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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 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\(^\circ\)C.

Short Answer

Expert verified
(a) Intensity: \(1.18 \times 10^{-12}\) W/m虏; (b) Sound Intensity Level: 0.70 dB; (c) Displacement Amplitude: \(5.8 \times 10^{-11}\) m.

Step by step solution

01

Calculate Intensity

The intensity \( I \) of a sound wave can be calculated using the formula:\[I = \frac{P^2}{2\rho v}\]where \( P \) is the pressure amplitude \( 6.0 \times 10^{-5} \) Pa, \( \rho \) is the density of air 1.21 kg/m\(^3\) at 20\(^\circ\)C, and \( v \) is the speed of sound in air 343 m/s at 20\(^\circ\)C.Substituting the values:\[I = \frac{(6.0 \times 10^{-5})^2}{2 \times 1.21 \times 343} \approx 1.18 \times 10^{-12} \text{ W/m}^2\]
02

Calculate 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.Substituting \( I = 1.18 \times 10^{-12} \text{ W/m}^2 \):\[L = 10 \log_{10} \left( \frac{1.18 \times 10^{-12}}{1.0 \times 10^{-12}} \right) \approx 0.70 \text{ dB}\]
03

Calculate Displacement Amplitude

The displacement amplitude \( s_m \) can be calculated using the formula:\[s_m = \frac{P}{\omega Z}\]where \( \omega = 2 \pi f \) is the angular frequency, and \( Z = \rho v \) is the acoustic impedance.Given \( f = 400 \) Hz, calculate \( \omega \):\[\omega = 2 \pi \times 400 = 800\pi \text{ rad/s}\]Calculate \( Z \):\[Z = 1.21 \times 343 = 415.03 \text{ kg/(m}^2\text{s)}\]Substitute these values into the formula:\[s_m = \frac{6.0 \times 10^{-5}}{800\pi \times 415.03} \approx 5.8 \times 10^{-11} \text{ m}\]

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Key Concepts

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

Understanding Sound Waves
Sound waves are vibrations that travel through the air, water, or any other medium. These waves are essentially oscillations of pressure in the medium, resulting from a source like a tuning fork or a speaker. As these pressure changes propagate through the medium, they create compressions and rarefactions.
  • Median Travel: Sound requires a medium to travel, as it cannot move through a vacuum. Examples of media include air, water, and solids.
  • Speed of Sound: The speed of sound varies based on the medium. In air, it averages around 343 m/s at room temperature (20掳C).
  • Frequency: This is the number of cycles of the wave that pass a point in one second, measured in Hertz (Hz).
Understanding these aspects of sound waves is crucial in studying their behavior and characteristics, like intensity and amplitude, which are important for applications such as hearing and audio technology.
Pressure Amplitude Insights
Pressure amplitude in sound waves refers to the maximum change in pressure caused by the wave's compressions and rarefactions. It is a measure of how strong or powerful the sound wave is.
  • Units and Measurement: Pressure amplitude is measured in Pascals (Pa).
  • Relation to Intensity: The higher the pressure amplitude, the more intense, or louder, the sound becomes.
  • Significance in Hearing: For a person with normal hearing, the faintest sound detectable at a specific frequency like 400 Hz corresponds to a minimum pressure amplitude, for example, 6.0 脳 10^-5 Pa in this scenario.
Given the importance of pressure amplitude, calculating sound intensity involves using it as a foundational parameter. It's also essential for understanding how we perceive different sounds.
The Concept of Displacement Amplitude
Displacement amplitude refers to the maximum distance a particle in the medium moves from its rest position as the sound wave passes through.
  • Understanding the Movement: As a sound wave travels, particles in the medium oscillate back and forth, with displacement amplitude indicating the extent of this movement.
  • Calculating Displacement: It is calculated using the formula \( s_m = \frac{P}{\omega Z} \), where \( P \) is the pressure amplitude, \( \omega \) is angular frequency, and \( Z \) is acoustic impedance.
  • Interplay of Quantities: Displacement amplitude is smaller for low pressure amplitudes and higher frequencies, meaning the actual movement of particles may be very subtle, yet critical for analyzing sound behavior.
Understanding displacement amplitude is particularly valuable in fields needing knowledge of particle motion, such as in the design of microphones and speakers, where precise replication of sound is key.

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Most popular questions from this chapter

At point \(A, 3.0 \mathrm{~m}\) from a small source of sound that is emitting uniformly in all directions, the sound intensity level is \(53 \mathrm{~dB}\). (a) What is the intensity of the sound at \(A ?\) (b) How far from the source must you go so that the intensity is one-fourth of what it was at \(A\) ? (c) How far must you go so that the sound intensity level is one-fourth of what it was at \(A ?\) (d) Does intensity obey the inverse-square law? What about sound intensity level?

The siren of a fire engine that is driving northward at 30.0 m/s emits a sound of frequency 2000 Hz. A truck in front of this fire engine is moving northward at 20.0 m/s. (a) What is the frequency of the siren's sound that the fire engine's driver hears reflected from the back of the truck? (b) What wavelength would this driver measure for these reflected sound waves?

The sound from a trumpet radiates uniformly in all directions in 20\(^\circ\)C air. At a distance of 5.00 m from the trumpet the sound intensity level is 52.0 dB. The frequency is 587 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 dB?

Singing in the Shower. A pipe closed at both ends can have standing waves inside of it, but you normally don't hear them because little of the sound can get out. But you \(can\) hear them if you are \(inside\) the pipe, such as someone singing in the shower. (a) Show that the wavelengths of standing waves in a pipe of length \(L\) that is closed at both ends are \(\lambda$$_n =\) \(2L /n\) and the frequencies are given by \({f_n}\) \(= n$$v/2L = nf$$_1\), where \(n =\) 1, 2, 3, . . . . (b) Modeling it as a pipe, find the frequency of the fundamental and the first two overtones for a shower 2.50 m tall. Are these frequencies audible?

In some applications of ultrasound, such as its use on cranial tissues, large reflections from the surrounding bones can produce standing waves. This is of concern because the large pressure amplitude in an antinode can damage tissues. For a frequency of 1.0 MHz, what is the distance between antinodes in tissue? (a) 0.38 mm; (b) 0.75 mm; (c) 1.5 mm; (d) 3.0 mm.
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