Problem 55
Question
The threshold of pain is generally taken to be around \(140 \mathrm{~dB}\). Find the intensity of sound \(I\) corresponding to \(140 \mathrm{~dB}\).
Step-by-Step Solution
Verified Answer
The intensity is 100 W/m².
1Step 1: Understand the Decibel Scale
The decibel (dB) scale is a logarithmic scale used to measure the intensity of sound. It is given by the formula: \( eta = 10 imes ext{log}_{10} rac{I}{I_0} \), where \( \beta \) is the sound level in decibels, \( I \) is the intensity of the sound, and \( I_0 = 10^{-12} \mathrm{~W/m^2} \) is the reference intensity.
2Step 2: Set Up the Equation for 140 dB
We know that the sound level \( \beta \) is \( 140 \mathrm{~dB} \). Substitute \( \beta = 140 \) into the equation: \( 140 = 10 \times ext{log}_{10} \frac{I}{10^{-12}} \).
3Step 3: Simplify the Equation
Divide both sides by 10 to simplify: \( 14 = ext{log}_{10} \frac{I}{10^{-12}} \).
4Step 4: Solve for Sound Intensity
To find \( I \), rewrite the equation in terms of \( I \): \( \frac{I}{10^{-12}} = 10^{14} \). Multiply both sides by \( 10^{-12} \): \( I = 10^{14} \times 10^{-12} \).
5Step 5: Calculate the Intensity
Simplify the expression: \( I = 10^{2} = 100 \mathrm{~W/m^2} \). This is the intensity corresponding to \( 140 \mathrm{~dB} \).
Key Concepts
Decibel ScaleLogarithmic ScaleReference Intensity
Decibel Scale
The decibel (dB) scale is a handy tool for expressing sound intensity levels. It's a logarithmic measure used primarily to compare the power levels of sound. Decibels are commonly used because they offer a more manageable range, translating large variations in sound intensity into smaller, more understandable numbers.
This scale is defined by the formula:
Using the decibel scale for calculations involves logarithmic values, making it ideal for measuring extremely high or low sound intensities without using cumbersome figures.
This scale is defined by the formula:
- \( \beta = 10 \times \log_{10} \frac{I}{I_0} \)
Using the decibel scale for calculations involves logarithmic values, making it ideal for measuring extremely high or low sound intensities without using cumbersome figures.
Logarithmic Scale
A logarithmic scale is particularly useful for plotting data that covers a large range of values. Here, it helps convert the wide range of possible sound intensities into a more comprehensible scale. This scale makes it easier to compare the intensity of sounds that are vastly different in magnitude.
On the decibel scale:
In context, the logarithmic scale transforms the threshold of pain (which is often at 140 dB) into a simple calculation involving exponents. This makes it far easier to conceptualize and work with than if we were dealing with the raw sound intensities directly.
On the decibel scale:
- The relationship is logarithmic, meaning it corresponds to powers of ten.
- A 10 dB increase is perceived as approximately doubling of the loudness.
In context, the logarithmic scale transforms the threshold of pain (which is often at 140 dB) into a simple calculation involving exponents. This makes it far easier to conceptualize and work with than if we were dealing with the raw sound intensities directly.
Reference Intensity
Reference intensity, noted as \( I_0 \), serves as a baseline for sound comparisons. It is critical in calculating decibel levels and is set at \( 10^{-12} \text{W/m}^2 \), representing the faintest sound that can be heard by a typical human ear. This value is fundamental as a lower limit in sound intensity measurements for decibel calculations.
- The reference intensity allows us to quantify how much more intense a sound is than this baseline.
- Sound intensity values are expressed relative to this low threshold, allowing for normalized comparisons.
Other exercises in this chapter
Problem 54
In Problems \(51-56\), use the laws of logarithms in Theorem \(6.2 .1\) to rewrite the given expression as one logarithm. $$ \ln \left(\frac{x}{y}\right)-2 \ln
View solution Problem 55
Find the \(x\) -intercepts of the graph of the given function. $$ f(x)=\frac{10}{2+e^{-2 x}}-1 $$
View solution Problem 55
In Problems \(51-56\), use the laws of logarithms in Theorem \(6.2 .1\) to rewrite the given expression as one logarithm. $$ \ln 5+\ln 5^{2}+\ln 5^{3}-\ln 5^{6}
View solution Problem 56
Find the \(x\) -intercepts of the graph of the given function. $$ f(x)=\frac{2^{x}-6+2^{3-x}}{x+2} $$
View solution