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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 approximately 6.02 decibels.

Step by step solution

01

Title

Step 1: Understand the Inverse Square Law for Sound IntensitySound intensity (I) is inversely proportional to the square of the distance (r) from the source:\[I \propto \frac{1}{r^2}\] When the distance doubles, the intensity becomes one-fourth.
02

Determine Initial and Final Intensity Levels

Step 2: Calculate Initial and Final IntensityLet's consider the initial intensity asI_1, and the new intensity when the distance is doubled asI_2. Using the inverse square law:\[I_2 = \frac{I_1}{4}\]
03

Calculate the Change in Decibels

Step 3: Use the Decibel Formula for Intensity Level ChangeThe change in decibel level (\Delta L) is calculated using the formula:\[\Delta L = 10 \cdot \log_{10}\left(\frac{I_2}{I_1}\right)\]Substitute for\frac{I_2}{I_1}:\[\Delta L = 10 \cdot \log_{10}\left(\frac{1}{4}\right) = 10 \cdot (-0.602) \approx -6.02\text{ dB}\]Therefore, the intensity level decreases by approximately 6.02 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 fundamental principle in physics that describes how a quantity like sound intensity diminishes with distance. Specifically, it tells us that sound intensity is inversely proportional to the square of the distance from the source. Think of it this way: if you double the distance from a sound source, the intensity isn't just halved; it's reduced to one-fourth. This is because the intensity at the further location has to spread out over a larger spherical area.
  • Sound spreads out uniformly in all directions from the source.
  • Distance increases, intensity decreases rapidly.
  • For a doubled distance, intensity changes by a factor of 1/4.
Understanding this principle is crucial in situations where accurate distance-based sound intensity calculations are required, such as in acoustics, engineering, and even astronomy.
Sound Intensity
Sound intensity is the amount of sound power that travels through a specific area. It is usually measured in watts per square meter ( ext{W/m}^2). Intensity isn't something that stays constant; it varies based on how far you are from the source of the sound.
  • Measured in watts per square meter ( ext{W/m}^2).
  • Varies with distance, becoming weaker as you move away from the source.
  • Influenced by the environment and the medium it travels through.
When considering sound intensity, always remember the effect of distance and how it interacts with various environments, like indoors where walls may reflect the sound.
Logarithms
Logarithms are a mathematical concept widely used in calculating changes in sound intensity levels. The decibel scale, which measures sound intensity, is logarithmic. This means it uses powers of ten to express ratios.
  • The formula for decibels is based on a logarithm: \[L = 10 \cdot \log_{10}\left(\frac{I}{I_0}\right)\]where \(L\) is the sound level in decibels, \(I\) is the intensity being measured, and \(I_0\) is a reference intensity.
  • A logarithmic scale is particularly useful in sound measurements because of the wide range of sound intensities the human ear can detect.
  • Helps transform multiplicative scales of sound intensity into more manageable additive scales.
Getting a grip on logarithms is essential for understanding how sound changes in different settings, especially when distances increase or decrease.
Physics Problems
Physics problems, particularly in sound, often require the application of several mathematical principles to reach a solution. These problems help in understanding how concepts like the Inverse Square Law, sound intensity, and logarithms intertwine.
  • Such problems develop critical thinking by requiring multiple steps and the application of various formulas.
  • Helps in understanding real-world applications, such as why sound might be quieter at a distance during a concert.
  • Solving problems step-by-step can illustrate how theoretical knowledge applies practically.
By practicing physics problems in sound, you embrace not only the calculations but also the reasoning behind sound behavior. This can be incredibly useful in a range of professions, including engineering and audio technology.

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Most popular questions from this chapter

When Gloria wears her hearing aid, the sound intensity level increases by 30.0 dB. By what factor does the sound intensity increase?

A hunter is standing on flat ground between two vertical cliffs that are directly opposite one another. He is closer to one cliff than to the other. He fires a gun and, after a while, hears three echoes. The second echo arrives 1.6 s after the first, and the third echo arrives 1.1 s after the second. Assuming that the speed of sound is 343 m/s and that there are no reflections of sound from the ground, find the distance between the cliffs.

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

A source of sound is located at the center of two concentric spheres, parts of which are shown in the drawing. The source emits sound uniformly in all directions. On the spheres are drawn three small patches that may or may not have equal areas. However, the same sound power passes through each patch. The source produces 2.3 W of sound power, and the radii of the concentric spheres are \(r_{\mathrm{A}}=0.60\) m and \(r_{\mathrm{B}}=0.80 \mathrm{m} .\) (a) Determine the sound intensity at each of the three patches. \((\mathrm{b})\) The sound power that passes through each of the patches is \(1.8 \times 10^{-3} \mathrm{W} .\) Find the area of each patch.

You are riding your bicycle directly away from a stationary source of sound and hear a frequency that is 1.0% lower than the emitted frequency. The speed of sound is 343 m/s. What is your speed?
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