Problem 41 - Algebra and Trigonometry

Question

A law of physics states that the intensity of sound is inversely proportional to the square of the distance \(d\) from the source: \(I=k / d^{2} .\) (a) Use this model and the equation $$B=10 \log \frac{I}{I_{0}}$$ (described in this section) to show that the decibel levels \(B_{1}\) and \(B_{B}\) at distances \(d\) and \(d_{2}\) from a sound source are related by the equation $$B_{2}=B_{1}+20 \log \frac{d_{1}}{d_{2}}$$ (b) The intensity level at a rock concert is 120 \(\mathrm{dB}\) at a distance 2 \(\mathrm{m}\) from the speakers. Find the intensity level at a distance of 10 \(\mathrm{m} .\)

Step-by-Step Solution

Answer

The intensity level at 10 m is approximately 106.02 dB.

Step 1: Express Intensity for Both Distances

Given the equation for intensity: \( I = \frac{k}{d^2} \). We write the intensity at distances \( d_1 \) and \( d_2 \) as:\[ I_1 = \frac{k}{d_1^2} \] and \[ I_2 = \frac{k}{d_2^2} \]

Step 2: Use the Decibel Formula for Both Distances

Plug the intensity expressions into the decibel formula \( B = 10 \log \frac{I}{I_0} \). Thus:\[ B_1 = 10 \log \frac{I_1}{I_0} = 10 \log \frac{k}{d_1^2 I_0} \]\[ B_2 = 10 \log \frac{I_2}{I_0} = 10 \log \frac{k}{d_2^2 I_0} \]

Step 3: Derive the Relationship Between \( B_1 \) and \( B_2 \)

Subtract the equations for \( B_1 \) and \( B_2 \):\[ B_2 - B_1 = 10 \log \frac{k}{d_2^2 I_0} - 10 \log \frac{k}{d_1^2 I_0} \]\[ B_2 - B_1 = 10 \log \left(\frac{k}{d_2^2 I_0} \cdot \frac{d_1^2 I_0}{k}\right) \]\[ B_2 - B_1 = 10 \log \frac{d_1^2}{d_2^2} \]\[ B_2 - B_1 = 20 \log \frac{d_1}{d_2} \]Thus, \( B_2 = B_1 + 20 \log \frac{d_1}{d_2} \).

Step 4: Apply the Formula to Find Intensity Level at 10 m

Given \( B_1 = 120 \) dB at \( d_1 = 2 \) m, we need to find \( B_2 \) at \( d_2 = 10 \) m. Using the relationship:\[ B_2 = B_1 + 20 \log \frac{d_1}{d_2} \]\[ B_2 = 120 + 20 \log \frac{2}{10} \]\[ B_2 = 120 + 20 \log 0.2 \]

Step 5: Calculate \( B_2 \) Using Logarithm

First compute \( \log 0.2 \):\[ \log 0.2 = \log \frac{1}{5} \approx -0.69897 \]Then compute:\[ B_2 = 120 + 20(-0.69897) \]\[ B_2 = 120 - 13.9794 \]\[ B_2 \approx 106.02 \]

Key Concepts

Sound Intensity

The concept of sound intensity is fundamental to understanding how sound propagates through space. Sound intensity refers to the amount of energy that a sound wave transports per unit area, perpendicular to the direction of the sound wave. The mathematical definition of sound intensity is \[ I = \frac{k}{d^2} \]where

• \( I \) represents the sound intensity,
• \( k \) is a constant specific to the sound source,
• \( d \) stands for the distance from the sound source.

The formula indicates an important property known as the inverse square law. This law states that as you move further away from the sound source, the intensity decreases proportionally to the square of the distance. For instance, doubling the distance from the source results in the sound intensity being reduced to one-fourth of its original level.

Decibels

Decibels (dB) are a logarithmic unit used to measure sound intensity levels. This system allows us to handle the vast range of sound intensities that humans can hear. In decibels, the sound intensity level \(B\) is calculated using the formula \(B = 10 \log \frac{I}{I_0}\)where:

• \(I\) is the sound intensity,
• \(I_0\) is the reference sound intensity (usually the threshold of hearing \(10^{-12}\) W/m\(^2\)).

The use of logarithms compresses the wide range of intensity values into a manageable scale. A small change in decibels represents a significant change in actual sound intensity. For instance, an increase of 10 dB represents ten times more sound intensity. In the context of the given exercise, the decibel level at different distances is calculated based on changing sound intensities.

Logarithms

Logarithms are mathematical tools used to transform multiplicative relationships into additive ones. The basic idea is that if you have two numbers and their logarithms, adding these logarithms is equal to the original numbers being multiplied together. In sound measurements, logarithms help us manage the large disparity in sound intensity values. For example, the power in the formula \[B = 10 \log \frac{I}{I_0}\]involves a logarithm base 10. It transforms the ratio of sound intensity \(I\) to the reference intensity \(I_0\) into a more intuitive scale. When comparing decibel levels at different distances, the formula \[B_2 = B_1 + 20 \log \frac{d_1}{d_2}\]uses logarithms to express the relationship between distance and intensity in terms of decibels. This formula stemmed from properties of logarithms that allow us to understand changes in sound levels beautifully and compactly over distances. Thus, logarithms are crucial in making complex relationships more comprehensible in sound intensity calculations.