Question

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

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

Step-by-step solution

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:\[\lambda = \frac{343 \text{ m/s}}{1000 \text{ Hz}} = 0.343 \text{ m}\]

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 \( s_m \):\[s_m = \frac{30}{1.42 \times 10^5 \times \frac{2\pi}{0.343}} \approx 1.2 \times 10^{-7} \text{ m}\]

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 \( v = 343 \text{ m/s} \):\[f = \frac{v}{\lambda} = \frac{343}{7.11} \approx 48.2 \text{ Hz}\]

Key concepts

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.