Problem 66 - College Algebra

Question

For the following exercises, solve for the indicated value, and graph the situation showing the solution point. The formula for measuring sound intensity in decibels \(D\) is defi ed by the equation \(D=10 \log \left(\frac{I}{I_{0}}\right)\), where \(I\) is the intensity of the sound in watts per square meter and \(I_{0}=10^{-12}\) is the lowest level of sound that the average person can hear. How many decibels are emitted from a jet plane with a sound intensity of \(8.3 \cdot 10^{2}\) watts per square meter?

Step-by-Step Solution

Answer

The sound intensity of the jet plane is approximately 149.19 decibels.

Step 1: Understand the formula

The formula to measure sound intensity in decibels is given by \( D = 10 \log \left( \frac{I}{I_0} \right) \). Here, \( I \) is the sound intensity in watts per square meter, and \( I_0 = 10^{-12} \) is the reference sound intensity level.

Step 2: Plug values into the formula

We have \( I = 8.3 \cdot 10^2 \) watts per square meter and \( I_0 = 10^{-12} \). Substitute these values into the formula: \(D = 10 \log \left( \frac{8.3 \cdot 10^2}{10^{-12}} \right) \).

Step 3: Simplify the fraction

Simplify the fraction inside the logarithm: \( \frac{8.3 \cdot 10^2}{10^{-12}} = 8.3 \cdot 10^{2+12} = 8.3 \cdot 10^{14} \).

Step 4: Calculate the logarithm

Calculate the logarithm: \( \log(8.3 \times 10^{14}) \). Use the properties of logarithms to split it: \( \log(8.3) + \log(10^{14}) = \log(8.3) + 14 \).

Step 5: Evaluate and finalize

Using a calculator, find \( \log(8.3) \), which is approximately 0.919. Therefore, \( \log(8.3 \times 10^{14}) = 0.919 + 14 = 14.919 \). Then, multiply by 10: \( D = 10 \times 14.919 = 149.19 \). Thus, the sound intensity in decibels is approximately 149.19.

Step 6: Plot the graph

On a graph, plot a point on a coordinate plane to represent the decibels. The x-axis can represent different scenarios (or types of sound), and the y-axis can represent sound intensity levels in decibels. For this problem, plot a point at the y-value 149.19 corresponding to a jet plane on the x-axis.

Key Concepts

Understanding Logarithms in Decibel Calculation

Logarithms are a crucial part of understanding how to calculate the intensity of sound in decibels. When we see the equation \(D = 10 \log \left( \frac{I}{I_0} \right)\), the log function helps us transform a large range of sound intensity values into a more manageable scale. This works because logarithms reduce multiplicative scales to additive ones. By taking the log, very large numbers become more usable.
In our decibel equation, the base 10 logarithm \(\log_{10}\) is used. The formula essentially tells us how many times you would multiply 10 to get your intensity compared to the quietest sound \(I_0\). Logarithms, especially in the context of sound, allow us to understand phenomena that span many orders of magnitude. This is exactly the case with measuring sound, which can vary dramatically in intensity.
To solve our exercise step, simplifying the fraction to \( \frac{8.3 \cdot 10^2}{10^{-12}} = 8.3 \cdot 10^{14} \) makes it easy to see how logarithms help. When processed through the logarithmic function, this large number is easier to comprehend and manage, becoming approximately \(14.919\) before multiplying by 10. This is the beauty of logarithms!

Intensity of Sound and Decibel Levels

The intensity of sound is an important aspect when calculating decibels. Sound intensity refers to the amount of sound energy passing through a unit area. It is typically measured in watts per square meter. In the context of our exercise, understanding sound intensity this way tells us how loud something is in a measurable way.
The reference level \(I_0 = 10^{-12}\) watts per square meter is the threshold of hearing, which is the quietest sound that the average human ear can detect. In comparison, the given intensity \(I = 8.3 \cdot 10^2\) watts per square meter for a jet plane is extremely loud, explaining the need to scale it logarithmically.
This measurement in decibels allows us to express sound intensity in a more understandable form. Particularly, a jet plane has a sound intensity that produces a decibel level of approximately 149.19. Knowing this helps in numerous fields such as acoustics, engineering, and even regulatory settings for noise pollution.

Graphing Solutions for Sound Intensity

Graphing the solution is a helpful way to visualize mathematical concepts and enhance understanding. In this exercise, once we have solved the equation, we graph the decibel level to comprehend it visually.
To plot the graph, we can set the x-axis for different sounds and the y-axis for decibel levels. This means that different points on the x-axis will represent various sound sources, from quiet to loud, like whispers or rock concerts. The calculated decibel value of a jet engine, 149.19, is placed as a point on this graph.
By doing so, one can easily compare this level with other common sounds.

• A whisper would be plotted much lower on the y-axis, reflecting its lower intensity.
• Thunder might be in between, but not as high as a jet engine.

Graphing thus helps turn abstract numbers into a picture that talks. It allows students to see where different sounds fall in relation to one another, highlighting the vast range of human hearing and the usefulness of decibels in everyday communication.