4} \) into the decibel formula: \( eta - eta_0 = 10 \log_{10} \left( \frac{\frac{I_0}{4}}{I_0} \right) = 10 \log_{10} \left( \frac{1}{4} \right) = 10 \log_{10} \left( 4^{-1} \right) = 10 \times (-1) \log_{10} (4) = -10 \log_{10} (4) \).
04

Approximate and Final Calculation

Using \( \log_{10} (4) \approx 0.6 \), we approximate \( -10 \log_{10} (4) \approx -10 \times 0.6 = -6 \) decibels. Hence, the change in the sound intensity level is a decrease of 6 decibels.

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Key Concepts

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

Decibels
Decibels are the unit used to measure sound intensity levels. They provide an easy-to-understand way of comparing different sound intensities. By using the decibel scale, we can express very large or small numbers in a more manageable form. The formula for calculating the intensity level of sound in decibels is given by:\[ \text{Sound Intensity Level (dB)} = 10 \log_{10} \left( \frac{I}{I_0} \right) \]where:
  • \( I \) is the intensity of the sound.
  • \( I_0 \) is a reference intensity, typically the threshold of hearing.
The reason decibels use a logarithmic scale is primarily due to the huge range of sound intensities that humans can perceive, from the faintest whisper to the roar of a jet engine. The logarithmic scale helps compress this wide range so it's easier to work with, and allows us to compare sound levels with a simple addition of decibels.
Inverse Square Law
The inverse square law is a principle that describes how sound intensity decreases as you move away from the source. Specifically, it implies that intensity is inversely proportional to the square of the distance from the source. Mathematically, if you double your distance from a sound source, the sound intensity becomes:\[ I = \frac{I_0}{4} \]This means the intensity drops to a quarter of its original value when the distance is doubled.This principle is crucial in understanding how sound diminishes over distance. It affects not just sound, but also light and other types of waves that spread out from a point source. This drop in intensity with increasing distance is why sounds appear fainter the further away you are.
Logarithmic Scale
The logarithmic scale is an effective tool used in various scientific disciplines, including acoustics. It's essential for measuring phenomena that have a wide range of values, like sound intensity. A logarithmic scale is constructed such that each unit increase on this scale represents a tenfold increase in the quantity being measured. This allows us to manage very small and very large numbers efficiently. For example:
  • The noise level from a quiet library might be about 30 decibels.
  • A busy street produces sound levels around 70 decibels.
  • A rock concert could hit around 110 decibels.
With a logarithmic scale, when sound intensity doubles, the decibel level doesn't double鈥攖his is due to the way the logarithmic function works. This means that a seemingly small shift in decibels can represent a substantial change in sound intensity, helping us better understand and manage noise in our environments.

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Most popular questions from this chapter

A loudspeaker in a parked car is producing sound whose frequency is \(20510 \mathrm{Hz} .\) A healthy young person with normal hearing is standing nearby on the sidewalk but cannot hear the sound because the frequency is too high. When the car is moving, however, this person can hear the sound. (a) Is the car moving toward or away from the person? Why? (b) If the speed of sound is \(343 \mathrm{m} / \mathrm{s},\) what is the minimum speed of the moving car?

In a discussion person A is talking 1.5 dB louder than person B, and person \(\mathrm{C}\) is talking \(2.7 \mathrm{dB}\) louder than person A. What is the ratio of the sound intensity of person \(\mathrm{C}\) to the sound intensity of person \(\mathrm{B} ?\)

A spider hangs from a strand of silk whose radius is \(4.0 \times 10^{-6} \mathrm{m}\). The density of the silk is \(1300 \mathrm{kg} / \mathrm{m}^{3} .\) When the spider moves, waves travel along the strand of silk at a speed of \(280 \mathrm{m} / \mathrm{s}\). Ignore the mass of the silk strand, and determine the mass of the spider.

The speed of a transverse wave on a string is \(450 \mathrm{m} / \mathrm{s},\) and the wavelength is \(0.18 \mathrm{m}\). The amplitude of the wave is \(2.0 \mathrm{mm}\). How much time is required for a particle of the string to move through a total distance of \(1.0 \mathrm{km} ?\)

A woman stands a distance \(d\) from a loud motor that emits sound uniformly in all directions. The sound intensity at her position is an uncomfortable \(3.2 \times 10^{-3} \mathrm{W} / \mathrm{m}^{2} .\) There are no reflections. At a position twice as far from the motor, what are (a) the sound intensity and (b) the sound intensity level relative to the threshold of hearing?
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