Chapter 16: Problem 14
(a) By what factor must the sound intensity be increased to raise the sound intensity level by 13.0 dB? (b) Explain why you don't need to know the original sound intensity.
Short Answer
Expert verified
The sound intensity must be increased by a factor of 20. The original intensity is unnecessary as the increase depends only on the change in dB level.
Step by step solution
01
Understanding the Problem
We need to determine by how many times the sound intensity must be increased to increase the sound intensity level by 13.0 dB. This involves understanding the relationship between intensity levels in decibels and sound intensity.
02
The Decibel Formula
The sound intensity level in decibels (dB) is calculated using the formula: \( L = 10 \log_{10} \left( \frac{I}{I_0} \right) \), where \( L \) is the sound level in decibels, \( I \) is the sound intensity, and \( I_0 \) is the reference intensity, typically \( 10^{-12} \text{ W/m}^2 \).
03
Relating Change in Levels to Intensity
When the intensity level increases by \( \Delta L = 13.0 \text{ dB} \), it can be expressed as: \[ \Delta L = 10 \log_{10} \left( \frac{I_{new}}{I_{old}} \right) = 13.0 \].We need to solve for \( \frac{I_{new}}{I_{old}} \).
04
Isolating the Intensity Ratio
Rearrange the equation to solve for the intensity ratio:\[ \log_{10} \left( \frac{I_{new}}{I_{old}} \right) = \frac{13.0}{10} \],which simplifies to\[ \log_{10} \left( \frac{I_{new}}{I_{old}} \right) = 1.3 \].
05
Calculating the Intensity Factor
To solve for \( \frac{I_{new}}{I_{old}} \), exponentiate both sides:\[ \frac{I_{new}}{I_{old}} = 10^{1.3} \].Calculate this value,\[ \frac{I_{new}}{I_{old}} \approx 19.95 \].The sound intensity must be increased by a factor of approximately 20.
06
Explanation of Independence from Original Intensity
The formula for decibels involves a factor where sound intensity is expressed as a ratio (\( \frac{I}{I_0} \)), meaning that any proportional increase in \( I \) results in the same change in decibels regardless of the starting intensity. Hence, we do not need to know the original sound intensity for this calculation.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Decibel Formula
The decibel formula is a key concept for understanding sound intensity levels. This formula allows you to calculate the sound level in decibels (dB) using the intensity of the sound compared to a reference intensity.
The formula is expressed as:
The formula is expressed as:
- \( L = 10 \log_{10} \left( \frac{I}{I_0} \right) \)
- \( L \) is the sound level in decibels
- \( I \) is the sound intensity
- \( I_0 \) is the reference intensity, typically \( 10^{-12} \text{ W/m}^2 \)
Sound Intensity Increase
Increasing sound intensity can significantly affect the sound intensity level measured in decibels. When you aim to increase the sound intensity level, like in our exercise where a 13.0 dB increase is required, this involves a substantial change in the energy or power of the sound.
But, how do we calculate this increase? By referring to the decibel formula, we identify the intensity increase necessary to achieve a specific dB level change. For a 13.0 dB increase, as calculated in the example:
But, how do we calculate this increase? By referring to the decibel formula, we identify the intensity increase necessary to achieve a specific dB level change. For a 13.0 dB increase, as calculated in the example:
- The intensity must be increased by approximately a factor of 20.
Intensity Ratio
The intensity ratio is significant when determining how much a sound's intensity needs to change to achieve a different decibel level. In our example, we examined how the intensity ratio, \( \frac{I_{new}}{I_{old}} \), helps solve for the sound intensity increase needed.
To find this ratio:
To find this ratio:
- We start from the equation derived from the decibel difference: \( \log_{10} \left( \frac{I_{new}}{I_{old}} \right) = 1.3 \).
- Exponentiating both sides, the intensity ratio is calculated as \( 10^{1.3} \), resulting in about 19.95.
Logarithmic Scale
A logarithmic scale is used in the decibel system to quantify sound levels. This scale is not linear, meaning that equal steps on the scale represent equal ratios rather than equal increments.
Here's why it's helpful:
Here's why it's helpful:
- It allows us to manage very large ranges of values in a compact format.
- A small increase in decibels represents a large increase in sound intensity.
- Helps to better reflect how humans perceive sound, as our ears are sensitive to logarithmic changes.