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What sound intensity is \(3.0 \mathrm{~dB}\) louder than a sound of intensity of \(10 \mu \mathrm{W} / \mathrm{cm}^{2}\) ?

Short Answer

Expert verified
The new sound intensity is 20 \(\mu \mathrm{W} / \mathrm{cm}^2\).

Step by step solution

01

Understanding Sound Intensity

The intensity level in decibels (dB) is a logarithmic measure of the sound's power relative to a reference intensity. A change of 3.0 dB means the intensity level is approximately doubled.
02

Convert Decibel Change to Intensity Ratio

Any increase of 3.0 dB means that the intensity is multiplied by approximately 2. Hence, if the original intensity is \(10 \mu \mathrm{W} / \mathrm{cm}^2\), the new intensity will be \(10 \mu \mathrm{W} / \mathrm{cm}^2 \times 2\).
03

Calculate the New Intensity

Multiply the original intensity by 2 to find the new intensity:\[10 \mu \mathrm{W} / \mathrm{cm}^2 \times 2 = 20 \mu \mathrm{W} / \mathrm{cm}^2\]
04

Verification

To ensure correctness, remember that each 3.0 dB increase approximately doubles the sound intensity, which matches with the calculation of doubling \(10 \mu \mathrm{W} / \mathrm{cm}^2\) to get \(20 \mu \mathrm{W} / \mathrm{cm}^2\).

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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 measure the intensity of sound, among other things. It's a way to express the amplitude of waves on a logarithmic scale, which takes into account the wide range of human hearing. Sound intensity in decibels is calculated using the formula:
\[L = 10 \log_{10}\left(\frac{I}{I_0}\right)\]
- **\(L\)** is the sound level in decibels.- **\(I\)** is the intensity of the sound in watts per square meter (or the given units, like microwatts per square centimeter).- **\(I_0\)** is the reference intensity, typically considered to be the threshold of hearing, \(1 \times 10^{-12} \text{W/m}^2\).
This logarithmic measure is advantageous because it translates the very large range of audible sound intensities into a more manageable scale. For example, a quiet whisper might be around 20 dB, while a jet engine could be around 120 dB.
Understanding decibels helps in comparing the perceived volume of different sources of sound or changes in sound levels as measured in exercises such as the one discussed.
Logarithmic measure
A logarithmic measure is crucial in the study of sound because sound intensity varies over an incredibly wide range. Logarithms help compress this range into measurable values. When we say something is measured 'logarithmically', we're using the powers of a base number to create a scale. In the case of sound, this base number is often 10.
The reason we use a logarithmic scale for sound is that the human ear perceives sound intensity logarithmically. This means that a sound needs to be many times more powerful to be perceived as "twice as loud" by an average listener.
- If the intensity is expressed in a linear scale, even small changes in sound intensity can be hard to represent and understand, hence the need for a logarithmic approach. - A logarithmic scale allows us to discuss very large or very small quantities with ease.
Given these qualities, decibels are a prime example of a logarithmic measure, where each increase of 10 dB represents a tenfold increase in intensity.
Intensity doubling
In the realm of sound measurement, a change in intensity that results in doubling the perceived loudness often correlates with an increase of about 3 decibels (dB). This means when you increase the sound intensity by a factor of two, the decibel level rises by 3 dB.
This relationship is a key concept in acoustics and can be expressed mathematically:
- When sound intensity doubles, the formula becomes:- \[10 \log_{10}(2) \approx 3 \text{ dB}\]
This shows that a sound at 10 dB will sound twice as loud at 13 dB, given the same frequency and contextual conditions. This principle is also applicable in several real-world scenarios, such as sound system enhancements, audio equipment calibration, and hearing protection design.
Understanding intensity doubling aids in grasping why certain sounds appear dramatically louder despite only slight increases on a linear scale, illustrating how human perception aligns with mathematical principles of logarithms.

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Most popular questions from this chapter

A car moving at \(20 \mathrm{~m} / \mathrm{s}\) with its horn blowing \((f=1200 \mathrm{~Hz})\) is chasing another car going at \(15 \mathrm{~m} / \mathrm{s}\) in the same direction. What is the apparent frequency of the horn as heard by the driver being chased? Take the speed of sound to be \(340 \mathrm{~m} / \mathrm{s}\). This is a Doppler problem. Draw the observer-to-source arrow; that's the positive direction (see Fig. 23-1). Both the source and the observer are moving in the negative direction. Hence, we use \(-v_{o}\) and \(-v_{s}\) $$ f_{o}=f_{s} \frac{v \pm v_{o}}{v \mp v_{s}}=(1200 \mathrm{~Hz}) \frac{340-15}{340-20}=1.22 \mathrm{kHz} $$ Because the source is approaching the observer, the latter will measure an increase in frequency.

An automobile moving at \(30.0 \mathrm{~m} / \mathrm{s}\) is approaching a factory whistle that has a frequency of \(500 \mathrm{~Hz}\). (a) If the speed of sound in air is \(340 \mathrm{~m} / \mathrm{s}\), what is the apparent frequency of the whistle as heard by the driver? \((b)\) Repeat for the case of the car leaving the factory at the same speed. This is a Doppler shift problem. Draw an arrow from observer to source; this is the positive direction. Here in part \((a)\) the observer is moving in the positive direction, and \(v_{s}=0\). Hence, use \(+v_{o}\) and \(s 0\) (a) \(f_{o}=f_{s} \frac{v \pm v_{o}}{v \mp v_{s}}=(500 \mathrm{~Hz}) \frac{340 \mathrm{~m} / \mathrm{s}+30.0 \mathrm{~m} / \mathrm{s}}{340 \mathrm{~m} / \mathrm{s}-0}=544 \mathrm{~Hz}\) With the car leaving in the negative direction use \(-v_{0}\) and (b) \(f_{o}=f_{s} \frac{v \pm v_{o}}{v \mp v_{s}}=(500 \mathrm{~Hz}) \frac{340 \mathrm{~m} / \mathrm{s}-30.0 \mathrm{~m} / \mathrm{s}}{340 \mathrm{~m} / \mathrm{s}-0}=456 \mathrm{~Hz}\)

To determine the speed of a harmonic oscillator, a beam of sound is sent along the line of the oscillator's motion. The sound, which is emitted at a frequency of \(8000.0 \mathrm{~Hz}\), is reflected straight back by the oscillator to a detector system. The detector observes that the reflected beam varies in frequency between the limits of \(8003.1\) \(\mathrm{Hz}\) and \(7996.9 \mathrm{~Hz}\). What is the maximum speed of the oscillator? Take the speed of sound to be \(340 \mathrm{~m} / \mathrm{s}\).

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 .

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

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