Problem 60 - Precalculus

Question

A whisper corresponds to \(1 \times 10^{-10} \mathrm{W} / \mathrm{m}^{2}\) (or \(20 \mathrm{dB}\) ) and a normal conversation is typically around \(1 \times 10^{-6} \mathrm{W} / \mathrm{m}^{2}\) (or \(60 \mathrm{dB}\) ). How many decibels are you exposed to if one friend is whispering in your ear, while the other one is talking at a normal level? (Hint: Add the sound pressures and convert to dB.)

Step-by-Step Solution

Answer

You're exposed to about 60 dB, same as the conversation alone.

Step 1: Understand Sound Intensity Levels

The sound intensity level in decibels (dB) is calculated using the formula \( L = 10 \log_{10} \left( \frac{I}{I_0} \right) \), where \( I \) is the intensity of the sound and \( I_0 \) is the reference intensity (typically \( 1 \times 10^{-12} \mathrm{W/m}^2 \) for air). In this problem, we are given two intensity levels: a whisper at \( 1 \times 10^{-10} \mathrm{W/m}^2 \) and normal conversation at \( 1 \times 10^{-6} \mathrm{W/m}^2 \).

Step 2: Convert Intensities to a Common Unit

We have intensities of \( 1 \times 10^{-10} \mathrm{W/m}^2 \) for the whisper and \( 1 \times 10^{-6} \mathrm{W/m}^2 \) for the normal conversation. We will use the sound intensity formula later, but first, we need to add these intensities together.

Step 3: Add Sound Intensities

To find the total sound intensity when both a whisper and a conversation occur simultaneously, add the two intensities: \( I_{total} = I_{whisper} + I_{conversation} = 1 \times 10^{-10} + 1 \times 10^{-6} = 1.0001 \times 10^{-6} \mathrm{W/m}^2 \).

Step 4: Calculate Total Decibels

Now, use the formula for sound intensity level: \[ L_{total} = 10 \log_{10} \left( \frac{I_{total}}{I_0} \right) = 10 \log_{10} \left( \frac{1.0001 \times 10^{-6}}{1 \times 10^{-12}} \right) \].\Now calculate: \( L_{total} = 10 \log_{10} (1.0001 \times 10^{6}) \approx 10 \times 6.00004 \approx 60.0004 \mathrm{dB} \).

Step 5: Interpret the Result

Adding a whisper to a normal conversation does not significantly increase the dB level noticeably because the conversation's intensity is much higher than that of a whisper. The total is approximately equal to the conversation's dB level due to the difference in intensities.

Key Concepts

Decibels (dB)

Decibels (dB) are the units used to measure sound intensity levels. They provide a way of expressing sound pressure levels in a logarithmic scale, which helps in dealing with a vast range of intensities in a manageable format. Decibels are not like linear units such as meters or grams. Instead, they represent a ratio of a particular value to a reference value.

The benefit of using decibels lies in their ability to simplify comparisons of sound levels. Since human ears perceive sound logarithmically rather than linearly, decibels make it easier to relate the physical intensity of sound to the perceived loudness. For example, a sound that is 10 times more intense than another sound is perceived as twice as loud to the human ear. This non-linear relationship is effectively managed by the decibel scale.

• One Key Point: An increase of 10 dB represents a tenfold increase in intensity, but only about twice the perceived loudness.
• Practical Example: A normal conversation at 60 dB is 1000 times (30 dB higher) more intense than a whisper at 30 dB.

Sound Intensity Level Formula

The Sound Intensity Level Formula is an essential tool for understanding and calculating sound levels. This formula is expressed as: \[ L = 10 \log_{10} \left( \frac{I}{I_0} \right) \]where \( L \) is the sound intensity level in decibels, \( I \) is the intensity of the sound in watts per square meter (\( W/m^2 \)), and \( I_0 \) is a reference intensity, usually taken as \( 1 \times 10^{-12} W/m^2 \) which is the threshold of hearing.

The function of this formula is to allow conversion of a wide range of sound intensities into a more comprehensible scale. By using the logarithm base 10, it compresses the scale, making it simpler to compare and manage.

• This formula indicates that an increase in intensity by a factor of 10 will add 10 dB to the intensity level.
• In practical use, applying this formula helps us understand scenarios such as the combined intensity levels when two sound sources are added together, as seen in the example problem involving a whisper and a normal conversation.

Logarithms in Sound Measurement

Logarithms are integral to sound measurement as they enable a practical approach to expressing large variances in sound intensities. In the context of sound, a logarithmic scale manages these vast differences by compressing them into a more manageable form.

When calculating sound levels in decibels, the use of logarithms allows one to express exponential changes in sound intensity in a linear way. This simplifies comparing sound levels. The logarithmic nature means smaller changes in decibel levels can represent large differences in actual sound power.

• Understanding Logarithms: In mathematics, a logarithm answers the question of how many times one number (the base) must be multiplied by itself to achieve another number (the value). For example, \( \log_{10}(100) = 2 \) because \( 10^2 = 100 \).
• Application in Sound: In sound measurement, \( 10^{1} \) is ten times the reference intensity, \( 10^{2} \) is 100 times, and so on, which translates to an increase in 10 dB for every power of 10 in sound intensity.
• Practical Example: The step in converting a whisper and conversation intensities to dB levels employs the use of logarithms to consolidate these two wildly different intensity levels into a single decibel measure that can be easily compared