/*! 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 26 A \(1.0 \mathrm{~W}\) point sour... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 17: Problem 26

A \(1.0 \mathrm{~W}\) point source emits sound waves isotropically. Assuming that the energy of the waves is conserved, find the intensity (a) \(1.0 \mathrm{~m}\) from the source and (b) \(2.5 \mathrm{~m}\) from the source.

Short Answer

Expert verified
Intensity is approximately 0.0796 W/m虏 at 1.0 m and 0.0127 W/m虏 at 2.5 m.

Step by step solution

01

Understanding Intensity

Intensity \(I\), is defined as the power \(P\) per unit area \(A\). For a point source emitting sound waves isotropically, the area through which the sound spreads is the surface area of a sphere centered around the source. So, the formula to calculate the intensity is:\[ I = \frac{P}{A} = \frac{P}{4 \pi r^2} \]where \(P = 1.0 \text{ W}\) is the power of the source, and \(r\) is the distance from the source.
02

Calculate Intensity at 1.0 m

To find the intensity at \(1.0 \text{ m}\), plug in \(r = 1.0 \text{ m}\) into the formula:\[ I = \frac{1.0 \text{ W}}{4 \pi (1.0 \text{ m})^2} \]Calculate \(4\pi\) first, which is approximately \(12.57\). Thus, \(I = \frac{1.0}{12.57} \approx 0.0796 \text{ W/m}^2\).
03

Calculate Intensity at 2.5 m

For \(r = 2.5 \text{ m}\), use the same intensity formula:\[ I = \frac{1.0 \text{ W}}{4 \pi (2.5 \text{ m})^2} \]Calculate \(r^2 = (2.5)^2 = 6.25\). So, \(4 \pi \times 6.25\) is approximately \(78.54\).Thus, \(I = \frac{1.0}{78.54} \approx 0.0127 \text{ W/m}^2\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Point Source
A point source refers to an idealized source of sound that emits energy equally in all directions. It's like imagining a tiny speaker floating in space. When you hear the term "point source," think of that spot as the origin of the sound waves. In reality, no physical source is a true point source, but many sound sources can be approximated as such when the listener is at a reasonable distance away. This simplification helps in calculations and understanding sound propagation.
  • Simplification: Real sound sources aren't point sources, but small sources in large environments act like them.
  • Sound Spread: From a point source, sound spreads out spherically in three dimensions.
Understanding a point source helps in calculating the wave鈥檚 behavior and intensity as it moves outward from the source.
Isotropic Emission
Isotropic emission is a concept where the sound waves are emitted uniformly in all directions from the source. Imagine an expanding bubble of sound鈥攁n iso-tropic bubble鈥攕preading out evenly. This means the intensity or power of the sound, at any point equidistant from the source, is the same.
  • Uniformity: The energy is distributed equally in every direction.
  • Conservation of Energy: Despite spreading out, energy is conserved; it doesn't dissipate or disappear.
When sound is emitted isotropically from a point source, it's easier to determine how intensity lessens as the distance from the source increases due to the evenly distributed spread of sound waves.
Power of Sound Waves
The power of sound waves is the rate at which energy is emitted from the source. In our scenario, this power is given as 1.0 Watts. Power describes how powerful our point source is in emitting sound energy.
  • Unit of measurement: Watts ( ext{W}) measures the rate of energy transfer.
  • Energy flow: It shows how much energy is transmitted via sound per second.
In calculations involving sound intensity, knowing the power helps determine how the sound spreads over a distance. The greater the power, the more intense the sound waves will be at a given distance.
Surface Area of a Sphere
When calculating sound intensity from a point source with isotropic emission, the sound spreads out over the surface of a sphere centered on the source. The surface area of this sphere (\(4 \pi r^2\)) is crucial to understanding how sound intensity decreases with distance.
  • Formula: \[A = 4 \pi r^2\]
  • Radius importance: The farther from the source, the larger the sphere, which causes sound to spread thinner.
As the distance (radius) increases, the same power of sound must cover more area, therefore, sound intensity decreases. This relationship explains why sound grows quieter as we move away from its source.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Diagnostic ultrasound of frequency \(4.50 \mathrm{MHz}\) is used to examine tumors in soft tissue. (a) What is the wavelength in air of such a sound wave? (b) If the speed of sound in tissue is \(1500 \mathrm{~m} / \mathrm{s}\), what is the wavelength of this wave in tissue?

When the door of the Chapel of the Mausoleum in Hamilton, Scotland, is slammed shut, the last echo heard by someone standing just inside the door reportedly comes 15 s later. (a) If that echo were due to a single reflection off a wall opposite the door, how far from the door is the wall? (b) If, instead, the wall is \(25.7 \mathrm{~m}\) away, how many reflections (back and forth) occur?

In pipe \(A\), the ratio of a particular harmonic frequency to the next lower harmonic frequency is \(1.2 .\) In pipe \(B,\) the ratio of a particular harmonic frequency to the next lower harmonic frequency is 1.4. How many open ends are in (a) pipe \(A\) and (b) pipe \(B ?\)

A sound source \(A\) and a reflecting surface \(B\) move directly toward each other. Relative to the air, the speed of source \(A\) is \(29.9 \mathrm{~m} / \mathrm{s},\) the speed of surface \(B\) is \(65.8 \mathrm{~m} / \mathrm{s},\) and the speed of sound is \(329 \mathrm{~m} / \mathrm{s}\). The source emits waves at frequency \(1200 \mathrm{~Hz}\) as measured in the source frame. In the reflector frame, what are the (a) frequency and (b) wavelength of the arriving sound waves? In the source frame, what are the (c) frequency and (d) wavelength of the sound waves reflected back to the source?

A listener at rest (with respect to the air and the ground) hears a signal of frequency \(f_{1}\) from a source moving toward him with a velocity of \(15 \mathrm{~m} / \mathrm{s},\) due east. If the listener then moves toward the approaching source with a velocity of \(25 \mathrm{~m} / \mathrm{s},\) due west, he hears a frequency \(f_{2}\) that differs from \(f_{1}\) by \(37 \mathrm{~Hz}\). What is the frequency of the source? (Take the speed of sound in air to be \(340 \mathrm{~m} / \mathrm{s}\).)
See all solutions

Recommended explanations on Physics Textbooks

Translational Dynamics

Read Explanation

Physics of Motion

Read Explanation

Fluids

Read Explanation

Waves Physics

Read Explanation

Work Energy and Power

Read Explanation

Magnetism and Electromagnetic Induction

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.