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Example 16.1 (Section 16.1) showed that for sound waves in air with frequency 1000 Hz, a displacement amplitude of 1.2 \(\times\) 10\(^{-8}\) m produces a pressure amplitude of 3.0 \(\times\) 10\(^{-2}\) Pa. (a) What is the wavelength of these waves? (b) For 1000-Hz waves in air, what displacement amplitude would be needed for the pressure amplitude to be at the pain threshold, which is 30 Pa? (c) For what wavelength and frequency will waves with a displacement amplitude of 1.2 \(\times\) 10\(^{-8}\) m produce a pressure amplitude of 1.5 \(\times\) 10\(^{-3}\) Pa?

Short Answer

Expert verified
(a) 0.343 m; (b) 1.2 × 10⁻⁷ m; (c) λ ≈ 7.11 m, f ≈ 48.2 Hz.

Step by step solution

01

Calculate the Wavelength for Part (a)

The wavelength \( \lambda \) of a wave is related to the speed of sound \( v \) and frequency \( f \) by the formula:\[\lambda = \frac{v}{f}\]Using the speed of sound in air \( v \approx 343 \text{ m/s} \) and the frequency \( f = 1000 \text{ Hz} \), the wavelength is calculated as:\[\lambda = \frac{343 \text{ m/s}}{1000 \text{ Hz}} = 0.343 \text{ m}\]
02

Find Displacement Amplitude for Pain Threshold Pressure for Part (b)

The relationship between pressure amplitude \( \Delta P \) and displacement amplitude \( s_m \) is given by:\[\Delta P = Bk s_m\]where \( B \) is the bulk modulus of elasticity for air (approximately \( 1.42 \times 10^5 \text{ Pa} \)) and \( k \) is the wave number \( k = \frac{2\pi}{\lambda} \). Using \( \Delta P = 30 \text{ Pa} \) and \( \lambda = 0.343 \text{ m} \):\[30 = 1.42 \times 10^5 \times \frac{2\pi}{0.343} \times s_m\]Solving for \( s_m \):\[s_m = \frac{30}{1.42 \times 10^5 \times \frac{2\pi}{0.343}} \approx 1.2 \times 10^{-7} \text{ m}\]
03

Determine Wavelength and Frequency for Part (c)

Given a displacement amplitude \( s_m = 1.2 \times 10^{-8} \text{ m} \) and pressure amplitude \( \Delta P = 1.5 \times 10^{-3} \text{ Pa} \), use:\[\Delta P = Bk s_m\]Solving for the wave number \( k \):\[k = \frac{1.5 \times 10^{-3}}{1.42 \times 10^5 \times 1.2 \times 10^{-8}} \approx 0.884 \, \text{m}^{-1}\]The wavelength is \( \lambda = \frac{2\pi}{k} \approx \frac{2\pi}{0.884} \approx 7.11 \, \text{m} \).To find frequency \( f \) with wavelength \( \lambda = 7.11 \text{ m} \) and speed \( v = 343 \text{ m/s} \):\[f = \frac{v}{\lambda} = \frac{343}{7.11} \approx 48.2 \text{ Hz}\]

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Key Concepts

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

Wavelength Calculation
Understanding how to calculate the wavelength of a sound wave is crucial in various physics applications. The wavelength refers to the distance between two consecutive points in phase on a wave, such as from one crest to the next. In sound waves traveling through air, we can determine the wavelength by using the formula:\[ \lambda = \frac{v}{f} \]where \( v \) is the speed of sound in air and \( f \) is the frequency of the wave. For instance, if the speed of sound in air is approximately 343 m/s and the frequency is 1000 Hz, the wavelength calculation would be:\[ \lambda = \frac{343 \text{ m/s}}{1000 \text{ Hz}} = 0.343 \text{ m} \]This formula shows the inverse relationship between frequency and wavelength, meaning that as frequency increases, the wavelength decreases when the speed of sound remains constant.
Pressure Amplitude
Pressure amplitude in sound waves represents the maximum change in pressure from the resting atmospheric pressure due to the wave's motion. It is a crucial measure of the wave's intensity and can directly affect how we perceive the loudness of the sound. The relationship between pressure amplitude \( \Delta P \) and displacement amplitude \( s_m \) can be expressed as:\[ \Delta P = Bk s_m \]Here, \( B \) signifies the bulk modulus of the medium, and \( k \) stands for the wave number, calculated as \( k = \frac{2\pi}{\lambda} \). If we consider an example where a sound wave in air has a displacement amplitude of \( 1.2 \times 10^{-8} \text{ m} \) producing a pressure amplitude of \( 3.0 \times 10^{-2} \text{ Pa} \), we can perceive how pressure amplitude increases with higher displacement amplitude, showing their proportional relationship.
Displacement Amplitude
Displacement amplitude indicates how far particles in the medium, like air molecules, are moved from their equilibrium position by the sound wave. It signifies the vibration's degree and directly correlates with the sound wave's pressure amplitude. The greater the displacement amplitude, the higher the potential pressure amplitude. When waves reach the threshold of pain, such as 30 Pa, you can calculate the required displacement amplitude \( s_m \) using:\[ s_m = \frac{\Delta P}{B k} \]In our calculation example, using \( \Delta P = 30 \text{ Pa} \) for pain threshold and bulk modulus \( B = 1.42 \times 10^5 \text{ Pa} \), we find the displacement amplitude for this intense pressure is roughly \( 1.2 \times 10^{-7} \text{ m} \). This illustrates the necessity for a significant displacement amplitude to create perceptibly high pressure levels in sound waves.
Frequency and Wavelength Relationship
The core relationship between frequency and wavelength is fundamental in understanding how sound waves behave. Frequency is the number of wave cycles per second, measured in Hertz (Hz), while the wavelength is the spatial length of each wave cycle. They are intertwined by the equation:\[ v = f \lambda \]As sound speed \( v \) in air is usually constant, an increase in frequency results in a decrease in wavelength, and vice versa. This means if you have a sound wave traveling through air at 343 m/s frequency 1000 Hz, it results in a wavelength of 0.343 m. Similarly, if the frequency is decreased to 48.2 Hz, as calculated for a specific condition, the wavelength is extended to 7.11 m.This dynamic showcases the inverse relationship, significant in sound wave applications like audio technologies and acoustics, affecting how sound is transmitted and perceived across different media variables.

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Most popular questions from this chapter

You have a stopped pipe of adjustable length close to a taut 62.0-cm, 7.25-g wire under a tension of 4110 N. You want to adjust the length of the pipe so that, when it produces sound at its fundamental frequency, this sound causes the wire to vibrate in its second \(overtone\) with very large amplitude. How long should the pipe be?

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?

A railroad train is traveling at 30.0 m/s in still air. The frequency of the note emitted by the train whistle is 352 Hz. What frequency is heard by a passenger on a train moving in the opposite direction to the first at 18.0 m/s and (a) approaching the first and (b) receding from the first?

A 75.0-cm-long wire of mass 5.625 g is tied at both ends and adjusted to a tension of 35.0 N. When it is vibrating in its second overtone, find (a) the frequency and wavelength at which it is vibrating and (b) the frequency and wavelength of the sound waves it is producing.

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?

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