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Chapter 17: Problem 35

A point source emits \(30.0\) 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) The sound intensity is approximately 5.97 x 10^{-5} W/m虏. (b) The power intercepted is approximately 4.48 x 10^{-9} W.

Step by step solution

01

Understand the Problem

We are given a point source that emits sound isotropically with a power of 30.0 W. The sound is intercepted by a microphone with an area of 0.750 cm虏, located 200 meters away from the source. We need to calculate the sound intensity at the microphone's location and the power intercepted by the microphone.
02

Convert Units

The area of the microphone is provided in cm虏. First, we need to convert this area into meters squared to keep consistent SI units. Given: 1 cm虏 = 1e-4 m虏 Thus, 0.750 cm虏 = 0.750 * 1e-4 m虏 = 0.000075 m虏.
03

Calculate Sound Intensity

Sound Intensity (I) is defined as the power (P) per unit area (A). For a point emitting sound isotropically, the intensity at a distance r can be calculated as: \[ I = \frac{P}{4 \pi r^2} \] Given:P = 30.0 W, r = 200 m. Substitute the values into the formula to find the intensity: \[ I = \frac{30.0}{4 \pi (200)^2} \] \[ I = \frac{30.0}{4 \pi \times 40000} \] \[ I = \frac{30.0}{502654.82} \] \[ I \approx 5.97 \times 10^{-5} \text{ W/m}^2 \]
04

Calculate Power Intercepted

The power intercepted by the microphone (P_{mic}) can be calculated using the sound intensity (I) at the microphone and the area (A) of the microphone. \[ P_{mic} = I \times A \] Substitute the intensity (from Step 3) and the converted area from Step 2: \[ P_{mic} = 5.97 \times 10^{-5} \times 0.000075 \] \[ P_{mic} \approx 4.48 \times 10^{-9} \text{ W} \]

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Key Concepts

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

Isotropic Sound Distribution
When sound is emitted isotropically from a source, it spreads out uniformly in all directions, much like ripples on a water surface. This distribution of sound can be visualized as a series of expanding concentric spheres surrounding the source. At any given point on these spheres, the intensity of sound will be equal, assuming no obstacles are present to disturb the sound waves. Understanding isotropic distribution is crucial because it forms the basis for calculating sound intensity at any distance from the source. The further you move from the source, the sound energy spreads over a larger area, reducing its intensity.
Point Source Emission
A point source is an idealized source that emits sound equally in all directions. In physics, this concept simplifies calculations by allowing the sound field to be symmetrical. For practical applications, this means treating everyday sound sources, like a speaker, as if they are small points that spread sound uniformly. The defining characteristic of a point source is its uniform emission, leading to a predictable decrease in intensity with distance. The assumption of a point source helps in deriving the formula for sound intensity:
  • Intensity decreases with the square of the distance from the source.
  • This behavior is described by the formula: \[ I = \frac{P}{4 \pi r^2} \] where \( I \) is the intensity, \( P \) is power, and \( r \) is the distance from the source.
Acoustic Power Calculation
Acoustic power refers to the total sound power emitted by the source in all directions. It is typically measured in watts (W). This is a fundamental quantity that helps determine how sound behaves at different distances. To calculate acoustic power intercepted by a small area, such as a microphone, you must first determine sound intensity, which reflects how much power flows through a unit area. In dealing with acoustic power, you often need:
  • The power emitted by the source (given in the problem).
  • Distance from the source to the point of measurement (needed to find intensity).
  • Area through which the sound passes (like the area of a microphone).
Once you have intensity, power intercepted by an area is calculated by multiplying intensity by the area of interest.
Unit Conversion in Physics
Unit conversion in physics is often necessary to maintain consistent units within a calculation. In sound-related problems, lengths and areas frequently require conversion to the International System of Units (SI). For example, converting area measurements from square centimeters to square meters. Here's how to perform a basic conversion:
  • Identify the units to be converted (e.g., \(\text{cm}^2 \) to \(\text{m}^2\)).
  • Apply the conversion factor: 1 \(\text{cm}^2 = 10^{-4} \text{m}^2\).
  • Multiply the given area by this factor to get the area in \(\text{m}^2\).
Correct unit conversion is essential to ensure the equations work as intended and the results are accurate. This attention to detail helps avoid errors in calculations which may otherwise lead to incorrect results.

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Most popular questions from this chapter

A \(1.0 \mathrm{~W}\) point source emits sound waves isotropically. Assuming that the energy of the waves is conserved, find the intensity (a) \(1.0 \mathrm{~m}\) from the source and (b) \(2.5 \mathrm{~m}\) from the source.

A point source that is stationary on an \(x\) axis emits a sinusoidal sound wave at a frequency of \(686 \mathrm{~Hz}\) and speed \(343 \mathrm{~m} / \mathrm{s}\). The wave travels radially outward from the source, causing air molecules to oscillate radially inward and outward. Let us define a wavefront as a line that connects points where the air molecules have the maximum, radially outward displacement. At any given instant, the wavefronts are concentric circles that are centered on the source. (a) Along \(x\), what is the adjacent wavefront separation? Next, the source moves along \(x\) at a speed of \(110 \mathrm{~m} / \mathrm{s}\). Along \(x\), what are the wavefront separations (b) in front of and (c) behind the source?

A continuous sinusoidal longitudinal wave is sent along a very long coiled spring from an attached oscillating source. The wave travels in the negative direction of an \(x\) axis; the source frequency is \(25 \mathrm{~Hz}\); at any instant the distance between successive points of maximum expansion in the spring is \(24 \mathrm{~cm}\); the maximum longitudinal displacement of a spring particle is \(0.30 \mathrm{~cm} ;\) and the particle at \(x=0\) has zero displacement at time \(t=0\). If the wave is written in the form \(s(x, t)=s_{m} \cos (k x \pm \omega t)\), what are (a) \(s_{m}\), (b) \(k\), (c) \(\omega\), (d) the wave speed, and (e) the correct choice of sign in front of \(\omega ?\) Incident

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?

Pipe \(A\), which is \(1.20 \mathrm{~m}\) 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 \(B\) ?
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