Two people are talking at a distance of \(3.0 \mathrm{~m}\) from where you are, and you measure the sound intensity as \(1.1 \cdot 10^{-7} \mathrm{~W} / \mathrm{m}^{2}\). Another student is \(4.0 \mathrm{~m}\) away from the talkers. What sound intensity does the other student measure?

Short Answer

Expert verified
Answer: Approximately 6.19 x 10^(-8) W/m^2.

Step by step solution

01

Understand the inverse square relationship

The inverse square relationship states that the intensity of sound is inversely proportional to the square of the distance from the source. Mathematically, this can be expressed as: \(I_1/I_2 = (d_2/d_1)^2\) where \(I_1\) and \(I_2\) are the sound intensities at distances \(d_1\) and \(d_2\) from the source, respectively.
02

Plug given values into the inverse square relationship equation

We are given that at the distance \(d_1 = 3.0\ \mathrm{m}\), the sound intensity \(I_1 = 1.1 \cdot 10^{-7} \mathrm{~W} / \mathrm{m}^{2}\). We are to find the sound intensity, \(I_2\), at a distance \(d_2 = 4.0\ \mathrm{m}\). So, we will plug these values into the inverse square relationship equation: \(\frac{1.1 \cdot 10^{-7} \mathrm{~W} / \mathrm{m}^{2}}{I_2} = \left(\frac{4.0\ \mathrm{m}}{3.0\ \mathrm{m}}\right)^2\)
03

Solve the equation for \(I_2\)

Now, we will isolate \(I_2\) by multiplying both sides of the equation by \(I_2\) and subsequently dividing by the square of the distance ratio: \(I_2 = \frac{1.1 \cdot 10^{-7} \mathrm{~W} / \mathrm{m}^{2}}{\left(\frac{4.0\ \mathrm{m}}{3.0\ \mathrm{m}}\right)^2}\)
04

Simplify and calculate the sound intensity, \(I_2\)

Simplify the equation and calculate the value of \(I_2\): \(I_2 = \frac{1.1 \cdot 10^{-7} \mathrm{~W} / \mathrm{m}^{2}}{\left(\frac{4}{3}\right)^2} = 6.1875 \cdot 10^{-8}\ \mathrm{W} / \mathrm{m}^{2} \) So, the sound intensity that the other student measures at \(4.0\ \mathrm{m}\) distance is approximately \(6.19 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2}\).

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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 pivotal concept when understanding how sound diminishes as it travels away from its source. This principle affirms that the intensity of sound (or light, for that instance) decreases in proportion to the square of the distance from the source. In simpler terms, if you double the distance from a sound source, the intensity doesn't merely halve; it actually becomes a quarter of what it was.

Mathematically, we can capture this relationship using the formula: \[I_1 / I_2 = (d_2 / d_1)^2\]where \(I_1\) represents the intensity at a closer distance \(d_1\), and \(I_2\) is the intensity at a farther distance \(d_2\). It's utilized in various fields, from acoustics to astrophysics, and is crucial for accurate sound level predictions in different environments. Appreciating this law can help you understand why a sound source may seem fainter as you move away and how sound levels can be manipulated in audio engineering.
Sound Propagation
Sound propagation involves the travel of sound waves through a medium, which can be a solid, liquid, or gas. These vibrations travel in waveforms, and their behavior is influenced by the medium's characteristics, such as its density and elasticity. As sound waves spread out, they experience attenuation, meaning they lose energy over distance, impacting the sound intensity that we ultimately perceive.

Another aspect of sound propagation to consider is the environment's effect on wave dispersion. In an open field, sound will spread out more than in a contained, echoic space. Understanding how sound waves propagate ensures better control when designing performance spaces, recording studios, and any setting where sound level is a significant concern. Learning how these waves behave in different scenarios allows us to both manipulate sound for desired effects and protect against noise pollution.
Intensity Level of Sound
The intensity level of sound, also known as sound level or acoustic intensity, is a measure of the power per unit area carried by a sound wave. It is typically measured in units of watts per square meter (\(W/m^2\)). What's interesting is that the human ear perceives sound intensity on a logarithmic scale, which is why intensity levels are often expressed in decibels (dB).

