/*! 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 56 Two sources of sound are located... [FREE SOLUTION] | 91Ó°ÊÓ

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Two sources of sound are located on the \(x\) axis, and each emits power uniformly in all directions. There are no reflections. One source is positioned at the origin and the other at \(x=+123 \mathrm{~m}\). The source at the origin emits four times as much power as the other source. Where on the \(x\) axis are the two sounds equal in intensity? Note that there are two answers.

Short Answer

Expert verified
The sounds are equal in intensity at \(x = 41\) m and \(x = 123\) m.

Step by step solution

01

Understanding Intensity Formula

The intensity of sound from a point source is given by the formula \( I = \frac{P}{4\pi r^2} \), where \(P\) is the power emitted by the source and \(r\) is the distance from the source.
02

Establish Power Ratios

Let the power of the source at the origin be \(P_1 = 4P\) and the power of the source at \(x = 123\mathrm{~m}\) be \(P_2 = P\), where \(P\) is the power of the second source.
03

Set Up Equations for Intensities Equal

We want to find points \(x\) on the x-axis such that the intensity from both sources are the same. So, \( \frac{4P}{4\pi x^2} = \frac{P}{4\pi (123-x)^2} \). Simplify by cancelling terms to get \( \frac{4}{x^2} = \frac{1}{(123-x)^2} \).
04

Solve the Equation

Solve \( \frac{4}{x^2} = \frac{1}{(123-x)^2} \). Cross-multiply to get \(4(123-x)^2 = x^2 \). This simplifies to \(4(123-x)^2 = x^2\), or \(4(123-x)^2 - x^2 = 0\). Notice there are quadratic findings to be made here.
05

Simplify and Use Quadratic Formula

Rewrite the equation from the prior step as a standard quadratic: Expand \((123-x)^2 = 15129 - 246x + x^2\). Then, \(4x^2 - 492x + 15129 - x^2 = 0\), simplify to \(3x^2 - 492x + 15129 = 0\).
06

Apply the Quadratic Formula

The quadratic equation is in the form \(ax^2 + bx + c = 0\), where \(a = 3\), \(b = -492\), and \(c = 15129\). Use the quadratic formula: \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). Substitute the values: \(x = \frac{492 \pm \sqrt{(-492)^2 - 4\times 3 \times 15129}}{6}\).
07

Compute the Solutions

Calculate the discriminant \(b^2 - 4ac\), which results in \(242064 - 181548 = 60516\). Taking the square root of 60516 yields 246. Compute \(x = \frac{492 \pm 246}{6}\). The two solutions are: \(x = \frac{738}{6}\) and \(x = \frac{246}{6}\).
08

Simplify Final Answers

Simplify the solutions: \(x = 41\) and \(x = 123\). These are the points on the x-axis where the intensities of the two sounds are equal.

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Key Concepts

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

Intensity Formula
The intensity of sound refers to the power per unit area carried by a sound wave. It is an important concept in physics as it helps us understand how sound energy disperses in space. For a point source, sound intensity decreases with distance, spreading uniformly in all directions. In other words, the farther you are from the source, the quieter the sound appears.
To calculate sound intensity from a point source, use the formula:
  • \( I = \frac{P}{4\pi r^2} \)
Here, \( I \) is the intensity, \( P \) is the power of the source, and \( r \) is the distance from the source. This formula shows that intensity is inversely proportional to the square of the distance from the point source, which means as distance increases, intensity decreases rapidly. The factor \( 4\pi \) accounts for the spherical distribution of sound energy around the source.
Quadratic Equation
A quadratic equation is a polynomial equation of the second degree, usually in the form \( ax^2 + bx + c = 0 \). Solving a quadratic equation involves finding the values of \( x \) that make this equation true.
In this exercise, after setting up the equation for equal intensities of the two sound sources, we derive a quadratic equation:
  • \( 3x^2 - 492x + 15129 = 0 \)
This equation is typical for problems that can be simplified into sections involving squares. The coefficients \( a = 3 \), \( b = -492 \), and \( c = 15129 \) were derived from the simplification of the original expressions.
The quadratic formula is a reliable method to solve quadratic equations, given as:
  • \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
By plugging in the coefficients, we calculate the potential values of \( x \) that yield the desired condition of equal sound intensities.
Point Source
A point source in physics represents an idealized source of energy that emits uniformly in all directions. Despite being a theoretical construct, it is useful for simplifying complex calculations in wave phenomena like sound and light. In this context, a point source allows us to consider the spherical distribution of energy emanating from a tiny, focused region, similar to how a firework releases light.
In the exercise, both sound sources are treated as point sources situated on the x-axis, making it easier to mathematically express and calculate their intensities. The source at the origin has four times the power of the source at \( x = 123 \mathrm{~m} \), which affects how their intensities diminish with distance. By applying the intensity formula, we see how the emitted power and distance from the source determine the intensity a listener perceives at any given point along the axis.
Sound Power Ratio
The sound power ratio describes the relationship between the emitted powers of two sound sources. It is a critical factor in assessing sound intensity at various distances. In our example, one source emits four times the power of the other, which significantly influences the intensity comparison between the two.
Consider the power ratio setup:
  • Source at origin: \( P_1 = 4P \)
  • Source at \( x = 123\mathrm{~m} \): \( P_2 = P \)
This setup informs how we equate intensities in the exercise to find positions along the x-axis where the intensities perceived from both sources are identical.
By utilizing the concept of power ratios, and relating them through sound intensities, it provides a clearer understanding of how power levels impact sound distribution and perception in a given space.

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Most popular questions from this chapter

Suppose that when a certain sound intensity level (in dB) triples, the sound intensity (in \(\mathrm{W} / \mathrm{m}^{2}\) ) also triples. Determine this sound intensity level.

Two wires are parallel, and one is directly above the other. Each has a length of \(50.0 \mathrm{~m}\) and a mass per unit length of \(0.020 \mathrm{~kg} / \mathrm{m}\). However, the tension in wire \(\mathrm{A}\) 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?

Humans can detect a difference in sound intensity levels as small as \(1.0 \mathrm{~dB}\). What is the ratio of the sound intensities?

Concept Questions A uniform rope of mass \(m\) and length \(L\) is hanging straight down from the ceiling. (a) Is the tension in the rope greater near the top or near the bottom of the rope, or is the tension the same everywhere along the rope? Why? (b) A small-amplitude transverse wave is sent up the rope from the bottom end. Is the speed of the wave greater near the bottom or near the top of the rope, or is the speed of the wave the same everywhere along the rope? Explain. (c) Consider a section of the rope between the bottom end and a point that is a distance \(y\) meters above the bottom. What is the weight of this section? Express your answer in terms of \(m, L, y,\) and \(g\) (the acceleration due to gravity). Problem (a) For the rope described in the Concept Questions, derive an expression that gives the speed of the wave on the rope in terms of the distance \(y\) above the bottom end and the acceleration \(g\) due to gravity. (b) Use the expression that you have derived to calculate the speeds at distances of \(0.50 \mathrm{~m}\) and \(2.0 \mathrm{~m}\) above the bottom end of the rope. Be sure that your answers are consistent with your answer to Concept Question (b).

A wave has the following properties: amplitude \(=0.37 \mathrm{~m}\), period \(=0.77 \mathrm{~s}\), wave speed \(=12 \mathrm{~m} / \mathrm{s} .\) The wave is traveling in the \(-x\) direction. What is the mathematical expression (similar to Equation 16.3 or 16.4 ) for the wave?

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