/*! 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 11 A noise-level meter reads the so... [FREE SOLUTION] | 91Ó°ÊÓ

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A noise-level meter reads the sound level in a room to be \(85.0 \mathrm{~dB}\). What is the sound intensity in the room? Sound level \((\beta)\), in \(\mathrm{dB}\), is given by \(\beta=10 \log _{10}\left(I / I_{0}\right)\) and here it equals \(85.0 \mathrm{~dB}\). Accordingly, $$ \begin{array}{l} \text { and }\\\ \begin{aligned} \beta &=10 \log _{10}\left(\frac{1}{1.00 \times 10^{-12} \mathrm{w} / \mathrm{m}^{2}}\right)=85.0 \mathrm{~dB} \\ \log _{10}\left(\frac{l}{1.00 \times 10^{-12} \mathrm{~W} / \mathrm{m}^{2}}\right) &=\frac{85.0}{10}=8.50 \\ \frac{I}{1.00 \times 10^{-12} \mathrm{~W} / \mathrm{m}^{2}} &=\text { antilog }_{10} 8.50=3.16 \times 10^{\mathrm{k}} \\ I &=\left(1.00 \times 10^{-12} \mathrm{~W} / \mathrm{m}^{2}\right)\left(3.16 \times 10^{8}\right)=3.16 \times 10^{-4} \mathrm{~W} / \mathrm{m}^{2} \end{aligned} \end{array} $$

Short Answer

Expert verified
The sound intensity is \(3.16 \times 10^{-4} \text{ W/m}^2\).

Step by step solution

01

Understand the relationship between decibel and intensity

The sound level in decibels (\(eta\)) is related to the sound intensity (\(I\)) by the formula:\[\beta = 10 \log_{10}\left( \frac{I}{I_{0}} \right)\]where \(I_0\) is the reference intensity, \(1.00 \times 10^{-12} \text{ W/m}^2\). Given \(\beta\) is \(85.0 \text{ dB}\).
02

Solve for the logarithm equation

To find \(I\), we first solve the equation for \(\log_{10}\left( \frac{I}{I_0} \right)\) by dividing both sides of the equation \(\beta = 10 \log_{10}\left( \frac{I}{I_0} \right)\) by 10:\[\log_{10}\left( \frac{I}{I_0} \right) = \frac{85.0}{10} = 8.50\]
03

Calculate the antilogarithm of the equation

The next step is to find the antilogarithm to solve for \(\frac{I}{I_0}\):\[\frac{I}{I_0} = 10^{8.50}\]Solving this, we find:\[\frac{I}{I_0} = 3.16 \times 10^{8}\]
04

Solve for intensity \(I\)

Using the result from Step 3, solve for \(I\):\[I = I_0 \times 3.16 \times 10^{8}\]Plug in the value of \(I_0\):\[I = \left(1.00 \times 10^{-12} \text{ W/m}^2\right) \times 3.16 \times 10^{8}\]\[I = 3.16 \times 10^{-4} \text{ W/m}^2\]

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Key Concepts

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

Decibels
Decibels (dB) are a unit of measurement used to express the intensity of sound. They offer a way to compare the power of different sound waves by using a logarithmic scale. This scale helps us deal with the vast range of sound intensities we encounter in the real world, from the quietest whisper to the loudest jet engine.

The decibel scale is logarithmic, meaning each increase of 10 dB represents a tenfold increase in intensity. Thus, a sound that is 10 dB louder than another is 10 times more intense.
  • A sound measured at 0 dB is at the threshold of hearing for the average person.
  • Every 10 dB increase roughly doubles the perceived loudness of the sound.
This logarithmic nature makes decibels incredibly useful when dealing with the varied and vast sound pressures found in our environment.
Sound Level
Sound level, denoted by \(\beta\), is a measure of the intensity of sound relative to a chosen reference intensity. In practical terms, it indicates how loud a sound is perceived.

The relationship between sound level and intensity is expressed by the equation:
\[\beta = 10 \log_{10}\left( \frac{I}{I_{0}} \right)\]
where:
  • \(\beta\) is the sound level in decibels.
  • \(I\) is the sound intensity being measured.
This formula allows us to calculate the level of sound from its intensity. The logarithmic connection helps compress the large range of sound intensities into a more manageable scale that can be practically used in sound measurements. By using this equation, we convert the intensity into the much easier to understand decibel form.
Reference Intensity
Reference intensity (\(I_0\)) is the baseline sound intensity that the sound level formula references. It provides a standard for comparing other sound intensities, making it central to the measure of sound levels in decibels.

This standard reference intensity is \(1.00 \times 10^{-12} \text{ W/m}^2\), representing the quietest sound humans can typically hear.
  • This is the threshold of hearing, the point at which ambient noise is barely perceptible.
  • All other sound intensities are compared to this reference.
Given this standard reference, when we know the sound level in decibels, we can precisely compute the actual intensity using the reference intensity as a baseline.

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Most popular questions from this chapter

Determine the speed of sound in carbon dioxide \((M=44 \mathrm{~kg} / \mathrm{kmo}\) \(y=1.30\) ) at a pressure of \(0.50\) atm and a temperature of \(400^{\circ} \mathrm{C}\).

An automobile moving at \(30.0 \mathrm{~m} / \mathrm{s}\) is approaching a factory whistle that has a frequency of \(500 \mathrm{~Hz}\). (a) If the speed of sound in air is \(340 \mathrm{~m} / \mathrm{s}\), what is the apparent frequency of the whistle as heard by the driver? (b) Repeat for the case of the car leaving the factory at the same speed. This is a Doppler shift problem. Draw an arrow from observer to source; this is the positive direction. Here in part \((a)\) the observer is moving in the positive direction, and \(u_{s}=0 .\) Hence, use \(+u_{o}\) and So (a) \(f_{f}=f_{s} \frac{v \pm v_{o}}{v \mp v_{s}}=(500 \mathrm{~Hz}) \frac{340 \mathrm{~m} / \mathrm{s}+30.0 \mathrm{~m} / \mathrm{s}}{340 \mathrm{~m} / \mathrm{s}-0}=544 \mathrm{~Hz}\) With the car leaving in the negative direction use \(-\mathrm{u}_{o}\) and (b) \(f_{n}=f_{0} \frac{v \pm v_{o}}{v \mp v_{3}}=(500 \mathrm{~Hz}) \frac{340 \mathrm{~m} / \mathrm{s}-30.0 \mathrm{~m} / \mathrm{s}}{340 \mathrm{~m} / \mathrm{s}-0}=456 \mathrm{~Hz}\)

One sound has an intensity level of \(75.0 \mathrm{~dB}\), while a second has an intensity level of \(72.0 \mathrm{~dB}\). What is the intensity level when the two sounds are combined?

At S.T.P., the speed of sound in air is \(331 \mathrm{~m} / \mathrm{s}\). Determine the speed of sound in hydrogen at S.T.P. if the specific gravity of hydrogen relative to air is \(0.0690\) and if \(y=1.40\) for both gases.

What is the speed of compression waves (sound waves) in water? The bulk modulus for water is \(2.2 \times 10^{9} \mathrm{~N} / \mathrm{m}^{2}\). $$ v=\sqrt{\frac{\text { Bulk modulus }}{\text { Density }}}=\sqrt{\frac{2.2 \times 10^{9} \mathrm{~N} / \mathrm{m}^{2}}{1000 \mathrm{~kg} / \mathrm{m}^{3}}}=1.5 \mathrm{~km} / \mathrm{s} $$

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