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Magazine Chapter 16: Problem 19
BIO 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 \(6.0 \times 10^{-5}\) 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) 3.0 脳 10^-12 W/m^2; (b) 4.77 dB; (c) 5.81 脳 10^-11 m.
Step by step solution
01
Calculate Sound Intensity
The intensity of a sound wave is calculated using the formula \(I = \frac{p^2}{2 \rho v}\), where \(p\) is the pressure amplitude, \(\rho\) is the density of air, and \(v\) is the speed of sound in air at the given temperature. At \(20^{\circ}\)C, \(\rho \approx 1.2041 \, \text{kg/m}^3\) and \(v \approx 343 \, \text{m/s}\). Substitute the given pressure amplitude \(p = 6.0 \times 10^{-5} \, \text{Pa}\). \[I = \frac{(6.0 \times 10^{-5})^2}{2 \times 1.2041 \times 343} \approx 3.0 \times 10^{-12} \, \text{W/m}^2\]
02
Calculate Sound Intensity Level
The sound intensity level \(L\) is calculated using the formula \(L = 10 \log_{10} \left( \frac{I}{I_0} \right)\), where \(I_0 = 10^{-12} \, \text{W/m}^2\) is the reference intensity. Substitute the calculated intensity \(I = 3.0 \times 10^{-12} \, \text{W/m}^2\). \[L = 10 \log_{10} \left( \frac{3.0 \times 10^{-12}}{10^{-12}} \right) \approx 4.77 \, \text{dB}\]
03
Calculate Displacement Amplitude
The displacement amplitude \(A\) of the sound wave can be found using \(A = \frac{p}{\omega Z}\), where \(\omega = 2 \pi f\) is the angular frequency and \(Z = \rho v\) is the acoustic impedance. \(f = 400 \, \text{Hz}\). First, calculate \(\omega\) and \(Z\): \(\omega = 2 \pi \times 400 \approx 2513.27\) and \(Z = 1.2041 \times 343 \approx 413.01\). Then \[A = \frac{6.0 \times 10^{-5}}{2513.27 \times 413.01} \approx 5.81 \times 10^{-11} \, \text{m}\]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Pressure Amplitude
Pressure amplitude is a measure of the maximum change in pressure caused by a sound wave as it travels through a medium. It indicates how compressed or rarefied the medium becomes with the passing of the wave.
Understanding pressure amplitude is crucial because it relates to the physical sensation of loudness鈥攁 higher pressure amplitude typically means a louder sound.
Understanding pressure amplitude is crucial because it relates to the physical sensation of loudness鈥攁 higher pressure amplitude typically means a louder sound.
- Pressure amplitude is measured in Pascals (Pa).
- It is a peak value, representing the furthest extent of compression or rarefaction of the air particles.
- In the exercise provided, the pressure amplitude of the sound is given as approximately \(6.0 \times 10^{-5}\) Pa.
Sound Intensity Level
Sound intensity level is a logarithmic measure of the intensity of sound. It gives us a sense of how loud a sound appears to be in comparison to a standard reference point.
This concept is very useful because it scales vast intensity ranges into manageable numbers, typically measured in decibels (dB).
This concept is very useful because it scales vast intensity ranges into manageable numbers, typically measured in decibels (dB).
- The formula used is \(L = 10 \log_{10} \left( \frac{I}{I_0} \right)\), where \(I\) is the intensity of the sound, and \(I_0 = 10^{-12} \, \text{W/m}^2\) is the reference intensity, which is typically the threshold of hearing.
- In our case, the calculated sound intensity level is approximately 4.77 dB.
- This level tells us how the sound compares to the quietest sound perceivable by the average human ear.
Displacement Amplitude
The displacement amplitude refers to the maximum distance that particles in a medium are displaced from their resting positions due to the passage of a sound wave. It captures the actual movement of particles as the wave energy propagates through the medium.
For sound waves, this is fundamental in understanding how sound travels through materials.
For sound waves, this is fundamental in understanding how sound travels through materials.
- The relationship between pressure and displacement can be described by the equation \(A = \frac{p}{\omega Z}\), where \(A\) is displacement amplitude, \(p\) is the pressure amplitude, \(\omega\) is the angular frequency, and \(Z\) is the acoustic impedance.
- For the given example, the displacement amplitude is approximately \(5.81 \times 10^{-11}\) m, indicating very slight movements of the particles at the sound's stated intensity.
- This concept is instrumental in fields like acoustics, helping to design better soundproofing and understand wave interactions with different materials.
Angular Frequency
Angular frequency relates to how quickly a wave oscillates in terms of radians per second. It is essential in connecting circular motion concepts to wave phenomena.
In the context of sound waves, it helps characterize the wave鈥檚 energy and behavior.
In the context of sound waves, it helps characterize the wave鈥檚 energy and behavior.
- Calculated using the formula \(\omega = 2 \pi f\), where \(f\) is the frequency of the wave.
- For instance, in our given problem, \(f = 400 \, \text{Hz}\), making \(\omega \approx 2513.27\, \text{radians/second}\).
- Higher angular frequencies indicate more rapid vibrations, contributing to higher pitch sounds.
Acoustic Impedance
Acoustic impedance is a measure of resistance a medium offers to the propagation of sound waves. It connects the sound pressure with particle velocity in the medium and is pivotal in studying sound transmission and reflection.
The concept is akin to electrical impedance but applied to sound waves.
The concept is akin to electrical impedance but applied to sound waves.
- Defined as \(Z = \rho v\), where \(\rho\) is the density of the medium and \(v\) is the speed of sound within it.
- In the given example, considering the air at approximately \(20^{\circ} \mathrm{C}\), \(Z \approx 413.01\, \text{kg/(m}^2\cdot \text{s)}\).
- Acoustic impedance influences how sound waves travel through different media, determining how much energy is reflected or absorbed upon hitting a surface. This is crucial for engineering fields like audio technology and architectural acoustics.
