Problem 38 - Algebra and Trigonometry

Question

The intensity of the sound of traffic at a busy intersection was measured at \(2.0 \times 10^{-5} \mathrm{W} / \mathrm{m}^{2}\). Find the intensity level in decibels.

Step-by-Step Solution

Answer

The intensity level is approximately 73.01 decibels.

Step 1: Understand the formula for intensity level in decibels

The intensity level in decibels (\(\beta\)) is calculated using the formula: \[ \beta = 10 \log_{10} \left(\frac{I}{I_0}\right) \] where \(I\) is the intensity of the sound, and \(I_0\) is the reference intensity, typically \(1.0 \times 10^{-12} \text{W/m}^2\).

Step 2: Substitute the known values into the formula

Given \(I = 2.0 \times 10^{-5} \text{W/m}^2\) and \(I_0 = 1.0 \times 10^{-12} \text{W/m}^2\), substitute these values into the formula: \[ \beta = 10 \log_{10} \left(\frac{2.0 \times 10^{-5}}{1.0 \times 10^{-12}}\right) \]

Step 3: Simplify the fraction inside the logarithm

Calculate the fraction \(\frac{2.0 \times 10^{-5}}{1.0 \times 10^{-12}}\): \[ \frac{2.0 \times 10^{-5}}{1.0 \times 10^{-12}} = 2.0 \times 10^7 \]

Step 4: Compute the logarithm

Calculate \(\log_{10} (2.0 \times 10^7)\): \[ \log_{10} (2.0 \times 10^7) = \log_{10} 2.0 + \log_{10} 10^7 = 0.3010 + 7 = 7.3010 \]

Step 5: Calculate the intensity level

Now, substitute the value from Step 4 back into the formula: \[ \beta = 10 \times 7.3010 = 73.01 \] Therefore, the intensity level is approximately 73.01 decibels.

Key Concepts

Decibels and Sound Intensity

Decibels are a unit used to measure the intensity of sound. This unit might seem complex, but it's essentially a way to express sound intensity levels on a scale that's easy to understand.

Sound intensity, measured in watts per square meter (\( \text{W/m}^2 \)), can vary by huge amounts. Decibels help make these variations more manageable. By converting sound intensity to decibels, we use a logarithmic scale, which allows us to handle both very large and very small numbers efficiently.

With this scale, every increase of 10 decibels represents a tenfold increase in intensity. For example, a sound that registers at 30 decibels is 10 times more intense than a sound at 20 decibels.

Understanding Logarithms

Logarithms are mathematical tools that help simplify complex multiplicative relationships.When we talk about logarithms in the context of sound, we are often dealing with the base 10 logarithms (\( \log_{10} \)). Here's a simple breakdown:

• If you take a number and divide it by a reference number, the result can be huge or tiny, depending on the scenario.
• Logarithms shrink these numbers into a more manageable form, making them easier to interpret.

In our formula for calculating decibels, \( \beta = 10 \log_{10} \left(\frac{I}{I_0}\right) \), the logarithm helps us express the ratio of \( I \) (the sound intensity) to \( I_0 \) (the reference intensity) in decibels.

Using logarithms, we simplify potentially lengthy calculations, as seen when handling the value \( 2.0 \times 10^7 \) in the original solution.

Reference Intensity in Decibel Calculations

Reference intensity is a crucial concept in understanding how decibels are calculated. This value, \( I_0 \), serves as the baseline or the quietest sound a typical human ear can hear.

Standard reference intensity is \( 1.0 \times 10^{-12} \, \text{W/m}^2 \). Without this reference point, making relative comparisons between different sound levels becomes challenging.

• By comparing actual sound intensity to \( I_0 \), we determine how much louder (\( I \)) is than the quietest sound.
• This comparison gives us a ratio that the logarithm converts into a decibel level.

Knowing the reference intensity is similar to knowing the base point of a measuring scale, allowing us to determine how loud or soft a particular sound is relative to our baseline measurement.