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Chapter 23: Problem 34

A sound has an intensity of \(5.0 \times 10^{-7} \mathrm{~W} / \mathrm{m}^{2}\). What is its intensity level?

Short Answer

Expert verified
The intensity level is approximately 57 dB.

Step by step solution

01

Understanding Intensity Level Formula

The intensity level, also known as the sound level, in decibels (dB) is calculated using the formula \( L = 10 \, \log_{10} \left(\frac{I}{I_0}\right) \), where \( I \) is the sound intensity and \( I_0 = 1.0 \times 10^{-12} \, \text{W/m}^2 \) is the reference intensity level.
02

Plug in the Given Intensity

Substitute the given intensity into the formula. Here, \( I = 5.0 \times 10^{-7} \, \text{W/m}^2 \). Therefore, the equation becomes \( L = 10 \, \log_{10} \left(\frac{5.0 \times 10^{-7}}{1.0 \times 10^{-12}}\right) \).
03

Calculate the Ratio

Calculate the ratio inside the logarithm: \( \frac{5.0 \times 10^{-7}}{1.0 \times 10^{-12}} = 5.0 \times 10^5 \). This simplifies the calculation for the logarithm.
04

Apply the Logarithm Function

Calculate the logarithm of the ratio: \( \log_{10}(5.0 \times 10^5) = \log_{10}(5.0) + \log_{10}(10^5) \). Here, \( \log_{10}(5.0) \approx 0.699 \) and \( \log_{10}(10^5) = 5 \). Adding these gives approximately \( 5.699 \).
05

Compute the Intensity Level

Multiply the result of the logarithm by 10 to find the intensity level: \( L = 10 \times 5.699 = 56.99 \). Thus, the intensity level of the sound is approximately 57 dB.

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Key Concepts

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

Sound Intensity
Sound intensity is a measure of how much energy a sound wave carries per unit area. It is usually measured in watts per square meter (W/m虏). In simpler terms, it tells us how powerful a sound is in a given space. When you listen to a loud sound, like a jet engine or a concert, the sound intensity is quite high. Conversely, a whisper has very low sound intensity.

The energy that sound waves carry isn't just sent out in one direction. Instead, sound spreads in all directions, making the intensity decrease with distance. That's why a sound seems quieter when you are further away from its source. It's important to understand that sound intensity is a physical quantity that can be measured and quantified.
Decibels
Decibels (dB) are a unit used to express the intensity level of sounds. They're a bit different from other units you might be used to, because decibels are based on a logarithmic scale. This means they measure ratios of power, not absolute quantities.

Using decibels is helpful because the human ear perceives sound intensity on a logarithmic scale. For instance, when you double the intensity of a sound, it does not sound twice as loud to us. Instead, a change of roughly 10 dB is needed for a sound to be perceived as approximately twice as loud.
  • A sound that is 20 dB higher than another is 100 times more intense.
  • An increase of 10 dB means the sound is 10 times more intense.
Understanding the use of decibels helps us represent large variations in sound intensity with more manageable numbers.
Logarithmic Scale
A logarithmic scale is a nonlinear scale often used for data that covers a large range of values. Instead of counting up linearly, each step on a logarithmic scale represents a multiplication by a certain factor.

In the context of sound intensity, the logarithmic scale helps to compress the wide range of sound intensities into a format that is easier to manage and understand. For example, increasing sound intensity from 1 W/m虏 to 10 W/m虏 doesn't translate to adding a fixed number on the logarithmic scale. Instead, it increases by a set factor (in this case, corresponding to a fixed increase in decibels).
  • This makes it easier to compare sounds of vastly different intensities.
  • It provides a clearer picture of relative loudness to the human ear.
Logarithmic scales are used in many fields, not just sound, to simplify the representation of exponential relationships.

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Most popular questions from this chapter

Two cars are heading straight at each other with the same speed. The horn of one \((f=3.0 \mathrm{kHz})\) is blowing, and is heard to have a frequency of \(3.4 \mathrm{kHz}\) by the people in the other car. Find the speed at which each car is moving if the speed of sound is \(340 \mathrm{~m} / \mathrm{s}\).

