Problem 56 - Precalculus with Calculus Previews

Question

The threshold of pain is generally taken to be around \(140 \mathrm{~dB}\). Find the intensity of sound \(I\) corresponding to \(140 \mathrm{~dB}\). $$ I=\frac{k}{d^{2}} $$ where \(k\) is the constant of proportionality. Suppose \(d_{1}\) and \(d_{2}\) are distances from a source of sound, and that the corresponding intensity levels of the sounds are \(b_{1}\) and \(b_{2}\). Use (11) in (10) to show that \(b_{1}\) and \(b_{2}\) are related by $$ b_{2}=b_{1}+20 \log _{10} \frac{d_{1}}{d_{2}} $$.

Step-by-Step Solution

Answer

The intensity corresponding to 140 dB is 100 W/m². The sound levels are related by \( b_2 = b_1 + 20 \log_{10} \frac{d_1}{d_2} \).

Step 1: Understand the Problem

We need to find the sound intensity in decibels (dB), given the threshold of pain is around 140 dB. Then, we have to show the relationship between two levels of sound intensity at different distances.

Step 2: Define the Relationship Between Intensity and Distance

The intensity of sound at a distance from the source is given by the equation \( I = \frac{k}{d^2} \), where \( k \) is a constant. The intensity level \( b \) in dB is related to intensity \( I \) by the formula \( b = 10 \log_{10} \left( \frac{I}{I_0} \right) \), where \( I_0 \) is the reference intensity.

Step 3: Use Intensity Level Formula to Find I for 140 dB

For 140 dB, since \( I_0 \) is usually taken as \( 10^{-12} \text{ W/m}^2 \), we use \( 140 = 10 \log_{10} \left( \frac{I}{10^{-12}} \right) \). From this, solve for \( I \):\[ 140 = 10 \log_{10} (I) + 120 \]\[ 20 = 10 \log_{10} (I) \]\[ 2 = \log_{10} (I) \]\[ I = 10^2 = 100 \text{ W/m}^2 \]

Step 4: Relate Intensity Levels at Different Distances

Given \( d_1 \) and \( d_2 \) with corresponding levels \( b_1 \) and \( b_2 \), use \( I_1 = \frac{k}{d_1^2} \) and \( I_2 = \frac{k}{d_2^2} \). Hence: \[ b_1 = 10 \log_{10} \left( \frac{I_1}{I_0} \right) = 10 \log_{10} \left( \frac{k}{d_1^2 I_0} \right) \]\[ b_2 = 10 \log_{10} \left( \frac{I_2}{I_0} \right) = 10 \log_{10} \left( \frac{k}{d_2^2 I_0} \right) \]

Step 5: Show the Relationship From Formulas

Subtract the equations for \( b_1 \) and \( b_2 \):\[ b_2 - b_1 = 10 \log_{10} \left( \frac{k}{d_2^2 I_0} \right) - 10 \log_{10} \left( \frac{k}{d_1^2 I_0} \right) \]Using properties of logarithms, this simplifies to:\[ b_2 - b_1 = 10 \log_{10} \left( \frac{d_1^2}{d_2^2} \right) \]\[ b_2 = b_1 + 20 \log_{10} \frac{d_1}{d_2} \]

Step 6: Conclusion

Thus, we have shown that the relationship between sound levels at different distances confirms the formula \( b_2 = b_1 + 20 \log_{10} \frac{d_1}{d_2} \).

Key Concepts

Decibel Scale

The decibel scale is a logarithmic measure used to express the intensity of sound. It's essential for quantifying how loud a sound is perceived by the human ear. The scale is relative and uses a reference intensity, typically the threshold of hearing set at \( I_0 = 10^{-12} \text{ W/m}^2 \). The formula to convert sound intensity \( I \) into decibels is:

• \( b = 10 \log_{10} \left( \frac{I}{I_0} \right) \)
This equation transforms the actual physical measurement of intensity into a more manageable number that reflects how we perceive sound; small changes in intensity magnitude can result in noticeable changes on the decibel scale.
When a sound has an intensity level of 140 dB, it is at the threshold of pain for most people. To find the actual intensity \( I \) corresponding to 140 dB, you rearrange the decibel formula to solve for \( I \). You'll find that much higher decibel values represent extremely intense sounds, capable of causing discomfort or damage to the human ear.

Inverse Square Law

The inverse square law describes how sound intensity decreases as the distance from the sound source increases. This principle holds true for any point source emitting sound waves uniformly in all directions, and it's mathematically described by the formula:

• \( I = \frac{k}{d^2} \)

Here, \( I \) is the sound intensity at distance \( d \), and \( k \) is known as the constant of proportionality.
This law implies that the intensity of sound decreases with the square of the distance from the source. For example, if you double the distance from the source, the intensity of sound reduces to a quarter of its original value. This principle is crucial for understanding why sounds get quieter as you move away from their source, and it helps calculate sound intensities at various distances for real-world applications, such as in setting up audio equipment.

Logarithmic Relationships

Logarithmic relationships are integral in understanding how sound intensities relate when considering different distances from a source. This is seen particularly in how the decibel levels change when the distance varies.
From the step-by-step solution, you learn that the difference in decibel level \( b_2 - b_1 \) when measuring sound at two different distances \( d_2 \) and \( d_1 \) from the source, is given by:

• \( b_2 = b_1 + 20 \log_{10} \frac{d_1}{d_2} \)

This relationship arises due to the logarithmic nature of the decibel scale.
It shows how doubling the distance can affect sound perceptions, providing a more intuitive understanding than simple arithmetic changes. Such a framework allows engineers and scientists to predict how modifications in environmental setups or listeners' positions will influence sound levels greatly, using the logarithmic properties to handle large intensities and distances efficiently.