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(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
To increase the sound intensity level by 13.0 dB, the intensity must be increased by a factor of about 19.95. The original intensity isn't needed as dB is a relative measure, focusing on change rather than absolute values.

Step by step solution

01

Understanding the Concept of Decibel

The decibel (dB) is a logarithmic unit that measures the intensity of sound. The formula for sound intensity in decibels is given by \( L = 10 \, \log_{10}\left(\frac{I}{I_0}\right) \), where \( L \) is the sound level in decibels, \( I \) is the intensity of the sound, and \( I_0 \) is the reference intensity, typically \( 10^{-12} \, \text{W/m}^2 \).
02

Determining Intensity Increase for a Given dB Increment

Given that sound intensity level increases by 13.0 dB, the change in the sound level can be expressed as \( \Delta L = 10 \, \log_{10}\left(\frac{I_2}{I_1}\right) = 13.0 \, \text{dB} \).
03

Solving for the Intensity Ratio

Rearrange the formula from Step 2 to solve for the ratio of intensities: \(\frac{I_2}{I_1} = 10^{\frac{13.0}{10}} \,.\) Calculate this value: \( 10^{1.3} \).
04

Calculating the Factor

Compute \( 10^{1.3} \) to find the factor by which the intensity must be increased: approximately \( 19.95 \).
05

Explaining Why Original Intensity Isn't Needed

The change in sound intensity level (in dB) depends only on the ratio \( \frac{I_2}{I_1} \), not the absolute values of \( I_1 \) or \( I_2 \). This is because decibels are a relative measure.

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Key Concepts

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

Decibel
The decibel (dB) is a unit used to express sound intensity levels. It serves as a measure to quantify the power or intensity of sound by comparing it to a reference level, which is carefully chosen to be the threshold of human hearing, typically set at \( 10^{-12} \, \text{W/m}^2 \). In mathematical terms, the sound intensity level in decibels is calculated using the formula:
  • \( L = 10 \log_{10}\left(\frac{I}{I_0}\right) \)
Here, \( L \) represents the sound level in decibels, \( I \) indicates the actual sound intensity, and \( I_0 \) is the reference intensity. This formula emphasizes that decibels measure the intensity in a logarithmic scale. By presenting sound in decibels, we are able to deal with the vast range of sound intensities we encounter in a manageable way.
The decibel scale is widely used not only for sounds in acoustics but also in electronics and communications due to its flexibility and functionality, making it a comprehensive tool for sound level measurement. By using decibels, it becomes easier to compare sound levels and understand their practical implications in real-world scenarios.
Logarithmic Scale
A logarithmic scale is a nonlinear scale that is essential for measuring quantities that span several orders of magnitude, such as sound intensity. Instead of increasing linearly, a logarithmic scale grows exponentially, which better suits our perception of sound. This means that each step on the scale represents a tenfold increase or decrease in intensity.
The use of logarithmic scales is rooted in how humans perceive changes in loudness. Our ears naturally respond to sound in a logarithmic manner, meaning that we perceive equal ratios of intensity as equal differences in loudness. Consequently, representing sound intensity using a logarithmic scale such as decibels helps align measurements with human auditory perception.
  • For instance, a 10 dB increase in sound level is perceived as roughly "twice as loud."
This unique feature makes the logarithmic scale ideal for sound level measurement, allowing for an intuitive understanding of variations in sound intensity, from the faintest whisper to the roar of a jet engine.
Intensity Ratio
The intensity ratio, in the context of sound levels, is a crucial concept that provides insights into how much sound intensity changes. It is essentially the ratio of two sound intensities \( I_2 \) and \( I_1 \), where \( I_2 \) is the final intensity after an increase, and \( I_1 \) is the initial intensity before the increase.
  • The formula representing the change in sound intensity level in decibels incorporates this ratio:\( \Delta L = 10 \log_{10}\left(\frac{I_2}{I_1}\right)\).
  • By rearranging this formula, the intensity ratio can be calculated as: \( \frac{I_2}{I_1} = 10^{\frac{\Delta L}{10}} \).
For example, if the sound intensity level increases by 13 dB, this indicates that the intensity ratio, or how much the intensity has increased, is \( \frac{I_2}{I_1} = 10^{1.3} \), which simplifies to approximately 19.95. This states that the intensity of the sound must be increased by nearly 20 times to achieve a 13 dB increase.
Understanding intensity ratios not only helps in academic exercises like transforming dB into factor increases, but it is also pivotal in industries where sound control and management are required, ensuring environments meet safety and comfort levels regarding noise.
Sound Level Measurement
Sound level measurement is the process of quantifying how intense a sound is. This measurement is typically done using decibels, which help us navigate the wide range of sound intensities humans can hear. By using sound level meters, we can measure and display the loudness of sounds in a meaningful way.
  • Sound level meters are devices equipped with microphones that capture sound and convert it into electrical signals.
  • These signals are then processed to provide a reading, often displayed in decibels, which reflects the sound intensity perceived.
One key aspect of sound level measurement is that it focuses on changes rather than absolute values. This means that when we measure sound levels, we are often interested in comparing intensities. The reason we can do this without knowing the original intensity is because the change in intensity is determined by the ratio of the new intensity to the old one (\( \frac{I_2}{I_1}\)).
Hence, the specific calculation does not require knowing the actual starting intensity, only how much it has changed in relation to its original value. This makes decibels a highly practical unit for measuring changes in sound level, as seen in many fields, including environmental monitoring, audio engineering, and noise abatement solutions.

