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At a distance of \(3.8 \mathrm{m}\) from a siren, the sound intensity is \(3.6 \times\) \(10^{-2} \mathrm{W} / \mathrm{m}^{2}\). Assuming that the siren radiates sound uniformly in all directions, find the total power radiated.

Short Answer

Expert verified
The total power radiated by the siren is approximately 6.53 Watts.

Step by step solution

01

Determine the Formula for Total Power

The total power radiated by a source that radiates sound uniformly in all directions can be calculated using the formula for intensity: \[ I = \frac{P}{A} \]where \( I \) is the intensity, \( P \) is the total power, and \( A \) is the surface area over which the sound is distributed. For a point source radiating uniformly, the area \( A \) is the surface area of a sphere, \( A = 4 \pi r^2 \), where \( r \) is the distance from the source.
02

Plug Values into the Power Formula

We know the intensity \( I = 3.6 \times 10^{-2} \mathrm{W/m^2} \) and the distance \( r = 3.8 \, \mathrm{m} \). Substitute these values into the formula: \[ I = \frac{P}{4 \pi r^2} \]Simplifying for \( P \), we get: \[ P = I \times 4 \pi r^2 \]
03

Calculate the Surface Area of the Sphere

First, calculate \( 4 \pi r^2 \) using the given distance \( r = 3.8 \, \mathrm{m} \):\[ 4 \pi (3.8)^2 = 4 \pi \times 14.44 = 181.45 \; \mathrm{m}^2 \] (using \( \pi \approx 3.14159 \)).
04

Calculate the Total Power Radiated

Now, calculate the power \( P \) using the intensity and the area:\[ P = 3.6 \times 10^{-2} \times 181.45 = 6.5322 \, \mathrm{W} \]Thus, the total power radiated by the siren is approximately 6.53 Watts.

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Key Concepts

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

Intensity Formula
Sound intensity is the amount of sound power that passes through a specific area. It's an important concept because it helps us understand how loud a sound is when it reaches us at a certain distance from the source. The formula for sound intensity is \[ I = \frac{P}{A} \]where:- \( I \) is the intensity,- \( P \) is the power of the sound source,- \( A \) is the area through which the sound spreads.

Knowing this formula allows us to calculate either the power of the sound if the intensity and area are known or to find the intensity if the power and area are available. This relationship is crucial, especially when you're dealing with sounds that radiate in all directions like in our siren problem.
Surface Area of a Sphere
In our problem, sound radiates uniformly in all directions, forming a spherical shape as it moves away from the source. To calculate the area over which the sound spreads, we use the formula for the surface area of a sphere:\[ A = 4\pi r^2 \]This formula tells us that the area depends on the square of the radius, \( r \), which is the distance from the sound source.

This concept is helpful because, in reality, when sound emanates from a point source, it behaves as if it is traveling over the surface of a sphere. By knowing the surface area, we can determine how the energy spreads out, getting weaker the further it gets from the source due to the larger area.
Point Source Radiation
A point source is an idealized concept used to describe any source that emits sound (or other types of waves) uniformly in all directions. Picture a single point from where the sound seems to spread out evenly like ripples in a pond or bursts from a tiny sphere.

This model simplifies calculations and predicts how intensity decreases with distance based on the inverse square law. This law states that the intensity decreases proportionally to the square of the distance from the source. Because of this, the intensity diminishes rapidly the farther you get, as the sound energy dissipates over a larger area.
Acoustics
Acoustics is the branch of physics concerned with the study of sound. It involves understanding how sound is produced, how it travels, and how it is perceived. In practical terms, acoustics help us in designing buildings, concert halls, and even simple spaces so that sound behaves in a desirable way.

Appreciating the basics of acoustics, like sound intensity and how it's measured, enables us to handle noise control and soundproofing. Moreover, for individuals designing sirens as in this exercise, it ensures that they're powerful enough to be heard at required distances, without being damagingly loud at close range.

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Most popular questions from this chapter

Two submarines are under water and approaching each other head-on. Sub A has a speed of \(12 \mathrm{m} / \mathrm{s}\) and sub \(\mathrm{B}\) has a speed of \(8 \mathrm{m} / \mathrm{s}\). Sub \(\mathrm{A}\) sends out a \(1550-\mathrm{Hz}\) sonar wave that travels at a speed of \(1522 \mathrm{m} / \mathrm{s}\). (a) What is the frequency detected by sub \(B ?\) (b) Part of the sonar wave is reflected from sub \(B\) and returns to sub \(A\). What frequency does sub A detect for this reflected wave?

A convertible moves toward you and then passes you; all the while, its loudspeakers are producing a sound. The speed of the car is a constant \(9.00 \mathrm{m} / \mathrm{s},\) and the speed of sound is \(343 \mathrm{m} / \mathrm{s} .\) What is the ratio of the frequency you hear while the car is approaching to the frequency you hear while the car is moving away?

A siren, mounted on a tower, emits a sound whose frequency is \(2140 \mathrm{Hz}\). A person is driving a car away from the tower at a speed of \(27.0 \mathrm{m} / \mathrm{s}\). As the figure illustrates, the sound reaches the person by two paths: the sound reflected from the building in front of the car, and the sound coming directly from the siren. The speed of sound is \(343 \mathrm{m} / \mathrm{s}\). Concepts: (i) One way that the Doppler effect can arise is that the wavelength of the sound changes. For either the direct or reflected sound, does the wavelength change? (ii) Why does the driver hear a frequency for the reflected sound that is different from \(2140 \mathrm{Hz},\) and is it greater or smaller than \(2140 \mathrm{Hz} ?\) (iii) Why does the driver hear a frequency for the direct sound that is different from \(2140 \mathrm{Hz}\), and is it greater or smaller than \(2140 \mathrm{Hz}\) ? Calculations: What frequency does the person hear for the (a) reflected and (b) direct sound?

A typical adult ear has a surface area of \(2.1 \times 10^{-3} \mathrm{m}^{2}\). The sound intensity during a normal conversation is about \(3.2 \times 10^{-6} \mathrm{W} / \mathrm{m}^{2}\) at the listener's ear. Assume that the sound strikes the surface of the ear perpendicularly. How much power is intercepted by the ear?

In a discussion person A is talking 1.5 dB louder than person B, and person C is talking \(2.7 \mathrm{dB}\) louder than person \(\mathrm{A}\). What is the ratio of the sound intensity of person C to the sound intensity of person B?

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