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Chapter 12: Problem 43

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?

Short Answer

Expert verified
You need to move to 0.15 meters away to hear at 60 dB.

Step by step solution

01

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.
02

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 \(10^{(60-20)/10} = 10^4 = 10000\).
03

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. \[ \text{New distance } = \frac{r}{\sqrt{10000}} = \frac{15 \text{ m}}{100} \]
04

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.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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:
  • \( I \propto \frac{1}{r^2} \)
where \( I \) is the intensity and \( r \) is the distance from the source. This relationship means that if you double the distance from the sound source, the intensity becomes a quarter of what it was at the original distance.

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:
  • \[ L = 10 \log_{10}\left(\frac{I}{I_0}\right) \]
In this equation, \( L \) represents the sound level in decibels, \( I \) is the measured intensity, and \( I_0 \) is the reference intensity. This logarithmic relationship means each 10 dB increase represents a tenfold increase in the intensity. Hence, a jump from 20 dB to 60 dB increases intensity by a factor of 10,000.
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:
  • \[ L = 10 \log_{10}\left(\frac{I}{I_0}\right) \]
Combining these concepts, we leverage both the inverse square law and the decibel formula to understand how to adjust one鈥檚 position relative to a sound source for better hearing. If the goal is to increase the decibel level from a whisper (20 dB) to a much louder level (60 dB), it's necessary to move significantly closer to the source. By calculating according to the inverse square law, this change in position involves reducing the initial distance by a factor resulting from evaluating the intensity ratio:
  • 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.

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Most popular questions from this chapter

\(\bullet\) A person is playing a small flute 10.75 \(\mathrm{cm}\) long, open at one end and closed at the other, near a taut string having a fundamental frequency of 600.0 \(\mathrm{Hz}\) . If the speed of sound is \(344.0 \mathrm{m} / \mathrm{s},\) for which harmonics of the flute will the string res- onate? In each case, which harmonic of the string is in resonance?

\(\cdot\) Two guitarists attempt to play the same note of wavelength 6.50 \(\mathrm{cm}\) at the same time, but one of the instruments is slightly out of tune and plays a note of wavelength 6.52 \(\mathrm{cm}\) instead. What is the frequency of the beat these musicians hear when they play together?

\(\bullet\) The equation describing a transverse wave on a string is $$y(x, t)=(1.50 \mathrm{mm}) \sin \left[\left(157 \mathrm{s}^{-1}\right) t-\left(41.9 \mathrm{m}^{-1}\right) x\right]$$ Find (a) the wavelength, frequency, and amplitude of this wave, (b) the speed and direction of motion of the wave, and (c) the transverse displacement of a point on the string when \(t=0.100\) s and at a position \(x=0.135 \mathrm{m} .\)

\(\cdot\) On the planet Arrakis, a male ornithoid is flying toward his stationary mate at 25.0 \(\mathrm{m} / \mathrm{s}\) while singing at a frequency of 1200 \(\mathrm{Hz} .\) If the female hears a tone of 1240 \(\mathrm{Hz}\) , what is the speed of sound in the atmosphere of Arrakis?

\(\cdot\) The role of the mouth in sound. The production of sound during speech or singing is a complicated process. Let's concentrate on the mouth. A typical depth for the human mouth is about \(8.0 \mathrm{cm},\) although this number can vary. (Check it against your own mouth.) We can model the mouth as an organ pipe that is open at the back of the throat. What are the wavelengths and frequencies of the first four harmonics you can produce if your mouth is (a) open, (b) closed? Use \(v=354 \mathrm{m} / \mathrm{s}\) .
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