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Chapter 16: Problem 19

For a person with normal hearing, the faintest sound that can be heard at a frequency of 400 \(\mathrm{Hz}\) has a pressure amplitude of about \(6.0 \times 10^{-5} \mathrm{Pa}\) . Calculate the (a) intensity; (b) sound intensity level; (c) displacement amplitude of this sound wave at \(20^{\circ} \mathrm{C}\) .

Short Answer

Expert verified
(a) Intensity is \(2.94 \times 10^{-12} \, \text{W/m}^2\); (b) Sound intensity level is \(4.68\, \text{dB}\); (c) Displacement amplitude is \(4.01 \times 10^{-11} \, \text{m}\).

Step by step solution

01

Calculate the Intensity

The intensity of a sound wave can be calculated using the formula:\[ I = \frac{(p_0)^2}{2\rho v} \]where \( p_0 = 6.0 \times 10^{-5} \, \text{Pa} \) is the pressure amplitude, \( \rho \) is the density of air, and \( v \) is the speed of sound in air.At \( 20^\circ \text{C} \), \( \rho = 1.21 \, \text{kg/m}^3 \) and \( v = 343 \, \text{m/s} \). Substituting these values, we have:\[ I = \frac{(6.0 \times 10^{-5})^2}{2 \times 1.21 \times 343} \approx 2.94 \times 10^{-12} \, \text{W/m}^2 \]
02

Calculate the Sound Intensity Level

The sound intensity level \( L \) in decibels (dB) is given by:\[ L = 10 \log_{10}\left(\frac{I}{I_0}\right) \]where \( I_0 = 1.0 \times 10^{-12} \, \text{W/m}^2 \) is the reference intensity level.Using the intensity calculated in Step 1, substitute into the formula:\[ L = 10 \log_{10}\left(\frac{2.94 \times 10^{-12}}{1.0 \times 10^{-12}}\right) \approx 4.68 \, \text{dB} \]
03

Calculate the Displacement Amplitude

The displacement amplitude \( s_0 \) can be calculated using:\[ s_0 = \frac{p_0}{\omega \cdot \rho \cdot v} \]where \( \omega = 2\pi f \) (angular frequency) and \( f = 400 \, \text{Hz} \).First, calculate the angular frequency:\[ \omega = 2 \pi \times 400 = 800\pi \, \text{rad/s} \]Then, substitute into the formula for \( s_0 \):\[ s_0 = \frac{6.0 \times 10^{-5}}{800\pi \times 1.21 \times 343} \approx 4.01 \times 10^{-11} \, \text{m} \]

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Key Concepts

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

Sound Intensity
Sound intensity refers to the power per unit area carried by a sound wave, and it's important for understanding how much energy is received by a surface from a sound source. Sound intensity is typically measured in watts per square meter (W/m\(^2\)). Unlike loudness, which is how we perceive sound, intensity is an objective measurement. The formula to calculate sound intensity is:\[ I = \frac{(p_0)^2}{2\rho v} \]Here, \( p_0 \) is the pressure amplitude, \( \rho \) is the air density, and \( v \) is the speed of sound. At 20掳C, air density is usually about 1.21 kg/m\(^3\), and the speed of sound is approximately 343 m/s.
  • To calculate sound intensity, square the pressure amplitude.
  • Divide by 2, multiply with air density and sound speed.
  • This gives the acoustic power per unit area, essential in acoustics for studying sound distribution.
Sound Intensity Level
Sound intensity level is a logarithmic measure of the intensity of a sound relative to a reference intensity. It is expressed in decibels (dB). This measurement helps us understand how loud a sound is perceived in relation to a baseline, which is usually the threshold of hearing.The formula for sound intensity level \( L \) is:\[ L = 10 \log_{10}\left(\frac{I}{I_0}\right) \]where \( I \) is the sound intensity, and \( I_0 = 1.0 \times 10^{-12} \, \text{W/m}^2 \) is the reference intensity often used for the threshold of hearing in air.
  • Decibel scale is logarithmic; small changes in intensity can mean large changes in dB.
  • This scale helps visualize how much louder one sound is compared to another.
  • Useful in various applications such as audio engineering and environmental noise monitoring.
Displacement Amplitude
Displacement amplitude in a sound wave measures the maximum distance that particles in the medium are displaced from their rest position due to a passing wave. It's a fundamental concept because it directly links to how energetic a sound wave is in terms of particle movement.The formula to calculate displacement amplitude \( s_0 \) is:\[ s_0 = \frac{p_0}{\omega \cdot \rho \cdot v} \]where \( p_0 \) is the pressure amplitude, \( \omega \) is the angular frequency calculated as \( 2\pi \times f \) with \( f \) being the frequency, \( \rho \) is the density of the medium, and \( v \) is the speed of sound.
  • Angular frequency \( \omega \) connects the wave's frequency with a circular motion concept, making calculations easier.
  • Displacement amplitude is directly influenced by both the frequency and medium properties.
  • Knowing this value helps in understanding how intense a sound feels and its potential impact on surroundings.

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Most popular questions from this chapter

(a) By what factor must the sound intensity be increased to raise the sound intensity level by 13.0 dB? (b) Explain why you don't need to know the original sound intensity.

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