This logarithmic perception means that small increases in the number of decibels can signify a significant surge in the intensity level. For instance, a 10 dB increase represents a tenfold increase in acoustic intensity. Thus, when two people are having a conversation at a measured intensity level from a specific distance, and then the observer changes their distance, the perceived intensity can change drastically due to the combined effect of the inverse square law and the logarithmic sensitivity of human hearing. This concept is critical for tasks such as setting appropriate noise regulations and designing sound equipment that delivers suitable output levels for human use.

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Most popular questions from this chapter

Two 100.0-W speakers, A and B, are separated by a distance \(D=3.6 \mathrm{~m} .\) The speakers emit in-phase sound waves at a frequency \(f=10,000.0 \mathrm{~Hz}\). Point \(P_{1}\) is located at \(x_{1}=4.50 \mathrm{~m}\) and \(y_{1}=0 \mathrm{~m} ;\) point \(P_{2}\) is located at \(x_{2}=4.50 \mathrm{~m}\) and \(y_{2}=-\Delta y .\) Neglecting speaker \(\mathrm{B}\), what is the intensity, \(I_{\mathrm{A} 1}\) (in \(\mathrm{W} / \mathrm{m}^{2}\) ), of the sound at point \(P_{1}\) due to speaker \(\mathrm{A}\) ? Assume that the sound from the speaker is emitted uniformly in all directions. What is this intensity in terms of decibels (sound level, \(\beta_{\mathrm{A} 1}\) )? When both speakers are turned on, there is a maximum in their combined intensities at \(P_{1} .\) As one moves toward \(P_{2},\) this intensity reaches a single minimum and then becomes maximized again at \(P_{2}\). How far is \(P_{2}\) from \(P_{1},\) that is, what is \(\Delta y ?\) You may assume that \(L \gg \Delta y\) and that \(D \gg \Delta y\), which will allow you to simplify the algebra by using \(\sqrt{a \pm b} \approx a^{1 / 2} \pm \frac{b}{2 a^{1 / 2}}\) when \(a \gg b\).

A college student is at a concert and really wants to hear the music, so she sits between two in-phase loudspeakers, which point toward each other and are \(50.0 \mathrm{~m}\) apart. The speakers emit sound at a frequency of \(490 .\) Hz. At the midpoint between the speakers, there will be constructive interference, and the music will be at its loudest. At what distance closest to the midpoint could she also sit to experience the loudest sound?

In a suspense-thriller movie, two submarines, \(X\) and Y, approach each other, traveling at \(10.0 \mathrm{~m} / \mathrm{s}\) and \(15.0 \mathrm{~m} / \mathrm{s}\), respectively. Submarine X "pings" submarine Y by sending a sonar wave of frequency \(2000.0 \mathrm{~Hz}\). Assume that the sound travels at \(1500.0 \mathrm{~m} / \mathrm{s}\) in the water. a) Determine the frequency of the sonar wave detected by submarine Y. b) What is the frequency detected by submarine \(X\) for the sonar wave reflected off submarine Y? c) Suppose the submarines barely miss each other and begin to move away from each other. What frequency does submarine Y detect from the pings sent by X? How much is the Doppler shift?

Compare the intensity of sound at the pain level, \(120 \mathrm{~dB}\), with that at the whisper level, \(20 \mathrm{~dB}\).

Two vehicles carrying speakers that produce a tone of frequency \(1000.0 \mathrm{~Hz}\) are moving directly toward each other. Vehicle \(\mathrm{A}\) is moving at \(10.00 \mathrm{~m} / \mathrm{s}\) and vehicle \(\mathrm{B}\) is moving at \(20.00 \mathrm{~m} / \mathrm{s}\). Assume the speed of sound in air is \(343.0 \mathrm{~m} / \mathrm{s}\), and find the frequencies that the driver of each vehicle hears.

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