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Chapter 17: Problem 28

Two sounds differ in sound level by \(1.00 \mathrm{~dB}\). What is the ratio of the greater intensity to the smaller intensity?

Short Answer

Expert verified
The ratio of the greater intensity to the smaller intensity is approximately 1.2589.

Step by step solution

01

Understanding Decibels

The decibel ( ext{dB}) is a logarithmic unit used to express the ratio of two values, typically sound intensities. A sound level difference of 1 dB implies that one sound is slightly more intense than the other.
02

Decibel Formula

The intensity level in decibels between two sounds is given by the formula: \[eta = 10 imes ext{log}_{10} rac{I_2}{I_1}\],where \(I_2\) is the greater intensity, \(I_1\) is the smaller intensity, and \(\beta\) is the difference in decibels, which, in this case, is 1.00 dB.
03

Setting Up the Equation

Substitute the given values into the decibel formula: \[10 imes ext{log}_{10} rac{I_2}{I_1} = 1.00\].Now, solve for \(\frac{I_2}{I_1}\).
04

Solving for Intensity Ratio

Divide both sides of the equation by 10: \[ ext{log}_{10} rac{I_2}{I_1} = 0.1\].Remove the logarithm by raising 10 to the power of both sides: \[ rac{I_2}{I_1} = 10^{0.1}\].
05

Calculate the Ratio

Calculate the value of \(10^{0.1}\): \[ rac{I_2}{I_1} \approx 1.2589\].Therefore, the ratio of the greater intensity to the smaller intensity is approximately 1.2589.

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Key Concepts

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

Decibels
Sound intensity is often measured using a unit called the decibel, or dB. This unit helps us to understand how loud a sound is by comparing it to a certain reference level.
The important thing to remember is that decibels are not like regular numbers. They represent a logarithmic scale. This means they can express very large or small ratios in a more compact form.
  • A difference of 1 dB means a small change in sound intensity.
  • Higher decibel values represent louder sounds.
  • Decibels measure relative sound levels, not absolute intensity.
This makes decibels a powerful tool for comparing sounds, whether in acoustics, engineering, or even everyday conversations.
Logarithmic Unit
Decibels use a logarithmic scale, which is different from what we use in everyday math. Instead of being linear, like adding 1 + 1 to get 2, the logarithmic scale deals with orders of magnitude or how many times you multiply one number to get another.

Here鈥檚 why it鈥檚 useful:
  • This scale can handle a wide range of sound intensities without needing large numbers.
  • Logarithms help convert multiplicative relationships into additive ones, making calculations simpler.
For sound, the formula is \[\beta = 10 \times \log_{10} \left(\frac{I_2}{I_1}\right)\]where \(\beta\) is the sound level difference in dB, and \(I_2\) and \(I_1\) are two different sound intensities. This representation helps in effectively communicating large disparities in sound.
Intensity Ratio
The intensity ratio is how we define the difference in loudness between two sounds. In the context of decibels:
  • An intensity ratio tells us how much louder or softer one sound is compared to another.
  • An intensity ratio of 1.2589 corresponds to a 1 dB difference.
Using the formula\[ \text{log}_{10} \left(\frac{I_2}{I_1}\right) = 0.1 \]and solving it gives the \(\frac{I_2}{I_1}\) which equals \(10^{0.1}\).
This result confirms that, for every 1 dB increase in sound level, the intensity increases by approximately 26%, indicating that a small change in sound level can equate to a noticeable change in intensity.
Sound Level Difference
Sound level difference quantifies how one sound's intensity compares to another's in terms of decibels. When we say there is a 1 dB difference, one sound is slightly louder than the other.
  • The formula \[ \beta = 10 \times \log_{10} \left(\frac{I_2}{I_1}\right) \] shows the relationship between the decibel difference and the intensity ratio.
  • The formula is solved for \(\frac{I_{2}}{I_{1}}\) by rearranging it and finding the exponential \(10^{0.1} \approx 1.2589\).
This ratio implies that even a small decibel change makes a significant difference in intensity. Understanding this concept is crucial because it explains how sounds can appear vastly different in loudness, even when the decibel difference is minimal.

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Most popular questions from this chapter

A siren emitting a sound of frequency \(1000 \mathrm{~Hz}\) moves away from you toward the face of a cliff at a speed of \(10 \mathrm{~m} / \mathrm{s}\). Take the speed of sound in air as \(330 \mathrm{~m} / \mathrm{s}\). (a) What is the frequency of the sound you hear coming directly from the siren? (b) What is the frequency of the sound you hear reflected off the cliff? (c) What is the beat frequency between the two sounds? Is it perceptible (less than \(20 \mathrm{~Hz}\) )?

You are standing at a distance \(D\) from an isotropic point source of sound. You walk \(50.0 \mathrm{~m}\) toward the source and observe that the intensity of the sound has doubled. Calculate the distance \(D\).

Tap a metal spoon inside a mug of water and note the frequency \(f_{i}\) you hear. Then add a spoonful of powder (say, chocolate mix or instant coffee) and tap again as you stir the powder. The frequency you hear has a lower value \(f_{s}\) because the tiny air bubbles released by the powder change the water's bulk modulus. As the bubbles reach the water surface and disappear, the frequency gradually shifts back to its initial value. During the effect, the bubbles don't appreciably change the water's density or volume or the sound's wavelength. Rather, they change the value of \(d V / d p-\) that is, the differential change in volume due to the differential change in the pressure caused by the sound wave in the water. If \(f_{s} / f_{i}=0.333\), what is the ratio \((d V / d p)_{s} /(d V / d p)_{i} ?\)

A point source emits \(30.0\) W of sound isotropically. A small microphone intercepts the sound in an area of \(0.750 \mathrm{~cm}^{2}, 200 \mathrm{~m}\) from the source. Calculate (a) the sound intensity there and (b) the power intercepted by the microphone.

A certain loudspeaker system emits sound isotropically with a frequency of \(2000 \mathrm{~Hz}\) and an intensity of \(0.960 \mathrm{~mW} / \mathrm{m}^{2}\) at a distance of \(6.10 \mathrm{~m}\). Assume that there are no reflections. (a) What is the intensity at \(30.0 \mathrm{~m}\) ? At \(6.10 \mathrm{~m}\), what are (b) the displacement amplitude and (c) the pressure amplitude?
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