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Chapter 23: Problem 53

$$ \begin{array}{l} \text { What sound intensity is } 3.0 \mathrm{~dB} \text { louder than a sound of intensity of }\\\ 10 \mu \mathrm{W} / \mathrm{cm}^{2} ? \end{array} $$

Short Answer

Expert verified
The intensity is approximately \( 20 \, \mu \mathrm{W/cm}^2 \).

Step by step solution

01

Understand the Decibel Scale

The decibel (dB) scale is a logarithmic scale used to measure sound intensity. A change of 10 dB represents a tenfold change in intensity. Any change in decibels can be related to a change in intensity using the formula: \[ I_2 = I_1 \times 10^{\frac{\Delta L}{10}} \]where \( I_2 \) is the final intensity, \( I_1 \) is the initial intensity, and \( \Delta L \) is the change in decibels.
02

Assign Known Variables

We know \( I_1 = 10 \, \mu \mathrm{W}/\mathrm{cm}^2 = 10 \times 10^{-6} \, \mathrm{W/cm}^2 \) and \( \Delta L = 3.0 \, \mathrm{dB} \). Our task is to find \( I_2 \).
03

Apply the Formula

Insert the known values into the formula:\[ I_2 = 10 \times 10^{-6} \times 10^{\frac{3.0}{10}} \]
04

Calculate Intermediate Value

First calculate \( 10^{\frac{3.0}{10}} \):\[ 10^{\frac{3.0}{10}} \approx 1.99526 \]
05

Find the New Intensity

Substitute the intermediate result into the formula to find \( I_2 \):\[ I_2 = 10 \times 10^{-6} \times 1.99526 \]Calculate the final result:\[ I_2 \approx 19.9526 \times 10^{-6} \, \mathrm{W/cm}^2 \]
06

Round to Significant Figures

Since the initial intensity was given to 2 significant figures, round the final result to the same. Thus, \( I_2 \approx 20 \times 10^{-6} \, \mathrm{W/cm}^2 \).

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Key Concepts

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

Decibel Scale
To understand sound measurements, it's essential to grasp the decibel (dB) scale. The decibel scale is utilized globally. It signifies sound intensity in a way that's easier for humans to interpret. Since human perception of sound is logarithmic, the decibel scale reflects this attribute rather than a linear one. This means an increase of 10 dB represents a tenfold increase in sound intensity. Importantly, it's used not just for sound but also for other measurements like electronic signals. The scale allows for the comparison of two quantities, utilizing the relationship between a reference value and the current measurement.
Logarithmic Scale
A logarithmic scale presents numbers in powers of ten rather than linearly. This type of scale is particularly powerful for measurements that cover a vast range of values, such as sound intensities. Logarithmic scales on graphs can make exponential relationships appear linear. This can simplify the interpretation of data. The core feature of this scale is that each unit increase in the scale results in a value that is a fixed multiple of the previous value. For example, if the intensity is multiplied, the logarithmic scale will increase correspondingly. This makes dealing with high variations, such as those found in sound levels, easier to manage.
Sound Measurement
When we talk about sound measurement, we focus on how effectively sound energy and pressure waves are evaluated. These measurements often include both amplitude and frequency aspects. The intensity of sound, measured in watts per square meter (\( \text{W/m}^2 \)), defines how much sound power passes through an area. Measurement tools often use microphones to sense sound waves, converting these to electrical signals that can be analyzed. For a practical approach, sound levels are often reported in decibels, making the wide range of sound capabilities in the environment easier to understand. Moreover, sound measurement is crucial not only in scientific applications but also in protecting hearing and understanding noise pollution.
Intensity Formula
The intensity formula connects sound intensity changes on a logarithmic scale, as expressed in decibels. The formula helps calculate how much the intensity of sound changes when expressed in different dB levels. The formula, \[ I_2 = I_1 \times 10^{\frac{\Delta L}{10}} \] allows us to find the new intensity (\( I_2\)), when the current intensity (\( I_1\)) and change in decibels (\( \Delta L\)) are known. If the initial intensity is known, calculating the new intensity provides valuable insight into how much more, or less, intense a sound becomes after a certain change in decibels, facilitating better sound management and control. This is critical in environments where sound quality and levels need constant monitoring, like in studios or public events.

