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The source of a sound wave has a power of \(1.00 \mu \mathrm{W}\). If it is a point source, (a) what is the intensity \(3.00 \mathrm{~m}\) away and (b) what is the sound level in decibels at that distance?

Short Answer

Expert verified
Intensity is \(8.84 \times 10^{-9} \mathrm{~W/m^2}\) and sound level is 39.5 dB.

Step by step solution

01

Understand the Given Values

We are given that the power of a sound source is \(1.00 \mu \mathrm{W}\), which is equal to \(1.00 \times 10^{-6} \mathrm{W}\). The distance from the source is \(3.00 \mathrm{~m}\). We need to find the intensity and sound level in decibels at this distance.
02

Calculate the Intensity

For a point source, intensity \(I\) is calculated using the formula: \[ I = \frac{P}{4 \pi r^2} \]where \(P\) is the power and \(r\) is the distance from the source.Plug in the values:\[ I = \frac{1.00 \times 10^{-6}}{4 \pi (3.00)^2} \]Calculate \(I\).
03

Solve for Intensity \(I\)

Calculate the denominator first:\[ 4 \pi (3.00)^2 = 4 \pi \times 9 = 36 \pi \approx 113.097 \]Now solve for \(I\):\[ I = \frac{1.00 \times 10^{-6}}{113.097} \approx 8.84 \times 10^{-9} \mathrm{~W/m^2} \]
04

Calculate the Sound Level in Decibels

The sound level (in decibels) \(L\) is calculated from intensity \(I\) using:\[ L = 10 \times \log_{10} \left( \frac{I}{I_0} \right) \]where \(I_0 = 1.00 \times 10^{-12} \mathrm{~W/m^2}\) is the reference intensity.Plug in the values:\[ L = 10 \times \log_{10} \left( \frac{8.84 \times 10^{-9}}{1.00 \times 10^{-12}} \right) \]
05

Solve for Sound Level \(L\)

Calculate the argument of the logarithm:\[ \frac{8.84 \times 10^{-9}}{1.00 \times 10^{-12}} = 8.84 \times 10^{3} \]Now calculate \(L\):\[ L = 10 \times \log_{10} (8.84 \times 10^{3}) \approx 10 \times (3.95) \approx 39.5 \text{ dB} \]

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Key Concepts

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

Sound Level
Sound level is a measure of the intensity or loudness of a sound and is often expressed in decibels (dB). This concept helps us to quantify the sounds we encounter in daily life, ranging from a whisper to a rock concert.
Sound level depends on the sound wave's intensity, which is the amount of energy or power that passes through a certain area. The higher the energy, the louder the sound.
Intensity and sound level are directly related, as we will explore in the decibels section.
Decibels
Decibels (dB) are a popular unit to measure sound level, providing a logarithmic scale that accommodates the wide range of sound intensities people can hear.
The formula to calculate sound level in decibels is given by:
  • \[ L = 10 \times \log_{10} \left( \frac{I}{I_0} \right) \]
where:
  • \(L\) is the sound level in decibels,
  • \(I\) is the intensity of the sound wave, and
  • \(I_0\) is the reference intensity, typically \(1.00 \times 10^{-12} \mathrm{~W/m^2}\).
Using this logarithmic scale is highly beneficial, as it allows us to compare vastly differing sound intensities, compressing them to a more manageable range. For example, going from quiet to loud might mean intensifying sound by thousands or even millions of times, but the decibel scale simplifies this to more comprehensible figures.
Point Source
A point source in acoustics refers to a sound source that emits sound waves radiating equally in all directions. Imagine if you dropped a stone in a pond. Water waves travel outwards equally from the point of impact. In the same way, sound from a point source travels uniformly in every direction in three-dimensional space.
The intensity of sound from a point source decreases as the distance from the source increases. The mathematical expression for intensity
  • \[ I = \frac{P}{4 \pi r^2} \]
shows this relationship, emphasizing that as distance \(r\) increases, intensity \(I\) reduces.
This is due to the sound energy spreading out over a larger area the further it goes from the source.
Logarithm
A logarithm is the power to which a number (the base) must be raised to obtain another number. In sound measurements, logarithms help to manage the vast range of audible sound intensities by changing these into smaller, more manageable numbers.
The decibel scale uses logarithms with base 10, written as \(\log_{10}\). Understanding logarithms is critical in acoustics, as they play a huge role in expressing the sound levels needed for human perception, fitting a broader possible intensity into a tighter numeric range.
Therefore, when you see sound levels expressed in decibels, you're seeing numbers that are derived from a logarithmic transformation that starts with a vast, sometimes incomprehensibly large range of sound intensities.

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Most popular questions from this chapter

What is the bulk modulus of oxygen if \(32.0 \mathrm{~g}\) of oxygen occupies \(22.4 \mathrm{~L}\) and the speed of sound in the oxygen is \(317 \mathrm{~m} / \mathrm{s} ?\)

Two sound waves with an amplitude of \(12 \mathrm{nm}\) and a wave- length of \(35 \mathrm{~cm}\) travel in the same direction through a long tube, with a phase difference of \(\pi / 3\) rad. What are the (a) amplitude and (b) wavelength of the net sound wave produced by their interference? If, instead, the sound waves travel through the tube in opposite directions, what are the (c) amplitude and (d) wavelength of the net wave?

Suppose a spherical loudspeaker emits sound isotropically at \(10 \mathrm{~W}\) into a room with completely absorbent walls, floor, and ceiling (an anechoic chamber). (a) What is the intensity of the sound at distance \(d=3.0 \mathrm{~m}\) from the center of the source? (b) What is the ratio of the wave amplitude at \(d=4.0 \mathrm{~m}\) to that at \(d=3.0 \mathrm{~m} ?\)

Diagnostic ultrasound of frequency \(4.50 \mathrm{MHz}\) is used to examine tumors in soft tissue. (a) What is the wavelength in air of such a sound wave? (b) If the speed of sound in tissue is \(1500 \mathrm{~m} / \mathrm{s}\), what is the wavelength of this wave in tissue?

A point source emits sound waves isotropically. The intensity of the waves \(2.50 \mathrm{~m}\) from the source is \(1.91 \times 10^{-4} \mathrm{~W} / \mathrm{m}^{2}\). Assuming that the energy of the waves is conserved, find the power of the source.

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