Problem 40 - Precalculus Mathematics for Calculus

Question

These exercises deal with logarithmic scales. The intensity of the sound of a subway train was measured at 98 dB. Find the intensity in \(\mathrm{W} / \mathrm{m}^{2}\).

Step-by-Step Solution

Answer

6.31 x 10^{-3} W/m^2.

Step 1: Understanding the Problem

We know that the decibel (dB) scale is a logarithmic scale used to measure sound intensity. The relationship between the intensity level in decibels, denoted as \( L \), and the actual intensity \( I \) in watts per square meter \( \mathrm{W/m}^2 \) follows the formula: \[ L = 10 \log_{10}\left( \frac{I}{I_0} \right) \]where \( I_0 = 1 \times 10^{-12} \, \mathrm{W/m}^2 \) is the reference intensity, the threshold of hearing.

Step 2: Rearranging the Formula

Rearrange the logarithmic formula to solve for the intensity \( I \) in terms of \( L \):\[ L = 10 \log_{10}\left( \frac{I}{I_0} \right) \] means that \[ \frac{I}{I_0} = 10^{L/10} \].Therefore, \[ I = I_0 \cdot 10^{L/10} \].

Step 3: Substitute the Given Values

Given that the intensity level \( L \) is \( 98 \, \text{dB} \), substitute \( L = 98 \) and \( I_0 = 1 \times 10^{-12} \, \mathrm{W/m}^2 \) into the equation:\[ I = 1 \times 10^{-12} \cdot 10^{98/10} \].

Step 4: Perform the Calculation

Calculate \( 10^{98/10} = 10^{9.8} \). Use a calculator to find that:\[ 10^{9.8} \approx 6.31 \times 10^9 \].Then,\[ I = 1 \times 10^{-12} \times 6.31 \times 10^9 = 6.31 \times 10^{-3} \text{ W/m}^2 \].

Step 5: Concluding the Solution

The intensity of the sound from the subway train, in terms of energy per area per unit time, is \( 6.31 \times 10^{-3} \, \mathrm{W/m}^2 \).

Key Concepts

Decibels

The term decibel (dB) is essential in measuring sound intensity. It relates to the perception of sound loudness by the human ear. The decibel is a logarithmic unit used to express the ratio between two values of a physical quantity, often power or intensity. In our context, it measures how intense a sound is compared to a reference level.
The formula relating the decibel to sound intensity is:

• \( L = 10 \log_{10}\left( \frac{I}{I_0} \right) \)

• \( L \) is the sound level in decibels.
• \( I \) is the intensity of the sound in watts per square meter.
• \( I_0 \) is a reference intensity, typically taken as \( 1 \times 10^{-12} \, \mathrm{W/m}^2 \), corresponding to the quietest sound a human can hear.

A significant advantage of the decibel scale is its ability to transmute vast numeric data ranges into manageable figures. A change by 10 dB represents a 10-fold change in intensity, making it easier to convey measurement changes.

Sound Intensity

Sound intensity refers to the power carried by sound waves per unit area in a direction perpendicular to that area. It's an objective quantity that can be measured with precision, unlike the subjective perception of loudness. Sound intensity depends on two main elements:

• The power of the sound source
• The inverse relationship with the area over which the sound spreads

The farther the sound travels from its source, the more spread out its energy becomes, resulting in lower intensity.
The intensity level in decibels can be converted from sound intensity by using the formula given earlier. Since the human ear detects a wide range of intensities, a logarithmic scale like decibels helps convert this vast range into a more relatable form.

Watts per Square Meter

The measurement of sound intensity uses units of watts per square meter (\( \mathrm{W/m}^2 \)). This unit describes how much power is supplied by the sound through a square meter of area per second. In the realm of acoustics, understanding this conversion is crucial as it underscores the energy aspect of sound.
Calculating sound intensity in \( \mathrm{W/m}^2 \) requires using the standard reference intensity \( I_0 = 1 \times 10^{-12} \, \mathrm{W/m}^2 \). It illustrates the threshold of human hearing. By knowing the decibel level, as shown in the exercise, you can calculate the sound intensity using the rearranged formula:

• \( I = I_0 \cdot 10^{L/10} \)

Once the math is done, you achieve a result in \( \mathrm{W/m}^2 \), offering a quantitative perspective of how loud a sound is in energetic terms on a per area basis.