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Chapter 16: Problem 24

(a) If two sounds differ by 5.00 dB, find the ratio of the intensity of the louder sound to that of the softer one. (b) If one sound is 100 times as intense as another, by how much do they differ in sound intensity level (in decibels)? (c) If you increase the volume of your stereo so that the intensity doubles, by how much does the sound intensity level increase?

Short Answer

Expert verified
(a) Ratio is approximately 3.16. (b) 20 dB difference. (c) Increases by 3.01 dB.

Step by step solution

01

Understanding Decibels Formula

The sound intensity level in decibels (dB), \( eta \), is defined by the formula: \( eta = 10 \log_{10} \left( \frac{I}{I_0} \right) \). Here, \( I \) is the sound intensity and \( I_0 \) is the reference intensity. This formula will help us find the intensity ratio and level differences.
02

Solving Part (a) for Intensity Ratio

Given a 5 dB difference, \( \beta_1 - \beta_2 = 5 \). Using the decibel formula: \( 10 \log_{10} \left( \frac{I_1}{I_0} \right) - 10 \log_{10} \left( \frac{I_2}{I_0} \right) = 5 \), which simplifies to \( \log_{10} \left( \frac{I_1}{I_2} \right) = 0.5 \). Thus, the ratio \( \frac{I_1}{I_2} = 10^{0.5} = \sqrt{10} \approx 3.16 \).
03

Solving Part (b) for Decibel Difference

If \( I_1 = 100I_2 \), using the same formula: \( \beta_1 - \beta_2 = 10 \log_{10} \left( \frac{I_1}{I_2} \right) = 10 \log_{10} (100) = 10 \times 2 = 20 \). Therefore, the two sounds differ by 20 dB.
04

Solving Part (c) for Doubling Intensity

When the intensity doubles, \( \frac{I_1}{I_2} = 2 \). The change in decibels is \( 10 \log_{10}(2) = 10 \times 0.301 = 3.01 \). Hence, the sound intensity level increases by approximately 3.01 dB.

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Key Concepts

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

Decibels
Decibels are a unit of measurement used to express sound intensity level. The name comes from the famous scientist Alexander Graham Bell. Decibels use a logarithmic scale to manage the wide range of sound intensities that the human ear can detect. This scale simplifies the comparison of different intensities because it expresses them with logarithms, which can make vast differences appear more manageable and comprehensible.
  • A small increase in decibels indicates a relatively large increase in intensity.
  • Decibels are a relative measurement, which means they compare the intensity of sound to a reference point, usually the threshold of hearing.
  • In real-world terms, a difference of 10 dB represents a tenfold increase in intensity, while a 20 dB increase represents a hundredfold increase.
In summary, using decibels helps translate the complexities of sound intensity into an easier-to-understand format that's friendly to our senses and daily experiences.
Sound Intensity
Sound intensity is an essential part of how we measure and understand sound in physical terms. It's defined as the power per unit area carried by a sound wave.
  • Sound intensity gives us a sense of how 'strong' or 'weak' a sound is.
  • It is measured in watts per square meter (W/m虏).
When dealing with sound intensity in practical scenarios, it's crucial to consider that sound spreads out in all directions from its source. This spreading diminishes intensity because the same energy is distributed over an increasingly larger area.
  • The human ear is sensitive to an enormous range of intensities, from the faintest whisper to the roar of a jet engine, which explains why measuring sound intensity can be challenging without the decibel scale.
Understanding sound intensity helps in fields ranging from audio engineering to environmental science where controlling sound levels is critical.
Intensity Ratio
The intensity ratio is a crucial concept when comparing the loudness of two sounds. This ratio compares the intensities of two sounds to determine how much louder one is compared to the other. It's often used in conjunction with the decibel scale to quantify changes in intensity.
  • The intensity ratio between two sounds is defined mathematically as \( \frac{I_1}{I_2} \), where \( I_1 \) and \( I_2 \) are the intensities of the two sounds.
  • This ratio is crucial for computing decibel differences between sounds.
For instance, if two sounds have an intensity ratio of 10, the ears would perceive the sound with higher intensity as much louder because it carries more energy. Using the intensity ratio can effectively measure how one sound's force compares to another, providing necessary insights in acoustics, sound design, and hearing research.
Logarithms
Logarithms are mathematical tools that help solve equations involving exponential increases or decreases, such as sound intensity. They are especially useful in the context of measuring sound, as the human perception of loudness isn't linear; it follows a logarithmic relationship.
  • Logarithms convert multiplicative relationships into additive ones, which is why a logarithmic scale is used for sound intensity.
  • Using a logarithm (base 10) in the decibel formula helps us manage wide intensity ranges.
  • In equations involving sound, \( \log_{10} \left( \frac{I}{I_0} \right) \) is used to calculate the decibel level \( \beta \).
By understanding and applying logarithms, one can translate the abstract notions of intensity into understandable figures, making it easier to study and apply sound principles in technology and nature. Logarithms ensure the huge differences in sound intensity levels can be effectively understood and communicated.

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