Problem 43
Question
Eavesdropping! You are trying to overhear a juicy conversation, but from your distance of \(15.0 \mathrm{m},\) it sounds like only an average whisper of 20.0 \(\mathrm{dB}\) . So you decide to move closer to give the conversation a sound level of 60.0 \(\mathrm{dB}\) instead. How close should you come?
Step-by-Step Solution
Verified Answer
You need to move to 0.15 meters away to hear at 60 dB.
1Step 1: Understand the concept of sound intensity level
The sound intensity level in decibels (dB) is given by the formula \[ L = 10 \log_{10}\left(\frac{I}{I_0}\right) \]where \( I \) is the sound intensity, and \( I_0 \) is the reference intensity (usually \( 10^{-12} \mathrm{W/m^2} \)). For every decrease of 10 dB, the sound intensity decreases by a factor of 10.
2Step 2: Establish the relationship between distance and sound intensity
Sound intensity \( I \) is inversely proportional to the square of the distance \( r \) from the source: \[ I \propto \frac{1}{r^2} \]Going from 20 dB to 60 dB implies that the intensity increases by a factor of \(10^{(60-20)/10} = 10^4 = 10000\).
3Step 3: Apply inverse square law to find new distance
According to the inverse square law, if intensity increases by a factor of 10000, the distance must decrease by a factor of the square root of 10000. \[ \text{New distance } = \frac{r}{\sqrt{10000}} = \frac{15 \text{ m}}{100} \]
4Step 4: Calculate the exact new distance
Using the inverse square law calculation from the previous step:\[ \text{New distance } = \frac{15 \text{ m}}{100} = 0.15 \text{ m} \]Thus, to hear the conversation at 60 dB, you must be \(0.15\) meters away.
Key Concepts
Inverse Square LawDecibel ScaleSound Intensity Formula
Inverse Square Law
The Inverse Square Law is a fundamental principle that applies to various physical phenomena, including sound intensity. This law states that the intensity of a sound (or any point source energy) decreases with the square of the distance from the source. In simpler terms, as you move further from the sound source, the sound becomes less intense quite rapidly.
Mathematically, we express this as:
When observing a change in sound intensity and distance, the Inverse Square Law helps predict how far one must move to achieve a desired sound level. For instance, moving closer to a sound source increases the intensity felt by the listener.
Mathematically, we express this as:
- \( I \propto \frac{1}{r^2} \)
When observing a change in sound intensity and distance, the Inverse Square Law helps predict how far one must move to achieve a desired sound level. For instance, moving closer to a sound source increases the intensity felt by the listener.
Decibel Scale
Sound intensity levels are often measured in decibels (dB), a unit that uses a logarithmic scale to express ratios of power.
The decibel scale simplifies large ranges of sound intensities into manageable numbers by using a reference intensity level. Normally, this reference level is \( 10^{-12} \mathrm{W/m^2} \), which is considered the threshold of hearing for the average human ear.
The formula used to calculate sound intensity in decibels is:
The decibel scale simplifies large ranges of sound intensities into manageable numbers by using a reference intensity level. Normally, this reference level is \( 10^{-12} \mathrm{W/m^2} \), which is considered the threshold of hearing for the average human ear.
The formula used to calculate sound intensity in decibels is:
- \[ L = 10 \log_{10}\left(\frac{I}{I_0}\right) \]
Sound Intensity Formula
Understanding the sound intensity formula is crucial for calculating the change in distance needed to reach a different decibel level.
The given formula, \[ I \propto \frac{1}{r^2} \], highlights how sound intensity decreases as distance from the source increases, which we further explore in the context of the decibel scale using:
The given formula, \[ I \propto \frac{1}{r^2} \], highlights how sound intensity decreases as distance from the source increases, which we further explore in the context of the decibel scale using:
- \[ L = 10 \log_{10}\left(\frac{I}{I_0}\right) \]
- A 10,000 times increase in intensity equates to moving 100 times closer.
- Thus, being 0.15 meters from the source will result in an increase to 60 dB.
Other exercises in this chapter
Problem 41
\(\bullet\) Find the noise level (in dB) of a sound having an intensity of (a) 0.000127 \(\mathrm{W} / \mathrm{m}^{2}\)(b) \(6.53 \times 10^{-10} \mathrm{W} / \
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\(\bullet\) The intensity due to a number of independent sound sources is the sum of the individual intensities. (a) When four quadruplets cry simultaneously, h
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\(\bullet(\) a) What is the sound intensity level in a car when the sound intensity is 0.500\(\mu \mathrm{W} / \mathrm{m}^{2} ?\) (b) What is the sound intensit
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