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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 脳 10鈦烩伔 m; (c) 位 鈮 7.11 m, f 鈮 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

The Sacramento City Council adopted a law to reduce the allowed sound intensity level of the much-despised leaf blowers from their current level of about 95 dB to 70 dB. With the new law, what is the ratio of the new allowed intensity to the previously allowed intensity?

Two swift canaries fly toward each other, each moving at 15.0 m/s relative to the ground, each warbling a note of frequency 1750 Hz. (a) What frequency note does each bird hear from the other one? (b) What wavelength will each canary measure for the note from the other one?

The sound source of a ship's sonar system operates at a frequency of 18.0 kHz. The speed of sound in water (assumed to be at a uniform 20\(^\circ\)C) is 1482 m/s. (a) What is the wavelength of the waves emitted by the source? (b) What is the difference in frequency between the directly radiated waves and the waves reflected from a whale traveling directly toward the ship at 4.95 m/s ? The ship is at rest in the water.

The frequency of the note F\(_4\) is 349 Hz. (a) If an organ pipe is open at one end and closed at the other, what length must it have for its fundamental mode to produce this note at 20.0\(^\circ\)C? (b) At what air temperature will the frequency be 370 Hz, corresponding to a rise in pitch from F to F-sharp? (Ignore the change in length of the pipe due to the temperature change.)

Many opera singers (and some pop singers) have a range of about 2\(\frac{1}{2}\) octaves or even greater. Suppose a soprano's range extends from A below middle C (frequency 220 Hz) up to E-flat above high C (frequency 1244 Hz). Although the vocal tract is quite complicated, we can model it as a resonating air column, like an organ pipe, that is open at the top and closed at the bottom. The column extends from the mouth down to the diaphragm in the chest cavity, and we can also assume that the lowest note is the fundamental. How long is this column of air if \(v =\) 354 m/s? Does your result seem reasonable, on the basis of observations of your own body?

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