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Chapter 11: Problem 11

If you double your distance from a sound source, how does the sound intensity change? What about the intensity level?

Short Answer

Expert verified
Doubling the distance reduces sound intensity to a quarter and decreases the intensity level by 6 dB.

Step by step solution

01

Understanding Sound Intensity

Sound intensity is defined as the power per unit area. It is often measured in watts per square meter ( ext{W/m}^2). If you have a sound source radiating equally in all directions, the intensity decreases as you move further from it because the same amount of energy is spread over a larger area.
02

Formula for Area and Distance

As sound spreads out from a source, it forms a sphere. The surface area of the sphere is given by the formula \( A = 4\pi r^2 \), where \( r \) is the distance from the source. When you double the distance, the new area becomes \( 4\pi (2r)^2 = 16\pi r^2 \), four times the original area.
03

Calculating New Intensity

Since intensity ( ext{I}) is inversely proportional to the area (\( A \)), doubling the distance decreases the intensity by a factor of four. Mathematically, \( ext{I}_{new} = rac{ ext{I}_{original}}{4} \).
04

Understanding Intensity Level

Intensity level, measured in decibels (dB), is calculated using the formula \( ext{L} = 10 imes ext{log}_{10} rac{ ext{I}}{ ext{I}_0} \), where \( ext{I}_0 \) is the reference intensity level, usually \( 10^{-12} ext{W/m}^2 \).
05

Calculating Change in Intensity Level

When intensity is reduced by a factor of 4, the change in intensity level, \( ext{L}_{change} \), is given by \( ext{L}_{change} = 10 imes ext{log}_{10} rac{1}{4} \). This calculates to \( -6 ext{dB} \). Therefore, doubling the distance decreases the intensity level by 6 decibels.

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Key Concepts

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

Inverse Square Law
The Inverse Square Law is a crucial principle that helps us understand how sound intensity behaves as a function of distance. Essentially, this law states that intensity is inversely proportional to the square of the distance from the sound source.
  • This means if you increase the distance from the sound source, the intensity decreases according to the square of that distance increase.
  • As an example, if you double the distance from the sound source, the intensity spreads over four times the area. This reduces the sound intensity by a factor of four.
This principle helps explain why sounds become quieter as you move away from the source. Every doubling of distance results in a quarter of the original intensity. This concept is essential for understanding phenomena in acoustics and is widely applicable in physics when dealing with waves and fields.
Intensity Level
Intensity level is a way to express the loudness of a sound using a logarithmic scale. This makes it easier to compare different sounds because human hearing perceives sound intensity logarithmically.Here's how intensity level works:
  • It's measured in decibels (dB), providing a more manageable range of numbers than using raw intensity values.
  • The formula for calculating the intensity level is:\[\text{L} = 10 \times \log_{10}\frac{\text{I}}{\text{I}_0}\]where \( \text{I} \) is the sound intensity and \( \text{I}_0 \) is the reference intensity, usually \( 10^{-12} \text{W/m}^2 \), which is the threshold of hearing.
  • This logarithmic relationship implies that a small change in intensity leads to a significant change in intensity level.
If the intensity of sound is reduced to a quarter of its original value, the intensity level decreases by approximately 6 dB. This scaling helps us understand and measure sound in a way that aligns better with human perception.
Decibels
Decibels are the units used to quantify sound intensity level, offering a scale that matches human auditory perception. Using decibels allows for a more intuitive sense of changes in sound levels, even though intensity changes can be massive. Here's why decibels are useful:
  • Decibels provide a relative measure of sound intensity compared to a standard reference point.
  • They are based on a logarithmic scale, making them well-suited to capture the wide range of sound intensities we encounter daily.
  • The log scale means that every 10 dB increase represents a tenfold increase in intensity, while every 10 dB decrease represents a tenfold reduction.
This system makes it easier to compare sounds of very different intensities. Understanding decibels is key to analyzing audio environments, designing sound systems, and protecting hearing health. When we say a sound is 6 dB lower, it means the intensity has decreased significantly, often missed if we only considered it linearly.

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Most popular questions from this chapter

The human eardrum is roughly circular with a diameter of approximately \(1.0 \mathrm{~cm} .\) Find the total power impinging on the eardrum when it's subject to \(85-\mathrm{dB}\) sound.

A water wave has frequency \(0.50 \mathrm{~Hz}\), amplitude \(0.35 \mathrm{~m}\), and wavelength \(2.6 \mathrm{~m}\). If the amplitude doubles, the wavelength becomes (a) \(5.2 \mathrm{~m} ;\) (b) \(2.6 \mathrm{~m} ;\) (c) \(1.3 \mathrm{~m} ;\) (d) \(1.0 \mathrm{~m}\)

At a rock concert you're \(15 \mathrm{~m}\) from the stage, where the intensity level is \(105 \mathrm{dB.}\) How far away should you move in order to reduce the intensity level to a slightly more bearable \(92 \mathrm{~dB}\) ?

An organ pipe is designed to play at \(22 \mathrm{~Hz},\) near the frequency threshold for human hearing. Find the pipe's length if it's open (a) at both ends and (b) at one end.

Person A can barely hear a sound at a particular frequency with an intensity level of \(2.4 \mathrm{dB.}\) Person \(\mathrm{B}\), who has hearing loss, can barely hear a \(9.4-\mathrm{dB}\) tone with the same frequency. Find the ratio of sound intensities at these two hearing thresholds.
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