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Chapter 17: Problem 35

A point source emits \(30.0 \mathrm{~W}\) of sound isotropically. A small microphone intercepts the sound in an area of \(0.750 \mathrm{~cm}^{2}, 200 \mathrm{~m}\) from the source. Calculate (a) the sound intensity there and (b) the power intercepted by the microphone.

Short Answer

Expert verified
(a) Sound intensity is approximately \(5.97 \times 10^{-5}\) W/m虏. (b) Intercepted power is approximately \(4.48 \times 10^{-9}\) W.

Step by step solution

01

Understanding Sound Intensity

Sound intensity is defined as the sound power per unit area. For a point source emitting sound isotropically (equally in all directions), the intensity at a given distance can be found using the formula: \(I = \frac{P}{A}\) where \(P\) is the power of the source and \(A\) is the area of a sphere with a radius equal to the distance from the source.
02

Calculating the Surface Area at Given Distance

The area \(A\) of a sphere at a distance of 200 meters from the source can be calculated using the formula for the surface area of a sphere: \[ A = 4\pi r^2 \]Substituting \(r = 200 \text{ m}\), we find:\[ A = 4 \times \pi \times (200)^2 \approx 502,654.82 \text{ m}^2 \]
03

Finding Sound Intensity

Now we calculate the sound intensity \(I\) using the formula \(I = \frac{P}{A}\), where \(P = 30.0 \text{ W}\) and \(A = 502,654.82 \text{ m}^2\):\[ I = \frac{30.0}{502,654.82} \approx 5.97 \times 10^{-5} \text{ W/m}^2 \]
04

Calculating Power Intercepted by the Microphone

The power intercepted by the microphone can be calculated using the formula: \[ P_m = I \times A_m \] where \(A_m\) is the area of the microphone. Given \(A_m = 0.750 \text{ cm}^2 = 0.750 \times 10^{-4} \text{ m}^2\), we get:\[ P_m = 5.97 \times 10^{-5} \times 0.750 \times 10^{-4} \approx 4.48 \times 10^{-9} \text{ W} \]

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Key Concepts

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

Isotropic Emission
When we talk about isotropic emission, we are considering a situation where a sound source, like a speaker or any point source, emits sound waves equally in all directions. Imagine this as creating a big bubble of sound around the source. This is highly idealized because in the real world, few sources emit sound perfectly isotropically.
However, for theoretical purposes, isotropic emission helps simplify calculations and understand how sound disperses over distances.
This type of emission means that the intensity of sound, which is the sound power passing through a unit area, will be evenly distributed across the surface of a sphere surrounding the source. This concept is critical when calculating how sound weakens with distance from the source due to the expansion of the sphere's surface area.
Point Source
A point source is essentially an ideal sound source that emits sound from a single point in space. Think of it like the center of a sphere from which sound radiates outward uniformly.
In reality, no sound source is a true point because all have some physical size, but the concept of a point source simplifies the math needed to calculate sound intensity and allows predictions of how sound spreads.
Understanding this concept helps depict how sound energy dissipates over distance, and it aids in setting up the initial conditions of sound intensity calculations.
Microphone Power Interception
When a microphone collects sound energy, it is important to calculate how much of the sound power it actually receives. This is known as intercepting power.
To find how much power a microphone intercepts, we first need to know the sound intensity at the microphone's location. Sound intensity can be calculated as the power of the sound source divided by the area over which it spreads. Once we have the intensity, we multiply it by the area of the microphone to find the power intercepted.
This involves considering the actual surface area of the microphone, which is typically very small, to see how much sound power is captured in that tiny part of the whole sound field.
Surface Area Calculation
Calculating the surface area is key when working out sound intensity. For a point source that emits sound in all directions, the sound travels over the surface of a sphere as it moves away from the source. Therefore, the area is not just any area but specifically the surface area of that sphere.
The formula for the surface area of a sphere is \(A = 4\pi r^2\), where \(r\) is the radius or the distance from the source. This formula helps us determine the area over which the sound power is distributed at any given distance.
  • In our problem, the distance to the microphone is 200 meters.
  • The sphere's surface area with this radius is \(4\pi(200)^2\).
  • This value is used in other calculations to find out, among other things, sound intensity.
Having this surface area allows us to plug into formulas to find out how the sound behaves across that distance.

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Most popular questions from this chapter

A well with vertical sides and water at the bottom resonates at \(7.00 \mathrm{~Hz}\) and at no lower frequency. The air-filled portion of the well acts as a tube with one closed end (at the bottom) and one open end (at the top). The air in the well has a density of \(1.10 \mathrm{~kg} / \mathrm{m}^{3}\) and a bulk modulus of \(1.33 \times 10^{5} \mathrm{~Pa}\). How far down in the well is the water surface?

An acoustic burglar alarm consists of a source emitting waves of frequency \(28.0 \mathrm{kHz}\). What is the beat frequency between the source waves and the waves reflected from an intruder walking at an average speed of \(0.950 \mathrm{~m} / \mathrm{s}\) directly away from the alarm?

The speed of sound in a certain metal is \(v_{m}\). One end of a long pipe of that metal of length \(L\) is struck a hard blow. A listener at the other end hears two sounds, one from the wave that travels along the pipe's metal wall and the other from the wave that travels through the air inside the pipe. (a) If \(v\) is the speed of sound in air, what is the time interval \(\Delta t\) between the arrivals of the two sounds at the listener's ear? (b) If \(\Delta t=1.00 \mathrm{~s}\) and the metal is steel, what is the length \(L ?\)

Two sound waves, from two different sources with the same frequency, \(540 \mathrm{~Hz}\), travel in the same direction at \(330 \mathrm{~m} / \mathrm{s}\). The sources are in phase. What is the phase difference of the waves at a point that is \(4.40 \mathrm{~m}\) from one source and \(4.00 \mathrm{~m}\) from the other?

A violin string \(30.0 \mathrm{~cm}\) long with linear density \(0.650 \mathrm{~g} / \mathrm{m}\) is placed near a loudspeaker that is fed by an audio oscillator of variable frequency. It is found that the string is set into oscillation only at the frequencies 880 and \(1320 \mathrm{~Hz}\) as the frequency of the oscillator is varied over the range \(500-1500 \mathrm{~Hz}\). What is the tension in the string?
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