/*! 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 99 You are standing at a distance \... [FREE SOLUTION] | 91Ó°ÊÓ

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You are standing at a distance \(D\) from an isotropic point source of sound. You walk \(50.0 \mathrm{~m}\) toward the source and observe that the intensity of the sound has doubled. Calculate the distance \(D\).

Short Answer

Expert verified
The distance \(D\) is approximately 170.71 meters.

Step by step solution

01

Understand the Problem

We know that the intensity of sound is related to the distance from the source. Our initial position is at distance \(D\) and when we move 50 meters closer, the intensity doubles. We need to find the initial distance \(D\).
02

Apply the Intensity Formula

The intensity \(I\) of sound from a point source is inversely proportional to the square of the distance \(r\) from the source: \(I \propto \frac{1}{r^2}\). So, if we denote the initial intensity by \(I_1 = \frac{C}{D^2}\), then after moving 50 meters closer, \(I_2 = \frac{C}{(D - 50)^2}\).
03

Set Up the Intensity Relation

Since the intensity doubles, \(I_2 = 2I_1\). Substituting from our equations: \(\frac{C}{(D - 50)^2} = 2 \times \frac{C}{D^2}\).
04

Simplify the Equation

Cancel \(C\) from both sides and set up the equation: \(\frac{1}{(D - 50)^2} = \frac{2}{D^2}\). Cross multiply to get: \(D^2 = 2(D - 50)^2\).
05

Solve the Quadratic Equation

Expand \((D - 50)^2 = D^2 - 100D + 2500\). Thus, the equation becomes \(D^2 = 2(D^2 - 100D + 2500)\). Simplify to get: \(D^2 = 2D^2 - 200D + 5000\). Rearrange to \(0 = D^2 - 200D + 5000\).
06

Use the Quadratic Formula

The quadratic formula is \(D = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) where \(a = 1\), \(b = -200\), \(c = 5000\). Substitute these into the formula: \(D = \frac{-(-200) \pm \sqrt{(-200)^2 - 4 \cdot 1 \cdot 5000}}{2 \cdot 1}\).
07

Calculate the Solution

This simplifies to \(D = \frac{200 \pm \sqrt{40000 - 20000}}{2} = \frac{200 \pm \sqrt{20000}}{2}\). Further simplifying, \(\sqrt{20000} = \sqrt{100 \cdot 200} = 10 \sqrt{200} \approx 141.42\). Thus, \(D = \frac{200 \pm 141.42}{2}\).
08

Choose the Physical Solution

Calculate both solutions: \(D_1 = \frac{200 + 141.42}{2} = 170.71\), \(D_2 = \frac{200 - 141.42}{2} = 29.29\). Since \(D\) must be greater than 50 (as you move closer by 50 meters), choose \(D = 170.71 \: \mathrm{m}\).

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Key Concepts

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

Inverse Square Law
The Inverse Square Law is a cornerstone concept in physics, especially when it comes to understanding how sound and light behave over distance. It states that the intensity of a physical quantity decreases proportionally to the square of the distance from the source of that quantity. This means that as you move further away from a sound source, the intensity of the sound decreases rapidly.

For sound intensity, this relationship can be expressed mathematically as: \[ I \propto \frac{1}{r^2} \] where \( I \) stands for the intensity, and \( r \) represents the distance from the source. This equation vividly illustrates that if you double the distance from the sound source, the intensity reduces to one-quarter of its original value. Thus, small changes in distance can greatly affect sound intensity, making this principle crucial in sound design, acoustics, and audio engineering.
Isotropic Point Source
An isotropic point source is a hypothetical concept used in physics to simplify calculations related to wave emissions, such as sound or light. "Isotropic" means that the source emits energy equally in all directions. This unidirectional assumption helps us model and understand how energy spreads out in a circle or sphere, depending on the dimensions being considered.