A tuning fork having a frequency of \(400 \mathrm{~Hz}\) (shown in Fig. 23-2) is moved away from an observer and toward a flat wall with a speed of \(2.0 \mathrm{~m} / \mathrm{s}\). What is the apparent frequency \((a)\) of the unreflected sound waves coming directly to the observer, and \((b)\) of the sound waves coming to the observer after reflection? ( \(c\) ) How many beats per second are heard? Assume the speed of sound in air to be \(340 \mathrm{~m} / \mathrm{s}\) (a) The fork, the source, is receding from the observer in the positive direction and so we use \(+v_{s} .\) It doesn't matter what the sign associated with \(v_{o}\) is since \(v_{e}=0\). $$ f_{o}=f_{s} \frac{v \pm v_{o}}{v \mp v_{s}}=(400 \mathrm{~Hz}) \frac{340 \mathrm{~m} / \mathrm{s}+0}{340 \mathrm{~m} / \mathrm{s}+2.0 \mathrm{~m} / \mathrm{s}}=397.7 \mathrm{~Hz}=398 \mathrm{~Hz} $$ The source is moving away from the observer and the frequency is properly shifted down from \(400 \mathrm{~Hz}\) to \(398 \mathrm{~Hz}\). (b) Think of the wall as a source that reflects sound of the same frequency as that which impinges upon it. The wave crests reaching the wall are closer together than normally because the fork is moving toward the wall. Therefore, the wall will appear as a stationary source emitting sound of a higher frequency than \(400 \mathrm{~Hz}\). due to the \(2.0-\mathrm{m} / \mathrm{s}\) motion of the fork. Alternatively we can think of the reflected wave as if it came from a source (the wall) moving at \(2.0 \mathrm{~m} / \mathrm{s}\) toward the observer. Hence, we enter \(-v_{x}:\) $$ f_{o}=f_{s} \frac{v \pm v_{o}}{v \mp v_{s}}=(400 \mathrm{~Hz}) \frac{340 \mathrm{~m} / \mathrm{s}+0}{340 \mathrm{~m} / \mathrm{s}-2.0 \mathrm{~m} / \mathrm{s}}=402.4 \mathrm{~Hz}=402 \mathrm{~Hz} $$ and the frequency is properly shifted up. (c) Beats per second \(=\) Difference between frequencies \(=(402.4-397.7) \mathrm{Hz}=4.7\) beats per second

Find the speed of sound in a diatomic ideal gas that has a density of \(3.50 \mathrm{~kg} / \mathrm{m}^{3}\) and a pressure of \(215 \mathrm{kPa}\) We know that \(v=\sqrt{\gamma R T / M}\) and can find the temperature from the pressure. Using the gas law \(P V=(m / M) R T\), $$ \frac{R T}{M}=P \frac{V}{m} $$ Moreover, \(\rho=m / V\), and so the expression for the speed becomes $$ v=\sqrt{\frac{\gamma P}{\rho}}=\sqrt{\frac{(1.40)\left(215 \times 10^{3} \mathrm{~Pa}\right)}{3.50 \mathrm{~kg} / \mathrm{m}^{3}}}=293 \mathrm{~m} / \mathrm{s} $$ We used the fact that \(\gamma \approx 1.40\) for a diatomic ideal gas, as discussed in Chapter 20 .

Determine the speed of sound in carbon diaxide \((M-44 \mathrm{~kg} / \mathrm{kmol}, \gamma=1.30)\) at a pressure of \(0.50 \mathrm{~atm}\) and a temperature of \(400^{\circ} \mathrm{C}\).

A locomotive moving at \(30.0 \mathrm{~m} / \mathrm{s}\) approaches and passes a person standing beside the track. Its whistle is emitting a note of frequency \(2.00 \mathrm{kHz}\). What frequency will the person hear \((a)\) as the train approaches and (b) as it recedes? The speed of sound is \(340 \mathrm{~m} / \mathrm{s}\).
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