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Chapter 17: Problem 19

A point source emits \(30.0 \mathrm{~W}\) of sound isotropically. A small microphone intercepts the sound in an area of \(0.750 \mathrm{~cm}^{2}, 180 \mathrm{~m}\) from the source. Calculate (a) the sound intensity there and (b) the power intercepted by the microphone.

Short Answer

Expert verified
The sound intensity is \(7.37 \times 10^{-5} \, \text{W/m}^2\) and the power intercepted is \(5.53 \times 10^{-9} \, \text{W}\).

Step by step solution

01

Understanding the Problem

The task requires us to calculate two things: (a) the sound intensity at a certain distance from a point source and (b) the sound power intercepted by a microphone. We have a sound source emitting sound power of 30.0 W isotropically (evenly in all directions), and a microphone 180 meters away with an area of 0.750 cm虏.
02

Converting Units

Convert the area of the microphone from cm虏 to m虏 for consistency in SI units.Since 1 cm虏 = 0.0001 m虏, we have:\[0.750 \, \text{cm}^2 = 0.750 \, \times 0.0001 \, \text{m}^2 = 7.5 \times 10^{-5} \, \text{m}^2\]
03

Calculate the Surface Area of the Sphere

Determine the surface area of the sphere with a radius equal to the distance from the source:\[A = 4 \pi r^2 = 4 \pi (180 \, \text{m})^2\]Calculate this:\[A = 4 \pi (32400 \, \text{m}^2) = 407150.6 \, \text{m}^2\]
04

Calculate Sound Intensity

The sound intensity \( I \) is given by the power of the source divided by the surface area of the sphere:\[I = \frac{P}{A} = \frac{30.0 \, \text{W}}{407150.6 \, \text{m}^2}\] Compute the intensity:\[I = 7.37 \times 10^{-5} \, \text{W/m}^2\]
05

Calculate Power Intercepted by the Microphone

Use the formula for power intercepted, which is the product of intensity and area,\[P = I \times A_m = 7.37 \times 10^{-5} \, \text{W/m}^2 \times 7.5 \times 10^{-5} \, \text{m}^2\]Calculate the power intercepted:\[P = 5.53 \times 10^{-9} \, \text{W}\]
06

Final Calculation Check

Verify the calculations by re-evaluating each formula. Ensure unit consistency and logical reasoning in each step.

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Key Concepts

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

Point Source
In acoustics, a point source is a theoretical model used to simplify sound emission problems. Imagine a sound source so small that it can be considered as a single point in space. This helps in analyzing how sound spreads in various environments.
Sound from a point source spreads uniformly in all directions, forming spherical wavefronts. This assumption makes it easier to calculate how sound diminishes over distances.
Real-world sources are not perfect point sources, but this model offers a simplified way to understand sound propagation in large open areas.
  • Useful for simple calculations.
  • Assumes uniform sound propagation.
Power Emission
Power emission is the amount of sound energy a source releases into the environment per unit time. For the given problem, the source emits power of 30.0 watts. This value indicates the total energy output from the sound source.
The key aspect in such problems is that this power is continuously radiated in all directions, dispersing over a larger area as distance increases. Power emission is fundamental to calculating sound intensity, as it remains constant regardless of distance.
  • Total output energy from the source.
  • Measured in watts.
Spherical Wavefront
As sound travels away from a point source, it forms a series of expanding spheres centered on the point source, known as spherical wavefronts. The surface area of these spheres increases as the radius (distance from the source) increases.
The formula for the surface area of a sphere (4蟺r虏) helps determine how sound energy spreads out. In the given problem, sound travels 180 meters from the source and occupies a sphere with considerable area.
  • Surface area increases with distance.
  • Intensity decreases as area increases due to constant power spread over a larger surface.
Isotropic Sound Emission
When sound is emitted isotropically, it spreads equally in all directions from the source. This uniform distribution makes it easier to predict sound intensity at any point around the source.
In the problem context, isotropic emission helps us calculate how much sound reaches the microphone placed 180 meters away. Knowing that the sound disperses uniformly allows us to use the total surface area of the sphere to find sound intensity.
  • Uniform sound distribution in all directions.
  • Assumes no directional preferences or obstacles.

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Most popular questions from this chapter

The pressure in a traveling sound wave is given by the equation $$ \Delta p=(2.00 \mathrm{~Pa}) \sin \pi\left[\left(0.900 \mathrm{~m}^{-1}\right) x-\left(450 \mathrm{~s}^{-1}\right) t\right] $$ Find the (a) pressure amplitude, (b) frequency, (c) wavelength, and (d) speed of the wave.

Organ pipe \(A\), with both ends open, has a fundamental frequency of \(425 \mathrm{~Hz}\). The fifth harmonic of organ pipe \(B\), with one end open, has the same frequency as the second harmonic of pipe \(A\). How long are (a) pipe \(A\) and (b) pipe \(B ?\)

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?

When the door of the Chapel of the Mausoleum in Hamilton, Scotland, is slammed shut, the last echo heard by someone standing just inside the door reportedly comes \(15 \mathrm{~s}\) later. (a) If that echo were due to a single reflection off a wall opposite the door, how far from the door is the wall? (b) If, instead, the wall is \(32.0 \mathrm{~m}\) away, how many reflections (back and forth) occur?

What is the intensity at radial distances (a) \(2.50 \mathrm{~m}\) and (b) \(6.00 \mathrm{~m}\) from an isotropic point source of sound that emits energy at the rate \(12.0 \mathrm{~W}\), assuming no energy absorption by the surrounding air?
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