For example, a small speaker suspended in free space could be considered an isotropic point source for sound waves. In real-world scenarios, achieving perfect isotropy is challenging; however, the concept is beneficial for theoretical analysis and foundational learning in acoustics and wave physics. This simplification allows us to apply laws like the Inverse Square Law more readily, as long as the distance is much greater than the size of the source itself.
Quadratic Equation
The quadratic equation is a fundamental mathematical tool used to solve for unknown values when equations are set in a certain form. It is generally expressed as:\[ ax^2 + bx + c = 0 \] where \( x \) is the unknown variable, and \( a \), \( b \), and \( c \) are constants with \( a eq 0 \). Solutions to quadratic equations can be found using the quadratic formula:\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] This formula allows us to determine the values of \( x \) that satisfy the equation. Quadratic equations are ubiquitous in physics, appearing in problems involving motion, optimization, and rates of change. In the context of sound, they help solve for distances or intensities when following physical laws like the Inverse Square Law and relate these measurements to real-world scenarios.
Sound Intensity Formula
The Sound Intensity Formula is used to quantify the power carried by sound waves per unit area. When dealing with point sources, the intensity is inversely related to the square of the distance from the point of emission. The formula is expressed as:\[ I = \frac{P}{4 \pi r^2} \] where \( I \) represents sound intensity, \( P \) is the acoustic power of the source, and \( r \) is the radius or distance from the source. When working with isotropic point sources, this formula helps us calculate how sound spreads evenly in all directions, diminishing in intensity the further it travels.

The Intensity of Sound helps engineers and scientists design systems for optimal acoustics and energy efficiency, as well as to understand phenomena like noise pollution and hearing safety. In many practical applications, this formula allows predicting sound levels at various distances from the source.

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Most popular questions from this chapter

A well with vertical sides and water at the bottom resonates at \(7.00 \mathrm{~Hz}\) and at no lower frequency. The air-filled portion of the well acts as a tube with one closed end (at the bottom) and one open end (at the top). The air in the well has a density of \(1.10 \mathrm{~kg} / \mathrm{m}^{3}\) and a bulk modulus of \(1.33 \times 10^{5} \mathrm{~Pa}\). How far down in the well is the water surface?

A stationary detector measures the frequency of a sound source that first moves at constant velocity directly toward the detector and then (after passing the detector) directly away from it. The emitted frequency is \(f\). During the approach the detected frequency is \(f_{\text {app }}^{\prime}\) and during the recession it is \(f_{\text {rec }}^{\prime}\). If \(\left(f_{\text {app }}^{\prime}-f_{\text {rec }}^{\prime}\right) / f=0.500,\) what is the ratio \(v_{s} / v\) of the speed of the source to the speed of sound?

Two sound waves with an amplitude of \(12 \mathrm{nm}\) and a wave- length of \(35 \mathrm{~cm}\) travel in the same direction through a long tube, with a phase difference of \(\pi / 3\) rad. What are the (a) amplitude and (b) wavelength of the net sound wave produced by their interference? If, instead, the sound waves travel through the tube in opposite directions, what are the (c) amplitude and (d) wavelength of the net wave?

Suppose a spherical loudspeaker emits sound isotropically at \(10 \mathrm{~W}\) into a room with completely absorbent walls, floor, and ceiling (an anechoic chamber). (a) What is the intensity of the sound at distance \(d=3.0 \mathrm{~m}\) from the center of the source? (b) What is the ratio of the wave amplitude at \(d=4.0 \mathrm{~m}\) to that at \(d=3.0 \mathrm{~m} ?\)

A siren emitting a sound of frequency \(1000 \mathrm{~Hz}\) moves away from you toward the face of a cliff at a speed of \(10 \mathrm{~m} / \mathrm{s}\). Take the speed of sound in air as \(330 \mathrm{~m} / \mathrm{s}\). (a) What is the frequency of the sound you hear coming directly from the siren? (b) What is the frequency of the sound you hear reflected off the cliff? (c) What is the beat frequency between the two sounds? Is it perceptible (less than \(20 \mathrm{~Hz}\) )?

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