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Chapter 14: Problem 99

Ten violins playing simultaneously with the same intensity combine to give an intensity level of \(70 \mathrm{~dB}\). (a) What is the intensity level of each violin? (b) If the number of violins is increased to 100 , will the intensity level be more than, less than, or equal to \(80 \mathrm{~dB}\) ? Explain.

Short Answer

Expert verified
The intensity level of each violin is 60 dB. With 100 violins, the intensity level is 80 dB.

Step by step solution

01

Understanding Sound Intensity Level

Sound intensity level in decibels (dB) can be related to intensity using the formula \( L = 10 \log_{10}( \frac{I}{I_0} ) \), where \( L \) is the sound level, \( I \) is the intensity, and \( I_0 \) is the reference intensity level of \( 10^{-12} \, \mathrm{W/m^2} \).
02

Finding Total Intensity for Ten Violins

Given that 10 violins produce a combined intensity level of \( 70 \, \mathrm{dB} \), use the formula to find total intensity: \[ 70 = 10 \log_{10}( \frac{I_{10}}{10^{-12}} ) \]. Solving for \( I_{10} \), we have: \[ I_{10} = 10^{-5} \, \mathrm{W/m^2} \].
03

Calculating Intensity of a Single Violin

Each violin produces an equal fraction of the total intensity \( I_{10} \). Therefore, the intensity of one violin \( I_1 \) is: \[ I_1 = \frac{I_{10}}{10} = 10^{-6} \, \mathrm{W/m^2} \].
04

Finding Sound Level of a Single Violin

The sound intensity level of one violin is: \[ L_1 = 10 \log_{10}( \frac{I_1}{10^{-12}} ) = 10 \log_{10} (10^6) = 10 \times 6 = 60 \, \mathrm{dB} \].
05

Assessing Sound Level of 100 Violins

Increasing the number of violins to 100 means the combined intensity \( I_{100} = 100 I_1 = 10^{-6} \times 100 = 10^{-4} \, \mathrm{W/m^2} \). The sound intensity level for 100 violins is: \[ L_{100} = 10 \log_{10}( \frac{10^{-4}}{10^{-12}} ) = 10 \times 8 = 80 \, \mathrm{dB} \].
06

Conclusion on 100 Violins

The sound level with 100 violins is exactly \( 80 \, \mathrm{dB} \), which matches the question requirement.

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Key Concepts

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

Decibels
Decibels (dB) are a unit used to measure sound intensity levels. It's important to understand that decibels are not like regular units of measurement, such as meters or grams. Instead, decibels are logarithmic, which helps us handle the wide range of human hearing. A change in 10 dB represents a tenfold change in sound intensity, which might sound complex but simplifies understanding large variations in sound levels. Here's how decibels work:
  • 10 dB increase: Indicates a sound perceived about twice as loud to the human ear.
  • 20 dB increase: Represents a sound perceived four times louder.
Moreover, Decibels give a relative measure of sound compared to a baseline reference. In many cases, this baseline is set at \(10^{-12} \, \mathrm{W/m^2}\), the quietest sound level a typical human can detect. Understanding decibels is key to grasping how sound intensity is measured and compared.
Intensity Level
Sound intensity level refers to the power per unit area, typically in watts per square meter (\(\mathrm{W/m^2}\)). Intensity level doesn't just consider how loud a sound is but relates it to a baseline via calculation. The formula to find intensity level in decibels is:\[L = 10 \log_{10}\left( \frac{I}{I_0} \right)\]where \(L\) is the intensity level in decibels, \(I\) is the intensity of the sound, and \(I_0\) is the reference intensity \(10^{-12} \, \mathrm{W/m^2}\). This scale demonstrates how even small changes in intensity can substantially alter perceived loudness.When multiple sound sources contribute to the overall intensity, each has a fraction of the total intensity. For example, if 10 violins create a specific intensity, one violin represents a tenth of this total intensity. Calculating individual contributions helps find the overall sound intensity level, whether we talk about a single instrument or an ensemble.
Logarithmic Scale
A logarithmic scale is used when dealing with quantities that vary over wide ranges, such as sound intensity. The reason we use a logarithmic scale for sound is to compress this wide range into manageable numbers. With sound, especially, the difference from the faintest to the loudest is vast, spanning trillions. Here's how a logarithmic scale impacts understanding:
  • Helps to manage large variances in data.
  • Converts multiplication relationships into addition, simplifying calculations.
For sound calculations, this means transforming the multiplication of intensities (like combining multiple instruments) into an addition of intensity levels (dB). For instance, doubling the number of violins adds 3 dB to the sound intensity level, equating to a perceived increase in loudness. The logarithmic scale also clarifies why an increase from 70 dB to 80 dB represents a tenfold increase in intensity, not merely a simple 10-unit increase. This perspective is vital to truly understanding changes in sound intensity and their physical implications.

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Most popular questions from this chapter

A particular jet engine produces a tone of \(495 \mathrm{~Hz}\). Suppose that one jet is at rest on the tarmac while a second identical jet flies overhead at \(82.5 \%\) of the speed of sound. The pilot of each jet listens to the sound produced by the engine of the other jet. (a) Calculate the frequency heard by the pilot in the moving jet. (b) Calculate the frequency heard by the pilot in the stationary jet, assuming the flying jet is moving toward it.

The wavelength of sound coming from a loudspeaker is doubled. Does the frequency of the sound waves increase, decrease, or stay the same? Explain. Does the speed of the sound waves increase, decrease, or stay the same? Explain.

Predict \& Explain (a) Is the beat frequency produced when a \(245-\mathrm{Hz}\) tone and a \(240-\mathrm{Hz}\) tone are played together greater than, less than, or equal to the beat frequency produced when a \(140-\mathrm{Hz}\) tone and a \(145-\mathrm{Hz}\) tone are played together? (b) Choose the best explanation from among the following: A. The beat frequency is determined by the difference between the frequencies and is independent of their actual values. B. The higher frequencies will produce a greater beat frequency. C. The percent change in frequency for \(240 \mathrm{~Hz}\) and \(245 \mathrm{~Hz}\) is less than for \(140 \mathrm{~Hz}\) and \(145 \mathrm{~Hz}\), resulting in a smaller beat frequency.

The frequency of a sound is doubled. Does the wave speed of this sound increase, decrease, or stay the same? Explain.

What area does a sound with an intensity of \(0.067 \mathrm{~W} / \mathrm{m}^{2}\) pass through if the power of the source is \(0.21 \mathrm{~W}\) ?
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