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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 \(400^{\circ} \mathrm{C}\).

When two tuning forks are sounded simultaneously, they produce one beat every \(0.30\) s. (a) By how much do their frequencies differ? (b) A tiny piece of chewing gum is placed on a prong of one fork. Now there is one beat every \(0.40 \mathrm{~s}\). Was this tuning fork the lower-or the higher-frequency fork? The number of beats per second equals the frequency difference. (a) Frequency difference \(=\frac{1}{0.30 \mathrm{~s}}=3.3 \mathrm{~Hz}\) (b) Frequency difference \(=\frac{1}{0.40 \mathrm{~s}}=2.5 \mathrm{~Hz}\) Adding gum to the prong increases its mass and thereby decreases its vibrational frequency. This lowering of frequency caused it to come closer to the frequency of the other fork. Hence, the fork in question had the higher frequency.

Suppose the intensity of sound increases by a multiplicative factor of \(10.0\), going from say \(1.0 \mathrm{~W} / \mathrm{m}^{2}\) to \(10 \mathrm{~W} / \mathrm{m}^{2}\) to \(100 \mathrm{~W} / \mathrm{m}^{2}\) and so on. By how much is the intensity level increased each time?

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, $$ \begin{array}{l} \text { and }\\\ \begin{aligned} \beta &=10 \log _{10}\left(\frac{1}{1.00 \times 10^{-12} \mathrm{w} / \mathrm{m}^{2}}\right)=85.0 \mathrm{~dB} \\ \log _{10}\left(\frac{l}{1.00 \times 10^{-12} \mathrm{~W} / \mathrm{m}^{2}}\right) &=\frac{85.0}{10}=8.50 \\ \frac{I}{1.00 \times 10^{-12} \mathrm{~W} / \mathrm{m}^{2}} &=\text { antilog }_{10} 8.50=3.16 \times 10^{\mathrm{k}} \\ I &=\left(1.00 \times 10^{-12} \mathrm{~W} / \mathrm{m}^{2}\right)\left(3.16 \times 10^{8}\right)=3.16 \times 10^{-4} \mathrm{~W} / \mathrm{m}^{2} \end{aligned} \end{array} $$

The two sound sources in vibrate in-phase. A loud sound is heard at \(P\) when \(L_{1}=L_{2}\). As \(L_{1}\) is slowly increased, the weakest sound is heard when \(L_{1}-L_{2}\) has the values \(20.0 \mathrm{~cm}, 60.0 \mathrm{~cm}\), and \(100 \mathrm{~cm}\). What is the frequency of the sound source if the speed of sound is \(340 \mathrm{~m} / \mathrm{s} ?\) The waves coming down directly from the fork toward the guy must be a little longer (more spaced) than the waves going up from the fork and back down from the wall, which have the same spacings. The weakest sound will be heard at \(P\) when a crest from \(S_{1}\) and a trough from \(S_{2}\) reach there at the same time. This will happen if \(L_{1}\) \(-L_{2}\) is \(\frac{1}{2} \lambda\), or \(\lambda+\frac{1}{2} \lambda\), or \(2 \lambda+\frac{1}{2} \lambda\), and so on. Hence, the increase in \(L_{1}\) between weakest sounds is \(\lambda\), and from the data we see that \(\lambda=\) \(0.400 \mathrm{~m}\). Then, from \(\lambda=\mathrm{v} / \mathrm{f}\), $$ f=\frac{v}{\lambda}=\frac{340 \mathrm{~m} / \mathrm{s}}{0.400 \mathrm{~m}}=850 \mathrm{~Hz} $$
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