Problem 83 - College Algebra

Question

The loudness level of a sound can be expressed by comparing the sound's intensity to the intensity of a sound barely audible to the human car. The formula $$ D=10\left(\log I-\log I_{0}\right) $$ describes the loudness level of a sound, \(D\), in decibels, where \(I\) is the intensity of the sound, in watts per meter". and \(I_{0}\) is the intensity of a sound barely audible to the human ear. a. Express the formula so that the expression in parentheses is written as a single logarithm. b. Use the form of the formula from part (a) to answer this question: If a sound has an intensity 100 times the intensity of a softer sound, how much larger on the decibel scale is the loudness level of the more intense sound?

Step-by-Step Solution

Answer

20 decibels

Step 1: Combine Logarithms

In order to express the formula so that the expression in parentheses is written as a single logarithm, we need to use the properties of logarithms. According to a property of logarithms, the difference between two logs with the same base equals to the log of the division between their arguments. It will transform the equation into \(D = 10 \log \left( \frac{I}{I_{0}} \right)\).

Step 2: Re-check the Form of the Formula

After applying the property of logarithms, the equation is simplified into a single logarithm. Therefore, the formula for loudness level is: \(D=10 \log \left( \frac{I}{I_{0}} \right)\). This equation tells us that the loudness level (D) in decibels is equal to 10 times the log of sound intensity (I) divided by the reference sound intensity (\(I_{0}\)).

Step 3: Apply the New Formula

Now let's use this formula to answer the second question. If a sound has an intensity (\(I\')) 100 times the intensity of a softer sound (\(I\)), then the increase in decibel level of the more intense sound can be found by \(\Delta D = D' - D = 10 \log \left( \frac{I'}{I_{0}} \right) - 10 \log \left( \frac{I}{I_{0}} \right) = 10 \log \left( \frac{I'/I}{I_0/I_0} \right)\) taking \(I' = 100I\), we get \(\Delta D = 10 \log(100) = 20 decibels\).

Step 4: Answer the Question

Therefore, if a sound has an intensity 100 times the intensity of a softer sound, the loudness level of the more intense sound is 20 decibels larger on the decibel scale.

Key Concepts

Sound Intensity

Sound intensity refers to the power carried by sound waves per unit area. It's a measure of the energy that passes through a surface as sound travels. Imagine someone talking loudly and softly. The loud talk generates more energy, hence more intensity, than the soft talk. - **Measured in**: watts per meter squared - **Varies with**: Distance from the sound source The greater the intensity, the louder the sound will seem. Sound intensity relates directly to how we perceive sound, although human perception is more complex due to factors like frequency and duration.

Loudness Level

Loudness level is how we perceive the sound's intensity. We don't perceive sound intensity linearly; instead, it follows a logarithmic scale. This means a sound 10 times as intense seems only twice as loud to us. - **Expressed in**: Decibels (dB) - **Human Perception**: Influenced by both intensity and frequency Loudness is subjective, but decibels provide a standardized way to measure it. This helps in comparing how different sounds feel to our ears.

Decibel Scale

The decibel scale is a logarithmic scale used to measure sound levels. Instead of directly measuring intensity, it operates on ratios and comparisons. The reference point is usually the smallest sound a typical human can hear. - **Base**: 10 (logarithmic base used) - **0 dB**: Represents the threshold of hearing A sound that is 10 times more intense than another is 10 dB higher on the decibel scale. This scale compresses large ranges of sound intensities into a manageable scale, allowing us to handle both very quiet and very loud sounds comfortably.

Logarithmic Properties

Logarithmic properties are key to understanding the relationships in sound measurements. Logarithms allow us to transform multiplication into addition, which simplifies many calculations.- **Key Property**: \[\log{a} - \log{b} = \log{\left(\frac{a}{b}\right)}\]- **Implication**: This means that comparing sound intensities can be simplified. Logarithms make it easier to compare huge differences in sound intensity by converting them into smaller, more manageable numbers. This is why logs are used extensively in acoustics and other scientific fields where large ranges are the norm.