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Chapter 16: Problem 60

CP The sound from a trumpet radiates uniformly in all directions in \(20^{\circ} \mathrm{C}\) air. At a distance of 5.00 \(\mathrm{m}\) from the trumpet the sound intensity level is 52.0 \(\mathrm{dB}\) . The frequency is 587 \(\mathrm{Hz}\) . (a) What is the pressure amplitude at this distance? (b) What is the displacement amplitude? (c) At what distance is the sound intensity level 30.0 \(\mathrm{dB}\) ?

Short Answer

Expert verified
(a) Pressure amplitude: 0.0122 Pa. (b) Displacement amplitude: 8.96 脳 10鈦宦光伆 m. (c) Distance: 63 m.

Step by step solution

01

Convert Intensity Level to Intensity

Given the sound intensity level \( \beta = 52.0 \, \mathrm{dB} \), we need to convert this to intensity using the formula:\[\beta = 10 \log_{10} \left( \frac{I}{I_0} \right) \]where \( I_0 = 1 \times 10^{-12} \, \mathrm{W/m^2} \). Solving for \( I \), we have:\[I = I_0 \times 10^{\beta/10} = 1 \times 10^{-12} \times 10^{5.2} \]\[ I \approx 1.58 \times 10^{-7} \, \mathrm{W/m^2} \].
02

Calculate Pressure Amplitude

The pressure amplitude \( p_{max} \) is related to intensity by the formula:\[I = \frac{p_{max}^2}{2 \rho v} \]where \( \rho \) is the air density (\( \approx 1.21 \, \mathrm{kg/m^3} \)) and \( v \) is the sound speed in air at \( 20^\circ \mathrm{C} \) (\( \approx 343 \, \mathrm{m/s} \)). Solving for \( p_{max} \):\[p_{max} = \sqrt{2 \rho v I} = \sqrt{2 \times 1.21 \times 343 \times 1.58 \times 10^{-7}} \]\[ p_{max} \approx 1.22 \times 10^{-2} \, \mathrm{Pa} \].
03

Calculate Displacement Amplitude

The displacement amplitude \( s_{max} \) is given by:\[s_{max} = \frac{p_{max}}{\rho v \omega} \]where \( \omega = 2 \pi f \) and \( f = 587 \, \mathrm{Hz} \). The angular frequency \( \omega \) is:\[\omega = 2 \pi \times 587 \]\[ \omega \approx 3686 \, \mathrm{rad/s} \].Thus, \( s_{max} \) is:\[s_{max} = \frac{1.22 \times 10^{-2}}{1.21 \times 343 \times 3686} \]\[ s_{max} \approx 8.96 \times 10^{-10} \, \mathrm{m} \].
04

Find Distance for 30.0 dB Intensity Level

Given \( \beta = 30.0 \, \mathrm{dB} \), find new intensity \( I' \):\[I' = I_0 \times 10^{3} = 1 \times 10^{-12} \times 10^3 \]\[ I' = 1 \times 10^{-9} \, \mathrm{W/m^2} \].Use \( I \propto \frac{1}{r^2} \) relationship to find the new distance \( r' \):\[\frac{I'}{I} = \left( \frac{r}{r'} \right)^2 \]\[r' = r \sqrt{\frac{I}{I'}} = 5 \sqrt{\frac{1.58 \times 10^{-7}}{1 \times 10^{-9}}} \]\[ r' \approx 63 \mathrm{m} \].

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Key Concepts

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

Pressure Amplitude
The concept of pressure amplitude is important when understanding how sound waves propagate through a medium, like air. The pressure amplitude refers to the maximum change in pressure occurring as a sound wave passes through the air. This change is measured in Pascals (Pa). In layman's terms, it essentially denotes how "strong" or "loud" the sound appears to a listener.

Sound intensity and pressure amplitude are related; you can calculate the pressure amplitude if you know the sound intensity. Using the formula:
  • \[ I = \frac{p_{max}^2}{2 \rho v} \],
where \(I\) is the intensity, \(p_{max}\) is the pressure amplitude, \(\rho\) is the density of air, and \(v\) is the speed of sound in air. By rearranging this formula, you can solve for the pressure amplitude.

This calculation shows us how energy is distributed across a sound wave. At a distance, the pressure amplitude diminishes, making the sound quieter.
Displacement Amplitude
In sound wave physics, displacement amplitude refers to the maximum distance that particles in the medium move from their rest position due to the disturbance of the sound wave. It's analogous to the "height" of the wave in sound, measured in meters (m).

To calculate displacement amplitude, we use the following formula:
  • \[ s_{max} = \frac{p_{max}}{\rho v \omega} \]
where \( s_{max} \) is the displacement amplitude, \( p_{max} \) is the pressure amplitude, \( \rho \) is the air density, \( v \) is the speed of sound, and \( \omega \) (angular frequency) is calculated as \( 2 \pi f \) with \( f \) being the frequency of the sound wave.

This calculation helps us understand the movement of air particles as sound waves travel through the air. Greater displacement amplitude results in louder sounds perceived by our ears.
Sound Intensity Level
The sound intensity level describes how intense or loud a sound is perceived, with measurements in decibels (dB). The intensity level relates to the amount of sound energy passing through a certain area per unit of time.

This level is calculated using the formula:
  • \[ \beta = 10 \log_{10} \left( \frac{I}{I_0} \right) \]
where \( \beta \) is the sound intensity level, \( I \) is the intensity, and \( I_0 \) is the reference intensity, typically \( 1 \times 10^{-12} \text{ W/m}^2 \).

Decibels provide a logarithmic scale to represent the vast range of sound intensities that can occur, enhancing our ability to compare different sound levels effectively. As you move further away from the source of sound, the intensity decreases, and thus the sound intensity level reduces as well.
Frequency of Sound Waves
The frequency of sound waves refers to the number of wave cycles passing a point per second, measured in Hertz (Hz). This measurement determines the pitch of the sound; higher frequency equates to a higher pitch, while lower frequency results in a lower pitch.

In sound wave calculations, frequency is critical because it impacts both the wavelength and the wave's speed in a particular medium. Essentially, knowing the frequency helps us understand the behavior of sound waves, particularly their interaction with various environments.
  • \( \omega = 2 \pi f \)
\( \omega \) is the angular frequency, which plays a key role in calculating displacement amplitude.

Higher frequencies tend to lose energy more rapidly in a medium like air, which is why high-pitched sounds don't travel as far as low-pitched ones.

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