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Chapter 23: Problem 37

A whisper has an intensity level of about \(15 \mathrm{~dB}\). What is the corresponding intensity of the sound?

Short Answer

Expert verified
The sound intensity is approximately \(3.162 \times 10^{-11} \text{ W/m}^{2}\).

Step by step solution

01

Understanding Decibels

Decibels (dB) is a logarithmic unit used to measure sound intensity. It is not the actual intensity but a measure of how intense a sound is compared to a reference intensity level.
02

Using the Decibel Intensity Formula

The formula to convert from decibels to intensity is given by:\[L = 10 imes ext{log}_{10}\left(\frac{I}{I_0}\right)\]where \(L\) is the intensity level in decibels, \(I\) is the sound intensity, and \(I_0 = 10^{-12} \text{W/m}^{2}\) is the reference intensity.
03

Re-arranging the Formula to Solve for Intensity

We want to find \(I\), so re-arranging the formula gives:\[I = I_0 \times 10^{\frac{L}{10}}\]
04

Substituting the Values

Given \(L = 15 \text{ dB}\), substitute into the formula:\[I = 10^{-12} \times 10^{1.5}\]Simplifying this gives:\[I = 10^{-12} \times 31.62 \]\[I \approx 3.162 imes 10^{-11} \text{ W/m}^{2}\]
05

Interpreting the Result

The intensity of a whisper is approximately \(3.162 \times 10^{-11} \text{ W/m}^{2}\), meaning it is slightly more intense than the reference intensity.

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Key Concepts

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

Decibels
Decibels are a unit of measurement used to express the intensity of a sound. This unit might sound a bit complex at first due to its logarithmic nature, but it is essentially a convenient way to express the vast range of sound intensities our ears can perceive.
When we hear sounds, we can range from the faintest whisper to the loudest rock concert. Measuring these intensities on a linear scale would be impractical due to the enormous range. This is where decibels come into play.
  • A small change in decibels can represent a large change in actual sound intensity.
  • The decibels scale is not linear but logarithmic, meaning each increase of 10 dB represents a tenfold increase in sound intensity.
Understanding decibels helps us communicate and compare sound levels effectively without dealing with huge numbers. When you see a sound level of 15 dB, like a whisper, it's telling us that this sound is significantly quieter than our typical sound exposure levels.
Logarithmic Units
The notion of logarithms might seem daunting, but they play a fundamental role in understanding and calculating sound intensities. When we say that decibels are measured using logarithmic units, it simply means we are translating sound intensity into a format that's easier to manage and comprehend.
Here's why logarithmic units are useful when it comes to sound intensity:
  • The human ear perceives sound intensity logarithmically, meaning we notice equal ratios of sound pressure changes rather than equal increments.
  • With logarithmic units, we can easily calculate relative changes in sound intensity using simple addition and subtraction rather than complex multiplications.
The logarithmic scale condenses a wide array of sound pressures into a manageable system, making it simpler to grasp and compare sound levels that our ears experience every day.
Reference Intensity
The concept of reference intensity is crucial when working with decibels and sound intensity measurement. Reference intensity serves as a baseline or starting point for comparing other sound intensities. The standard reference intensity used in calculating decibels is denoted as \(I_0\) and is equal to \(10^{-12} \text{W/m}^2\). This reference level is the quietest sound a typical human ear can detect.
Here's why reference intensity matters:
  • It provides a consistent point of comparison for determining intensity levels of different sounds.
  • The use of \(I_0\) helps standardize how we measure and interpret sound levels across different environments.
In practice, when you hear that a whisper has an intensity of 15 dB, it means the sound is 15 decibels more intense than this reference intensity. Understanding this baseline allows you to attribute meaningfulness to the dB level of any sound and approach the concept of sound intensities with clarity.

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Most popular questions from this chapter

Find the speed of sound in a diatomic ideal gas that has a density of \(3.50 \mathrm{~kg} / \mathrm{m}^{3}\) and a pressure of \(215 \mathrm{kPa}\) We know that \(v=\sqrt{\gamma R T / M}\) and can find the temperature from the pressure. Using the gas law \(P V=(m / M) R T\), $$ \frac{R T}{M}=P \frac{V}{m} $$ Moreover, \(\rho=m / V\), and so the expression for the speed becomes $$ v=\sqrt{\frac{\gamma P}{\rho}}=\sqrt{\frac{(1.40)\left(215 \times 10^{3} \mathrm{~Pa}\right)}{3.50 \mathrm{~kg} / \mathrm{m}^{3}}}=293 \mathrm{~m} / \mathrm{s} $$ We used the fact that \(\gamma \approx 1.40\) for a diatomic ideal gas, as discussed in Chapter 20 .

An increase in pressure of \(100 \mathrm{kPa}\) causes a certain volume of water to decrease by \(5 \times 10^{-3}\) percent of its original volume. ( \(a\) ) What is the bulk modulus of water? (b) What is the speed of sound (compression waves) in water?

A tuning fork having a frequency of \(400 \mathrm{~Hz}\) (shown in Fig. 23-2) is moved away from an observer and toward a flat wall with a speed of \(2.0 \mathrm{~m} / \mathrm{s}\). What is the apparent frequency \((a)\) of the unreflected sound waves coming directly to the observer, and \((b)\) of the sound waves coming to the observer after reflection? ( \(c\) ) How many beats per second are heard? Assume the speed of sound in air to be \(340 \mathrm{~m} / \mathrm{s}\) (a) The fork, the source, is receding from the observer in the positive direction and so we use \(+v_{s} .\) It doesn't matter what the sign associated with \(v_{o}\) is since \(v_{e}=0\). $$ f_{o}=f_{s} \frac{v \pm v_{o}}{v \mp v_{s}}=(400 \mathrm{~Hz}) \frac{340 \mathrm{~m} / \mathrm{s}+0}{340 \mathrm{~m} / \mathrm{s}+2.0 \mathrm{~m} / \mathrm{s}}=397.7 \mathrm{~Hz}=398 \mathrm{~Hz} $$ The source is moving away from the observer and the frequency is properly shifted down from \(400 \mathrm{~Hz}\) to \(398 \mathrm{~Hz}\). (b) Think of the wall as a source that reflects sound of the same frequency as that which impinges upon it. The wave crests reaching the wall are closer together than normally because the fork is moving toward the wall. Therefore, the wall will appear as a stationary source emitting sound of a higher frequency than \(400 \mathrm{~Hz}\). due to the \(2.0-\mathrm{m} / \mathrm{s}\) motion of the fork. Alternatively we can think of the reflected wave as if it came from a source (the wall) moving at \(2.0 \mathrm{~m} / \mathrm{s}\) toward the observer. Hence, we enter \(-v_{x}:\) $$ f_{o}=f_{s} \frac{v \pm v_{o}}{v \mp v_{s}}=(400 \mathrm{~Hz}) \frac{340 \mathrm{~m} / \mathrm{s}+0}{340 \mathrm{~m} / \mathrm{s}-2.0 \mathrm{~m} / \mathrm{s}}=402.4 \mathrm{~Hz}=402 \mathrm{~Hz} $$ and the frequency is properly shifted up. (c) Beats per second \(=\) Difference between frequencies \(=(402.4-397.7) \mathrm{Hz}=4.7\) beats per second

A rock band might easily produce a sound level of \(107 \mathrm{~dB}\) in a room. To two significant figures, what is the sound intensity at \(107 \mathrm{~dB}\) ?

What sound intensity is \(3.0 \mathrm{~dB}\) louder than a sound of intensity of \(10 \mu \mathrm{W} / \mathrm{cm}^{2}\) ?
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