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Chapter 16: Problem 18

The intensity of sound from a point source is \(1 \cdot 0 \times 10^{-8} \mathrm{~W} \mathrm{~m}^{-2}\) at a distance of \(5 \cdot 0 \mathrm{~m}\) from the source. What will be the intensity at a distance of \(25 \mathrm{~m}\) from the source?

Short Answer

Expert verified
The intensity at 25 m is \(4.0 \times 10^{-10} \text{ W/m}^2\).

Step by step solution

01

Understand the Problem

We need to find the intensity of sound from a point source at a different distance from the source. The initial intensity is given, and the formula for intensity from a point source is needed.
02

Identify Relevant Formula

The intensity of sound from a point source is given by the inverse square law:\[I = \frac{P}{4\pi r^2}\]where \(I\) is the intensity, \(P\) is the power of the source, and \(r\) is the distance from the source.
03

Apply the Inverse Square Law

The relationship between the initial intensity \(I_1\) at distance \(r_1\) and the new intensity \(I_2\) at distance \(r_2\) is:\[I_1 \cdot r_1^2 = I_2 \cdot r_2^2\]
04

Plug in Known Values

We know \(I_1 = 1.0 \times 10^{-8} \text{ W/m}^2\), \(r_1 = 5.0\text{ m}\), and \(r_2 = 25\text{ m}\). Substitute these values into the equation:\[(1.0 \times 10^{-8}) \cdot (5.0)^2 = I_2 \cdot (25)^2\]which simplifies to:\(1.0 \times 25 = I_2 \cdot 625\).
05

Solve for New Intensity \(I_2\)

Rearrange the equation from Step 4 to solve for \(I_2\):\[I_2 = \frac{1.0 \times 25}{625}\]Calculate \(I_2\):\(I_2 = \frac{25 \times 10^{-8}}{625} = 4.0 \times 10^{-10} \text{ W/m}^2\).

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Key Concepts

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

Sound Intensity
Sound intensity refers to the energy carried by sound waves per unit area in a direction perpendicular to that area. It measures how loud a sound is perceived at a given distance from the source.
Sound intensity is often expressed in watts per square meter (W/m虏). This concept is crucial because it directly relates to what we commonly perceive as the loudness of a sound. When dealing with sound coming from everyday objects or environments, the intensity can vary significantly because it depends on various factors like the type of source and the distance from it. The formula for calculating sound intensity is derived from dividing the power output of the sound source by the surface area of the sphere over which the sound energy is spread. In mathematical terms:
  • Sound intensity \( I \) is defined as \( I = \frac{P}{A} \), where \( P \) is the power in watts, and \( A \) is the area over which it is spread.
  • For a sound radiating out from a point source, this area becomes the surface area of a sphere, \( A = 4\pi r^2 \).
Point Source
A point source of sound is an idealized single, small location from which sound radiates uniformly in all directions. In reality, no source is a perfect point, but the concept helps simplify problems, assuming uniform distance measurement from the "center" of emission.
Point sources are often theoretical constructs used for analyzing sound properties or behavior. Despite their simplicity, they are foundational in fields such as acoustics and physics. When considering a point source:
  • The sound propagates symmetrically in three dimensions forming a spherical wavefront.
  • This simplicity allows for use of the inverse square law to predict how sound intensity changes with distance.
The consistent radial symmetry and distance uniformity theoretically make calculations straightforward, ensuring accuracy for academic problem-solving.
Distance and Intensity Relationship
The relationship between distance and sound intensity is a fundamental concept in acoustics governed by the inverse square law. This law explains how sound intensity decreases with increasing distance from the sound's point source.
The inverse square law basically states that intensity is inversely proportional to the square of the distance from the source. As one moves away from a point source, sound spreads over a larger area, and thus intensity diminishes.Mathematically, the relationship can be described as:
  • \( I_1 \cdot r_1^2 = I_2 \cdot r_2^2 \)
  • Where \( I_1 \) and \( I_2 \) are sound intensities at distances \( r_1 \) and \( r_2 \)
This equation helps solve problems involving changing distances between source and receiver, predicting the loudness of sound or informing design decisions in sound-sensitive environments like auditoriums and recording studios. Understanding this relationship is also crucial for applications in audio engineering, environmental noise assessment, and more.

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Most popular questions from this chapter

If the sound level in a room is increased from \(50 \mathrm{~dB}\) to \(60 \mathrm{~dB}\), by what factor is the pressure amplitude increased?

A tuning fork of unknown frequency makes 5 beats per second with another tuning fork which can cause a closed organ pipe of length \(40 \mathrm{~cm}\) to vibrate in its fundamental mode. The beat frequency decreases when the first tuning fork is slightly loaded with wax. Find its original frequency. The speed of sound in air is \(320 \mathrm{~m} / \mathrm{s}\)

Two audio speakers are kept some distance apart and are driven by the same amplifier system. A person is sitting at a place \(6 \cdot 0 \mathrm{~m}\) from one of the speakers and \(6 \cdot 4 \mathrm{~m}\) from the other. If the sound signal is continuously varied from \(500 \mathrm{~Hz}\) to \(5000 \mathrm{~Hz}\), what are the frequencies for which there is a destructive interference at the place of the listener? Speed of sound in air \(=320 \mathrm{~m} \mathrm{~s}^{-1}\).

A violin player riding on a slow train plays a \(440 \mathrm{~Hz}\) note. Another violin player standing near the track plays the same note. When the two are close by and the train approaches the person on the ground, he hears \(4 \cdot 0\) beats per second. The speed of sound in air= \(340 \mathrm{~m} / \mathrm{s}\). (a) Calculate the speed of the train. (b) What beat frequency is heard by the player in the train?

A person standing on a road sends a sound signal to the driver of a car going away from him at a speed of \(72 \mathrm{~km} \mathrm{~h}^{-1}\). The signal travelling at \(330 \mathrm{~m} \mathrm{~s}^{-1}\) in air and having a frequency of \(1600 \mathrm{~Hz}\) gets reflected from the body of the car and returns. Find the frequency of the reflected signal as heard by the person.
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