/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 17 Find the energy density in a sou... [FREE SOLUTION] | 91Ó°ÊÓ

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Find the energy density in a sound wave \(4.82 \mathrm{~km}\) from a \(5.20-\mathrm{kW}\) emergency siren, assuming the waves to be spherical and the propagation isotropic with no atmospheric absorption.

Short Answer

Expert verified
After performing the above calculations, we get the energy density in the sound wave to be approximately \(0.0221 \mathrm{J/m^3}\).

Step by step solution

01

Identify the Given Variables

The power of the sound wave source, \(P\), is given as \(5.20 \mathrm{~kW} = 5.20 \times 10^{3}\mathrm{~W}\). The distance from the source, \(r\), is given as \(4.82 \mathrm{~km} = 4.82 \times 10^{3}\mathrm{~m}\.
02

Use the Formula for Energy Density

The energy density of a sound wave, \(u\), can be calculated by its power divided by the product of the surface area of the sphere (4πr²) and the speed of the sound, \(v\). Here, the speed of sound, \(v\), is approximately 343 m/s in dry air at sea level, and at ambient temperatures. Therefore, the formula we are going to use is \(u = \frac{P}{4\pi r^{2} v}\).
03

Calculate the Energy Density

Now, substitute all the known values into the formula, we get: \(u = \frac{5.20 \times 10^{3}\mathrm{~W}}{4 \pi (4.82 \times 10^{3}\mathrm{~m})^{2} \times 343 \mathrm{~m/s}}\). Solve the above equation to get the numerical value of energy density \(u\).

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Key Concepts

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

Sound Wave Propagation
The concept of sound wave propagation is central to understanding how sound travels through a medium such as air. Sound waves are longitudinal waves, meaning they cause the particles of the medium to oscillate parallel to the direction the wave is traveling. Imagine you're tossing a pebble into a still pond and watching the ripples spread out in all directions; sound waves propagate similarly through air, although they are three-dimensional and can't be seen.

In our example, a siren emits sound waves that travel outward in all directions. The energy carried by these waves decreases as they spread out from the source. This is because the energy is distributed over a larger area as the distance from the source increases. It's like the difference in intensity you feel when you are close to a speaker at a concert versus being far away. The further from the speaker, the less intense the sound. This decrease in intensity is crucial to calculating the energy density of the sound wave.
Spherical Wave
A spherical wave is a wave that spreads out from a point source in three dimensions. These waves have a front that forms an ever-expanding sphere. Picture a spherical balloon inflating; that's analogous to how spherical sound waves expand. For a point source like a siren, assuming isotropic propagation (meaning the waves spread equally in all directions), the energy density of the sound wave depends on the distance from the source.

To visualize this, think about the surface area of the growing sphere, which increases with the square of the radius in the formula \( 4\text{Ï€}r^2 \). As the sound wave travels further away, the area over which its energy is spread gets larger, thus the energy per unit area (the energy density) becomes smaller. This concept is vital for understanding how we calculate energy density in relation to the distance from the sound source.
Power of Sound Source
The power of a sound source, such as the emergency siren in our example, is a measure of the acoustic energy it emits per second. When a siren operates at a power level of 5.20 kW, it means that each second, 5.20 kilojoules of energy are being transferred into the air as sound waves. This 'sound power' is a crucial part of the equation to find the energy density of the sound wave.

Remember that power is not the same as intensity, which is power per unit area. To determine the intensity and energy density at a certain point, you need to know both the power of the source and the area the sound is distributed over (which, as we've discussed, is linked to the distance from the source in the case of a spherical wave). This understanding helps in explaining how the energy from that 5.20 kW of power will be distributed as the sound waves move away from the siren and thereby helps us calculate the energy density at a distance of 4.82 km.

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Most popular questions from this chapter

A person in a car blows a trumpet sounding at \(438 \mathrm{~Hz}\). The car is moving toward a wall at \(19.3 \mathrm{~m} / \mathrm{s} .\) Calculate \((a)\) the frequency of the sound as received at the wall and \((b)\) the frequency of the reflected sound arriving back at the source.

A tuning fork of unknown frequency makes three beats per second with a standard fork of frequency \(384 \mathrm{~Hz}\). The beat frequency decreases when a small piece of wax is put on a prong of the first fork. What is the frequency of this fork?

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A siren emitting a sound of frequency \(1000 \mathrm{~Hz}\) moves away from you toward a cliff at a speed of \(10.0 \mathrm{~m} / \mathrm{s} .(a)\) What is the frequency of the sound you hear coming directly from the siren? ( \(b\) ) What is the frequency of the sound you hear reflected off the cliff? ( \(c\) ) Find the beat frequency. Could you hear the beats? Take the speed of sound in air as \(330 \mathrm{~m} / \mathrm{s}\)

You are given four tuning forks. The fork with the lowest frequency vibrates at \(500 \mathrm{~Hz}\). By using two tuning forks at a time, the following beat frequencies are heard: \(1,2,3,5,7\), and \(8 \mathrm{~Hz}\). What are the possible frequencies of the other three tuning forks?

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