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Chapter 16: Problem 58

A middle-aged man typically has poorer hearing than a middle-aged woman. In one case a woman can just begin to hear a musical tone, while a man can just begin to hear the tone only when its intensity level is increased by \(7.8 \mathrm{~dB}\) relative to that for the woman. What is the ratio of the sound intensity just detected by the man to that just detected by the woman?

Short Answer

Expert verified
The intensity ratio is approximately 6.03.

Step by step solution

01

Understand the Decibel Scale

The decibel (dB) scale is a logarithmic scale that describes the ratio of two sound intensity levels. The formula for this ratio in decibels is \( L = 10 \log_{10} \left( \frac{I}{I_0} \right) \), where \( L \) is the level in decibels, \( I \) is the sound intensity, and \( I_0 \) is the reference intensity. In our problem, we're dealing with a difference of 7.8 dB.
02

Set Up the Expression for Intensity Ratio

We know that the difference in decibel level between the man's and woman's hearing is 7.8 dB. Using the formula, this means: \( 7.8 = 10 \log_{10} \left( \frac{I_m}{I_w} \right) \), where \( I_m \) is the intensity just detected by the man and \( I_w \) is the intensity just detected by the woman.
03

Solve for the Intensity Ratio

Rearrange the equation from Step 2 to solve for the ratio of intensities. First, divide both sides by 10: \( 0.78 = \log_{10} \left( \frac{I_m}{I_w} \right) \). Then, exponentiate both sides to remove the logarithm: \( 10^{0.78} = \frac{I_m}{I_w} \).
04

Calculate the Intensity Ratio

Calculate \( 10^{0.78} \). This value gives us the ratio of the intensities: \( 10^{0.78} \approx 6.03 \).
05

Conclusion

The sound intensity just detected by the man is approximately 6.03 times greater than that detected by the woman.

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Key Concepts

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

Decibel Scale
The decibel scale is a way of expressing the ratio of sound intensities in a convenient manner. Because sound intensity can vary over incredibly wide ranges, a linear scale would not be practical. The decibel (dB) scale solves this by using a logarithmic approach, allowing us to measure sound in a manageable range of numbers. The decibel formula helps us relate intensity levels: \[ L = 10 \log_{10} \left( \frac{I}{I_0} \right) \] where:
  • \( L \) is the decibel level.
  • \( I \) is the actual intensity.
  • \( I_0 \) is the reference intensity (usually the threshold of hearing in air, set at \(10^{-12} \text{ W/m}^2\)).
By converting intensity ratios into decibels, we can easily compare different sound intensities in a way that reflects how humans perceive them.
Intensity Ratio
An important concept when dealing with sound intensities is the intensity ratio. This refers to comparing the intensity that one individual, like our middle-aged man, needs to hear a sound to the intensity needed by another, such as our woman. For example:
  • When the intensity at which the man hears a sound is compared to the woman's, it is expressed as a ratio \( \frac{I_m}{I_w} \).
  • The logarithmic nature of the decibel scale simplifies solving for these intensity ratios, once you have the decibel difference.
In our example problem, this ratio is found to be \(10^{0.78}\), which tells us how much more intense the sound needs to be for the man to perceive it the same way.
Hearing Differences
Hearing ability can differ significantly between individuals, and factors such as age or even gender may play a role. Biologically, women often have slightly better hearing sensitivity compared to men, especially as they age. This means that the sound intensity that a woman can detect might need to be louder for a man to hear equally well. When calculating the intensity levels different people need to hear the same sound, such difference is often measured in decibels. A 7.8-dB increase for the man, in the given example, indicates he needs a sound over six times stronger than what the woman requires to detect it. These differences are not uncommon and highlight why understanding sound intensity levels is crucial for designing auditory devices, like hearing aids.
Logarithmic Scale
A logarithmic scale is essential in handling values that span many orders of magnitude, such as sound intensities. Regular linear scales would be unwieldy and not convenient for practical use. The decibel scale is a classical use of a logarithmic scale, transforming multiplicative relationships into additive ones. For example, a tenfold increase in sound intensity changes the decibel count by a fixed amount (10 dB).
  • This makes understanding and comparing sound levels much more intuitive.
  • It reflects human perception more closely since humans perceive sound intensity logarithmically.
In real-world applications, employing a logarithmic scale helps us efficiently manage and interpret large variations in data.

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Most popular questions from this chapter

Refer to at for one approach to this type of problem. Two trucks travel at the same speed. They are far apart on adjacent lanes and approach each other essentially head-on. One driver hears the horn of the other truck at a frequency that is \(1.14\) times the frequency he hears when the trucks are stationary. The speed of sound is \(343 \mathrm{~m} / \mathrm{s}\). At what speed is each truck moving?

A transverse wave is traveling on a string. The displacement \(y\) of a particle from its equilibrium position is given by \(y=(0.021 \mathrm{~m}) \sin (25 t-2.0 x) .\) Note that the phase angle \(25 t-2.0 x\) is in radians, \(t\) is in seconds, and \(x\) is in meters. The linear density of the string is \(1.6 \times 10^{-2} \mathrm{~kg} / \mathrm{m} .\) What is the tension in the string?

Consult Multiple-Concept Example 11 in order to review a model for solving this type of problem. A car is accelerating while its horn is blowing. Just after the car passes a stationary person, the person hears a frequency of \(966.0 \mathrm{~Hz}\). Fourteen seconds later, the frequency heard by the person has decreased to \(912.0 \mathrm{~Hz}\). When the car is stationary, its horn emits a sound whose frequency is \(1.00 \times 10^{3} \mathrm{~Hz}\). The speed of sound is \(343 \mathrm{~m} / \mathrm{s}\). What is the acceleration of the car?

Review Conceptual Example 3 before starting this problem. A horizontal wire is under a tension of \(315 \mathrm{~N}\) and has a mass per unit length of \(6.50 \times 10^{-3} \mathrm{~kg} / \mathrm{m} .\) A transverse wave with an amplitude of \(2.50 \mathrm{~mm}\) and a frequency of \(585 \mathrm{~Hz}\) is traveling on this wire. As the wave passes, a particle of the wire moves up and down in simple harmonic motion. Obtain (a) the speed of the wave and (b) the maximum speed with which the particle moves up and down.

Two wires are parallel, and one is directly above the other. Each has a length of \(50.0 \mathrm{~m}\) and a mass per unit length of \(0.020 \mathrm{~kg} / \mathrm{m}\). However, the tension in wire \(\mathrm{A}\) is \(6.00 \times 10^{2} \mathrm{~N}\), and the tension in wire \(\mathrm{B}\) is \(3.00 \times 10^{2} \mathrm{~N}\). Transverse wave pulses are generated simultaneously, one at the left end of wire \(A\) and one at the right end of wire B. The pulses travel toward each other. How much time does it take until the pulses pass each other?
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