Chapter 16: Problem 15
You are trying to overhear a juicy conversation, but from your distance of 15.0 m, it sounds like only an average whisper of 20.0 dB. How close should you move to the chatterboxes for the sound level to be 60.0 dB?
Short Answer
Expert verified
Move to 0.15 meters away for the sound level to be 60 dB.
Step by step solution
01
Convert Decibels to Intensity Ratio
The sound level in decibels (dB) is a logarithmic measure of the ratio of the intensity of a sound wave to a reference intensity. We can use the formula \( \beta = 10 \cdot \log_{10} \left( \frac{I}{I_0} \right) \), where \( \beta \) is the sound level in decibels, \( I \) is the intensity of the sound, and \( I_0 \) is the reference intensity (\( 10^{-12} \text{ W/m}^2 \)). We first calculate the intensity ratio for both 20.0 dB and 60.0 dB.
02
Calculate Intensities at Both Distances
For 20 dB: \( 20 = 10 \cdot \log_{10} \left( \frac{I_1}{I_0} \right) \). Solving gives \( \frac{I_1}{I_0} = 10^2 = 100 \); hence, \( I_1 = 100I_0 \).For 60 dB: \( 60 = 10 \cdot \log_{10} \left( \frac{I_2}{I_0} \right) \). Solving gives \( \frac{I_2}{I_0} = 10^6 = 1,000,000 \); hence, \( I_2 = 1,000,000I_0 \).
03
Use Inverse Square Law
Sound intensity is inversely proportional to the square of the distance from the source: \( I \propto \frac{1}{r^2} \). Given \( I_1 = 100I_0 \) at 15.0 m, and we want \( I_2 = 1,000,000I_0 \), we use the relationship \( \frac{I_1}{I_2} = \left( \frac{r_2}{r_1} \right)^2 \).
04
Solve for the New Distance
Plug the values into the inverse square relationship: \( \frac{100I_0}{1,000,000I_0} = \left( \frac{r_2}{15} \right)^2 \). This simplifies to \( \frac{1}{10,000} = \left( \frac{r_2}{15} \right)^2 \). Solving for \( r_2 \), we get \( r_2 = 15 \times \frac{1}{100} = 0.15 \text{ meters} \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Decibel Scale
Sound levels are measured using the decibel (dB) scale, which is a logarithmic way to describe a ratio. The decibel scale helps us understand how loud a sound is compared to a reference sound. The reference intensity, usually denoted as \( I_0 \), is set at the threshold of hearing, which is \( 10^{-12} \text{ W/m}^2 \). This threshold represents the quietest sound that a healthy human ear can detect.
To calculate the intensity level of a sound in decibels, we use the formula:
To calculate the intensity level of a sound in decibels, we use the formula:
- \( \beta = 10 \cdot \log_{10} \left( \frac{I}{I_0} \right) \)
Inverse Square Law
The inverse square law is a fundamental principle that describes how sound intensity decreases with distance. According to this law, the intensity of sound is inversely proportional to the square of the distance from the sound source. Mathematically, we express this relationship as:
- \( I \propto \frac{1}{r^2} \)
- \( \frac{I_1}{I_2} = \left( \frac{r_2}{r_1} \right)^2 \)
Sound Intensity
Sound intensity is a measure of the power carried by sound waves per unit area, usually expressed in watts per square meter (\( \text{W/m}^2 \)). It indicates how much sound energy passes through a particular area and is a key factor in determining how loud a sound is perceived by a listener.
In our example, we are given two sound levels: 20 dB and 60 dB. To find the sound intensity at these levels, we first use the decibel formula:
In our example, we are given two sound levels: 20 dB and 60 dB. To find the sound intensity at these levels, we first use the decibel formula:
- At 20 dB: \( I_1 = 100 I_0 \)
- At 60 dB: \( I_2 = 1,000,000 I_0 \)