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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 is approximately 6.52 watts.

Step by step solution

01

Understand the Problem

The problem involves finding the total power radiated by a siren, given the sound intensity at a specific distance. Intensity is given, and power needs to be calculated.
02

Use the Intensity Formula

Sound intensity ( I ) is defined as power (P) per unit area. For a point source radiating uniformly in all directions, the sound intensity at a distance (r) is given by: \[ I = \frac{P}{4\pi r^2} \] where \( 4\pi r^2 \) is the surface area of a sphere with radius \( r \).
03

Relate Given Values to the Formula

We know that: \( r = 3.8 \, \text{m} \) and \( I = 3.6 \times 10^{-2} \, \text{W/m}^2 \). Plug these values into the formula: \[ 3.6 \times 10^{-2} = \frac{P}{4\pi (3.8)^2} \]
04

Solve for Power \(P\)

Rearrange the formula to solve for \(P\): \[ P = I \times 4\pi r^2 \] Substitute the known values: \[ P = 3.6 \times 10^{-2} \times 4 \pi (3.8)^2 \]
05

Calculate the Numerical Answer

Calculate \((3.8)^2\): \[ (3.8)^2 = 14.44 \] Then, compute \(4\pi (3.8)^2\): \[ 4 \pi \times 14.44 \approx 181.02 \] Finally, calculate \(P\): \[ P = 3.6 \times 10^{-2} \times 181.02 \approx 6.52 \text{ W} \]
06

Confirm and Interpret the Results

The total power radiated by the siren is approximately \(6.52 \, \text{W}\). This is reasonable given the initial intensity and distance.

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Key Concepts

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

Spherical Waves
When sound emanates from a source like a siren, it often spreads out in the form of spherical waves. Imagine a balloon expanding in all directions, with the siren at its center. This is akin to how sound spreads uniformly in every direction when there are no obstacles in the way. The surface area on which the sound intensity spreads increases as the distance from the source increases, forming a spherical pattern.

In mathematical terms, if a sound source radiates power uniformly, the intensity of the sound at a distance is determined by the surface area of a sphere with radius equal to that distance. The formula for the intensity, which is the power per unit area, is:
  • \[ I = \frac{P}{4\pi r^2} \]
Where:
  • \( I \) is the sound intensity.
  • \( P \) is the total power radiated by the source.
  • \( r \) is the distance from the sound source.
This relationship highlights how intensity decreases with the square of the distance. As you move farther away, the energy spreads over a larger spherical area, resulting in lower intensity.
Sound Power
Sound power refers to the total energy radiated by a sound source per unit time. It is a measure of the source's ability to produce sound and is independent of the medium through which it travels. Unlike sound intensity, which decreases with distance from the source, sound power remains constant as long as the source's output remains unchanged.

In the context of the problem, where the sound power is distributed uniformly, we can calculate it by rearranging the intensity formula. Given that the sound intensity and the distance are known, the total sound power can be derived using:
  • \[ P = I \times 4\pi r^2 \]
This equation helps in understanding how much total power a sound source like a siren emits. It is essential when considering how a sound might be heard at various distances or how it may be designed to meet regulatory standards regarding noise pollution. By focusing on total power, engineers and designers can ensure that devices function optimally across different settings.
Acoustic Radiation Pattern
An acoustic radiation pattern describes how sound is emitted from a source into the surrounding space. For many practical sound sources, like sirens, an assumption of uniform radiation leads to understanding that they behave as omnidirectional emitters.

This means the sound is distributed evenly in all directions as if the sound source sits at the center of an imaginary sphere. While this simplifies calculations and expectations, real-world tools may show variations due to factors like obstacles, reflections, and frequency changes.
  • This form of radiation can be greatly beneficial, ensuring effective coverage in alarms and public announcements.
  • Designing with an understanding of radiation patterns helps optimize sound systems for maximum efficacy.
By understanding the radiation pattern of a sound source, we can better predict how it will act in different environments—vital for sound design and audio engineering, where the pattern influences sound quality and reach.

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Most popular questions from this chapter

From a vantage point very close to the track at a stock car race, you hear the sound emitted by a moving car. You detect a frequency that is 0.86 times as small as the frequency emitted by the car when it is stationary. The speed of sound is 343 m/s. What is the speed of the car?

Two wires are parallel, and one is directly above the other. Each has a length of 50.0 m and a mass per unit length of 0.020 kg/m. However, the tension in wire \(A\) is \(6.00 \times 10^{2} \mathrm{N},\) and the tension in wire \(\mathrm{B}\) is \(3.00 \times 10^{2} \mathrm{N}\) . Transverse wave pulses are generated simultaneously, one at the left end of wire A and one at the right end of wire B. The pulses travel toward each other. How much time does it take until the pulses pass each other?

The security alarm on a parked car goes off and produces a frequency of 960 Hz. The speed of sound is 343 m/s. As you drive toward this parked car, pass it, and drive away, you observe the frequency to change by 95 Hz. At what speed are you driving?

A source emits sound uniformly in all directions. A radial line is drawn from this source. On this line, determine the positions of two points, 1.00 m apart, such that the intensity level at one point is 2.00 dB greater than the intensity level at the other.

Using an intensity of \(1 \times 10^{-12} \mathrm{W} / \mathrm{m}^{2}\) as a reference, the threshold of hearing for an average young person is 0 \(\mathrm{dB}\) . Person 1 and person \(2,\) who are not average, have thresholds of hearing that are \(\beta_{1}=-8.00 \mathrm{dB}\) and \(\beta_{2}=+12.0 \mathrm{dB}\) . What is the ratio \(I_{1} / I_{2}\) of the sound intensity \(I_{1}\) when person 1 hears the sound at his own threshold of hearing compared to the sound intensity \(I_{2}\) when person 2 hears the sound at his own threshold of hearing?

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