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Most popular questions from this chapter

A swimming duck paddles the water with its feet once cvery 1.6 \(\mathrm{s}\) , producing surface waves with this period. The duck is moving at constant speed in a pond where the speed of surface waves is 0.32 \(\mathrm{m} / \mathrm{s}\) , and the crests of the waves ahead of the duck are spaced 0.12 \(\mathrm{m}\) apart. (a) What is the duck's speed? (b) How far apart are the crests behind the duck?

You blow across the open mouth of an empty test tube and produce the fundamental standing wave of the air column inside the test tube. The speed of sound in air is 344 \(\mathrm{m} / \mathrm{s}\) and the test tube acts as a stopped pipe. (a) If the length of the air column in the test tube is 14.0 \(\mathrm{cm}\) , what is the frequency of this standing wave? (b) What is the frequency of the nfundamental standing wave in the air column if the test tube is half filled with water?

The fundamental frequency of a pipe that is open at both ends is 594 Hz (a) How long is this pipe? If one end is now closed, find (b) the wavelength and (c) the frequency of the new fundamental.

Example 16.1 (Section 16.1\()\) showed that for sound waves in air with frequency 1000 \(\mathrm{Hz}\) , a displacement amplitude of\( 1.2 \times 10^{-8} \mathrm{m}\) produces a pressure amplitude of \(3.0 \times 10^{-2} \mathrm{Pa}\) . Water at \(20^{\circ} \mathrm{C}\) has a bulk modulus of \(2.2 \times 10^{9} \mathrm{Pa}\) , and the speed of sound in water at this temperature is 1480 \(\mathrm{m} / \mathrm{s}\) . For \(1000-\mathrm{Hz}\) sound waves in \(20^{\circ} \mathrm{C}\) water, what displacement amplitude is produced if the pressure amplitude is \(3.0 \times 10^{-2}\) Pa? Explain why your answer is much less than \(1.2 \times 10^{-8} \mathrm{Pa}\) .

Singing in the Showen. A pipe closed at both ends can have standing waves inside of it, but you normally don't hear them because little of the sound can get out. But you can hear them if you are inside the pipe, such as someone singing in the shower. (a) Show that the wavelengths of standing waves in a pipe of length \(L\) that is closed at both ends are \(\lambda_{n}=2 L / n\) and the frequencies are given by \(f_{n}=n v / 2 L=n f_{1},\) where \(n=1,2,3, \ldots\) (b) Modeling it as a pipe, find the frequency of the fundamental and the first two overtones for a shower 2.50 \(\mathrm{m}\) tall. Are these frequencies audible?

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