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The source of a sound wave has a power of \(3.00 \mu \mathrm{W}\). If it is a point source, (a) what is the intensity \(4.20 \mathrm{~m}\) away and (b) what is the sound level in decibels at that distance?

Short Answer

Expert verified
The intensity is approximately 1.35 \times 10^{-8} W/m^2, and the sound level is 41.3 dB.

Step by step solution

01

Understand Power and Intensity

The power of the sound source is given as \(3.00 \mu \mathrm{W}\), which is \(3.00 \times 10^{-6} \mathrm{W}\). Intensity is defined as the power per unit area. For a point source spreading spherically, the intensity \(I\) at a distance \(r\) is calculated using the formula: \[ I = \frac{P}{4\pi r^2} \] where \(P\) is the power and \(r\) is the distance.
02

Calculate Intensity at 4.20 m

Substitute the known values into the intensity formula where \(P = 3.00 \times 10^{-6} \mathrm{W}\) and \(r = 4.20 \mathrm{~m}\):\[ I = \frac{3.00 \times 10^{-6}}{4\pi (4.20)^2} \]Calculate the denominator:\[ 4\pi (4.20)^2 = 4\pi \times 17.64 \approx 221.67 \]Hence, the intensity is:\[ I = \frac{3.00 \times 10^{-6}}{221.67} \approx 1.35 \times 10^{-8} \mathrm{~W/m^2} \]
03

Understand Sound Level in Decibels

The sound level \(L\) in decibels (dB) is calculated using the formula:\[ L = 10 \cdot \log_{10}\left(\frac{I}{I_0}\right) \]where \(I\) is the intensity calculated earlier and \(I_0 = 1 \times 10^{-12} \mathrm{W/m^2}\) is the reference intensity level.
04

Calculate Sound Level at 4.20 m

Substitute the intensity \(I = 1.35 \times 10^{-8} \mathrm{~W/m^2}\) into the decibel formula:\[ L = 10 \cdot \log_{10}\left(\frac{1.35 \times 10^{-8}}{1 \times 10^{-12}}\right) \]Calculate the ratio:\[ \frac{1.35 \times 10^{-8}}{1 \times 10^{-12}} = 1.35 \times 10^{4} \]Hence, the sound level is:\[ L = 10 \cdot \log_{10}(1.35 \times 10^{4}) \approx 10 \cdot 4.13 = 41.3 \mathrm{~dB} \]

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Key Concepts

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

Sound Power
Sound power is the total energy emitted by a sound source per second, measured in watts (W). It represents the acoustic energy output of the source, which can be thought of as how loud a sound source inherently is before considering how it spreads in the environment.
For instance, in our exercise, the source has a sound power of \(3.00 \mu \mathrm{W}\), or \(3.00 \times 10^{-6} \mathrm{W}\). This tiny amount of energy can still produce an audible sound because of how sound waves dramatically decrease in intensity as they spread out in space. The power of a sound source remains constant but how we perceive its loudness differs based on our position relative to the source.
Thus, understanding sound power is fundamental when dealing with sound waves, as it forms the basis upon which other effects like intensity and sound level are calculated.
Sound Level in Decibels
Sound level in decibels (dB) is a logarithmic measurement of the intensity of sound, compared to a reference intensity level. This measurement provides a way to express sound levels in terms that reflect how humans perceive differences in loudness.
To calculate the sound level in decibels, use the formula:
\[ L = 10 \cdot \log_{10}\left(\frac{I}{I_0}\right) \]
where \(I\) is the intensity of the sound wave, and \(I_0 = 1 \times 10^{-12} \mathrm{W/m^2}\) is the reference intensity level, representing the threshold of hearing for humans. In the example, the sound level was calculated as \(41.3 \mathrm{~dB}\).
Understanding decibels is crucial because it explains why a small increase in intensity results in a larger perceivable change in sound level. This logarithmic nature is why each increase of 10 dB represents a tenfold increase in perceived loudness.
Point Source Sound Waves
When a sound source is a point source, it emits sound waves uniformly in all directions, forming a spherical wavefront. This type of source simplifies calculations because we can assume symmetry in the way sound spreads out over distance.
For a point source, as distance from the source increases, the intensity dramatically decreases because the same amount of energy is spread over a larger area. The formula for intensity from a point source is:
\[ I = \frac{P}{4\pi r^2} \]
where \(P\) is the sound power and \(r\) is the distance from the source. The calculation shows that intensity decreases with the square of the distance, which is known as the inverse square law.
This explanation helps us understand why we need to be specific about distance when measuring sound intensity and why two identical sound sources can seem so different depending on how far away they are. A sound source's power doesn't change, but the intensity can widely vary based on distance.

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Most popular questions from this chapter

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 \(\left(f_{\text {app }}^{\prime}-f_{\text {rec }}^{\prime}\right) / f=\) \(0.200\), what is the ratio \(v_{s} / v\) of the speed of the source to the speed of sound?

The crest of a Parasaurolophus dinosaur skull is shaped somewhat like a trombone and contains a nasal passage in the form of a long, bent tube open at both ends. The dinosaur may have used the passage to produce sound by setting up the fundamental mode in it. (a) If the nasal passage in a certain Parasaurolophus fossil is \(1.8 \mathrm{~m}\) long, what frequency would have been produced? (b) If that dinosaur could be recreated (as in Jurassic Park), would a person with a hearing range of \(60 \mathrm{~Hz}\) to \(20 \mathrm{kHz}\) be able to hear that fundamental mode and, if so, would the sound be high or low frequency? Fossil skulls that contain shorter nasal passages are thought to be those of the female Parasaurolophus. (c) Would that make the female's fundamental frequency higher or lower than the male's?

Pipe \(A\), which is \(1.80 \mathrm{~m}\) long and open at both ends, oscillates at its third lowest harmonic frequency. It is filled with air for which the speed of sound is \(343 \mathrm{~m} / \mathrm{s}\). Pipe \(B\), which is closed at one end, oscillates at its second lowest harmonic frequency. This frequency of \(B\) happens to match the frequency of \(A\). An \(x\) axis extends along the interior of \(B\), with \(x=0\) at the closed end. (a) How many nodes are along that axis? What are the (b) smallest and (c) second smallest value of \(x\) locating those nodes? (d) What is the fundamental frequency of \(B\) ?

Approximately a third of people with normal hearing have ears that continuously emit a low-intensity sound outward through the ear canal. A person with such spontaneous otoacoustic emission is rarely aware of the sound, except perhaps in a noise-free environment, but occasionally the emission is loud enough to be heard by someone else nearby. In one observation, the sound wave had a frequency of \(1200 \mathrm{~Hz}\) and a pressure amplitude of \(2.50 \times 10^{-3} \mathrm{~Pa}\). What were (a) the displacement amplitude and (b) the intensity of the wave emitted by the ear?

A bat is flitting about in a cave, navigating via ultrasonic bleeps. Assume that the sound emission frequency of the bat is \(39000 \mathrm{~Hz}\). During one fast swoop directly toward a flat wall surface, the bat is moving at \(0.020\) times the speed of sound in air. What frequency does the bat hear reflected off the wall?

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