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Chapter 15: Problem 27

Energy Output. By measurement you determine that sound waves are spreading out equally in all directions from a point source and that the intensity is \(0.026 \mathrm{~W} / \mathrm{m}^{2}\) at a distance of \(4.3 \mathrm{~m}\) from the source. (a) What is the intensity at a distance of \(3.1 \mathrm{~m}\) from the source? (b) How much sound energy does the source emit in one hour if its power output remains constant?

Short Answer

Expert verified
a) The intensity at a distance of 3.1 m from the source is 0.048 \, \mathrm{W} / \mathrm{m}^{2}. b) The sound energy emitted by the source in one hour is 128332.33 J.

Step by step solution

01

Apply the Inverse Square Law

To find the new intensity at a distance of 3.1 m, apply the inverse square law. This law states that the intensity of a wave is inversely proportional to the square of the distance from the source, i.e., \(I_2 = I_1 * (d_1 / d_2)^2\), where \(I_1\) is the initial intensity, \(d_1\) the initial distance, \(I_2\) the final intensity, and \(d_2\) the final distance. Substituting the given values, \(I_2 = 0.026 * (4.3 / 3.1)^2\).
02

Calculate the New Intensity

Now, perform the calculation for \(I_2\). So, \(I_2 = 0.026 * (4.3 / 3.1)^2 = 0.048 \, \mathrm{W} / \mathrm{m}^{2}\).
03

Apply Energy Time Relation

For the second part, the energy emitted by the source in one hour is calculated by using the formula for power \(P = E / t\), where \(P\) is the power, \(E\) the total energy, and \(t\) the time period. Given that power output remains constant, the power is equal to the initial intensity times the area of the sphere at the initial distance \(P = I_1 * 4 * \pi * d_1^2 = 0.026 * 4 * \pi * 4.3^2\). As power is energy per unit time, and we are asked to find the energy in one hour, we can find the energy by multiplying the power by the time in seconds \(E = P * t = P * 3600\).
04

Calculate the Total Energy

Finally, substitute the values into the formula and calculate the total energy. So, \(E = P * 3600 = 0.026 * 4 * \pi * 4.3^2 * 3600 = 128332.33 \, \mathrm{J}\).

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Key Concepts

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

Sound Waves
Sound waves are vibrations that travel through the air or another medium and can be heard when they reach a person's or animal's ear. When an object vibrates, it causes the surrounding particles to vibrate as well, which, in turn, causes the adjacent particles to vibrate. This chain effect carries the sound wave through the medium.

Sound waves are longitudinal waves, meaning the vibrations occur in the same direction as the wave is moving. These waves are characterized by their frequency and wavelength, which determine the pitch and tone of the sound. In a three-dimensional space, such as when emanating from a point source, sound waves radiate outwards in an expanding sphere.
Power Output Calculation
The power output of a sound source can be determined by examining the intensity of the sound waves it emits. Intensity refers to the power per unit area carried by a wave, and is measured in watts per square meter (\(\mathrm{W/m^2}\)).

To calculate the power output (\(P\)), we can use the initial intensity (\(I_1\)) and the surface area over which the sound wave spreads at a certain distance from the source. For a point source emitting sound equally in all directions, this surface area is the surface of a sphere with a radius equal to that distance: \(A = 4\pi r^2\). Hence, the power output formula is:\[P = I_1 脙聴 A = I_1 脙聴 4\pi r^2\], where\(I_1\) is the intensity at the radius \(r\).
Energy-Time Relation
The energy-time relation in sound physics is instrumental for calculating the amount of energy emitted by a source over a period of time. Power (\(P\)) is defined as the rate at which energy (\(E\)) is emitted or converted per unit of time (\(t\)). Thus, the formula for power in terms of energy and time is given by:\[P = \frac{E}{t}\].

Consequently, if we want to find the total energy emitted over a specific time period, we can rearrange the formula to find energy: \[E = P \times t\]. By knowing the power output of a source and the duration for which it has been emitting energy, we can calculate the total energy output during that time.
Inverse Square Law Application
The inverse square law is fundamental in understanding how the intensity of sound waves changes with distance from its source. It states that the intensity (\(I\)) of a sound wave is inversely proportional to the square of the distance (\(d\)) from the source, mathematically expressed as \[I \propto \frac{1}{d^2}\].

When applying this law to calculate a new intensity (\(I_2\)) at a different distance (\(d_2\)), we use the following formula:\[I_2 = I_1 \times \left(\frac{d_1}{d_2}\right)^2\], where \(I_1\) is the original intensity at distance \(d_1\). In essence, the inverse square law illustrates that as you move away from a sound source, the intensity decreases quickly; specifically, if you double the distance, the intensity becomes one-fourth.

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Most popular questions from this chapter

A small bead of mass \(4.00 \mathrm{~g}\) is attached to a horizontal string. Transverse waves of amplitude \(A=0.800 \mathrm{~cm}\) and frequency \(f=20.0 \mathrm{~Hz}\) are set up on the string. Assume the mass of the bead is small enough that the bead doesn't alter the wave motion. During the wave motion, what is the maximum vertical force that the string exerts on the bead?

A wire with mass \(40.0 \mathrm{~g}\) is stretched so that its ends are tied down at points \(80.0 \mathrm{~cm}\) apart. The wire vibrates in its fundamental mode with frequency \(60.0 \mathrm{~Hz}\) and with an amplitude at the antinodes of \(0.300 \mathrm{~cm}\). (a) What is the speed of propagation of transverse waves in the wire? (b) Compute the tension in the wire. (c) Find the maximum transverse velocity and acceleration of particles in the wire.

A horizontal wire is tied to supports at each end and vibrates in its second- overtone standing wave. The tension in the wire is \(5.00 \mathrm{~N}\), and the node-to-node distance in the standing wave is \(6.28 \mathrm{~cm}\). (a) What is the length of the wire? (b) A point at an antinode of the standing wave on the wire travels from its maximum upward displacement to its maximum downward displacement in \(8.40 \mathrm{~ms}\). What is the wire's mass?

Standing waves on a wire are described by Eq. (15.28), with \(A_{\mathrm{SW}}=2.50 \mathrm{~mm}, \omega=942 \mathrm{rad} / \mathrm{s},\) and \(k=0.750 \pi \mathrm{rad} / \mathrm{m} .\) The left end of the wire is at \(x=0 .\) At what distances from the left end are (a) the nodes of the standing wave and (b) the antinodes of the standing wave?

For a violin, estimate the length of the portions of the strings that are free to vibrate. (a) The frequency of the note played by the open E5 string vibrating in its fundamental standing wave is 659 Hz. Use your estimate of the length to calculate the wave speed for the transverse waves on the string. (b) The vibrating string produces sound waves in air with the same frequency as that of the string. Use \(344 \mathrm{~m} / \mathrm{s}\) for the speed of sound in air and calculate the wavelength of the E5 note in air. Which is larger: the wavelength on the string or the wavelength in air? (c) Repeat parts (a) and (b) for a bass viol, which is typically played by a person standing up. Start your calculation by estimating the length of the bass viol string that is free to vibrate. The G2 string produces a note with frequency \(98 \mathrm{~Hz}\) when vibrating in its fundamental standing wave.
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