Problem 34 - Calculus Early Transcendentals: Pearson New International Edition

Question

The loudness of sound is measured in decibels in honor of Alexander Graham Bell (1847-1922), inventor of the telephone. If the variation in pressure is \(P\) pounds per square inch, then the loudness \(L\) in decibels is $$ L=20 \log _{10}(121.3 P) $$ Find the variation in pressure caused by music at 115 decibels.

Step-by-Step Solution

Answer

The variation in pressure is approximately 463.68 psi.

Step 1: Setup the given equation

We are given the equation for loudness in decibels \( L = 20 \log_{10}(121.3P) \). We need to find the value of \( P \) when \( L = 115 \) decibels. Substitute \( L = 115 \) into the equation:\[ 115 = 20 \log_{10}(121.3P) \]

Step 2: Isolate the logarithmic expression

To isolate the logarithmic part of the equation, divide both sides by 20:\[ \frac{115}{20} = \log_{10}(121.3P) \] Simplify the left side:\[ \log_{10}(121.3P) = 5.75 \]

Step 3: Convert from log form to exponential form

Convert the logarithmic equation \( \log_{10}(121.3P) = 5.75 \) to its equivalent exponential form:\[ 121.3P = 10^{5.75} \]

Step 4: Calculate the value of the exponential expression

Calculate \( 10^{5.75} \) using a calculator:\[ 10^{5.75} \approx 56234.13 \]

Step 5: Solve for P

Now that we have \( 121.3P = 56234.13 \), divide both sides by 121.3 to solve for \( P \):\[ P = \frac{56234.13}{121.3} \approx 463.68 \]

Step 6: Concluding the solution

Thus, the variation in pressure \( P \), which corresponds to 115 decibels, is approximately 463.68 pounds per square inch.

Key Concepts

Logarithmic Functions

Understanding logarithmic functions is essential when dealing with sound intensity and decibels. Logarithms are the inverse operations of exponentiation, which means they help us find the exponent. In our problem, the formula for sound intensity in decibels uses a base-10 logarithmic function, represented as \( \log_{10} \). When you see \( L = 20 \log_{10}(121.3P) \), it means we are using logarithms to scale pressure variations into perceptible sound levels ordinate to human hearing.

• The logarithm, \( \log_{10}(x) \), answers the question: What power does 10 need to be raised to get \( x \)?
• This transformation is crucial for sound measurements because human ears perceive sound intensity on a logarithmic scale rather than a linear one.

In this specific exercise, we used logarithms to translate a given decibel level back into a pressure variation, making it easier to handle in real-world applications.

Pressure Variation

Pressure variation is a key component in the study of acoustics. It reflects the fluctuations in air pressure associated with sound waves. This phenomenon is what we perceive as sound and can be expressed in units like pounds per square inch (psi).In this exercise, the relationship between decibels and pressure is articulated using the formula \( L=20 \log_{10}(121.3P) \). Here, \( P \) stands for the pressure variation, which we need to find. By solving the equation for \( P \), students understand how pressure variations translate into a specific decibel level.

• To find the pressure, start by inputting the given decibel level into the formula.
• Rearrange and simplify the equation to isolate \( P \).
• Recognize the inverse relationship between logarithms and exponentials which allows us to solve for pressure variation.

Understanding pressure variation helps students grasp how different sound levels affect the environment and how we can measure them effectively.

Exponential Equations

Exponential equations are vital to reverse the logarithmic transformations done in our sound intensity formula. Once the logarithmic expression, \( \log_{10}(121.3P) = 5.75 \), is isolated, the goal is to convert it into an exponential form.Exponential equations typically involve a constant base raised to a variable exponent. For our problem, when we say \( 121.3P = 10^{5.75} \), we express \( P \) as being associated with an exponential function of base 10:

• Understand that converting logs to exponentials involves finding \( 121.3P = 10^{5.75} \).
• This conversion allows us to use a calculator to compute the value of the expression \( 10^{5.75} \).
• By solving \( P \) from this equation, students learn to translate logarithmic results back to linear measures, such as psi.

Grasping how exponential equations solve such logarithmic problems enriches learning by connecting abstract mathematical operations with tangible sound pressure measurements.