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Chapter 17: Problem 75

A certain loudspeaker system emits sound isotropically with a frequency of \(2000 \mathrm{~Hz}\) and an intensity of \(0.960 \mathrm{~mW} / \mathrm{m}^{2}\) at a distance of \(6.10 \mathrm{~m}\). Assume that there are no reflections. (a) What is the intensity at \(30.0 \mathrm{~m}\) ? At \(6.10 \mathrm{~m}\), what are (b) the displacement amplitude and (c) the pressure amplitude?

Short Answer

Expert verified
(a) 0.0396 mW/m^2; (b) 5.41 碌m; (c) 29.3 N/m^2.

Step by step solution

01

Understanding Intensity and Distance Relationship

Intensity of sound decreases as distance from source increases. The relationship given by the inverse square law is:\[ I_2 = I_1 \left( \frac{r_1}{r_2} \right)^2 \]where \( I_1 \) is the initial intensity, \( I_2 \) is the final intensity, \( r_1 \) is the initial distance, and \( r_2 \) is the final distance.
02

Calculating Intensity at 30.0 meters

Using the values \( I_1 = 0.960 \text{ mW/m}^2 \), \( r_1 = 6.10 \text{ m} \), and \( r_2 = 30.0 \text{ m} \), we substitute into the formula:\[ I_2 = 0.960 \times \left( \frac{6.10}{30.0} \right)^2 \]Calculating gives:\[ I_2 = 0.960 \times \left( 0.2033 \right)^2 = 0.960 \times 0.0413 = 0.0396 \text{ mW/m}^2 \]
03

Understanding Displacement Amplitude

The displacement amplitude \( s_m \) relates to intensity \( I \) by:\[ s_m = \sqrt{\frac{2I}{\rho v (2\pi f)^2}} \]where \( \rho \) is the density of air (approximately \(1.21 \text{ kg/m}^3\)), \( v \) is the speed of sound in air (approximately \(343 \text{ m/s}\)), and \( f \) is the frequency \(2000 \text{ Hz}\).
04

Calculating Displacement Amplitude

Substitute values into the displacement amplitude formula:\[ s_m = \sqrt{\frac{2 \times 0.960}{1.21 \times 343 \times (2\pi \times 2000)^2}} \]Perform the calculations:\( (2\pi \times 2000)^2 = 1.58 \times 10^8 \), and continue:\[ s_m = \sqrt{\frac{1.92}{1.21 \times 343 \times 1.58 \times 10^8}} = \sqrt{\frac{1.92}{6.55 \times 10^{10}}} \approx \sqrt{2.93 \times 10^{-11}} \approx 5.41 \times 10^{-6} \text{ m} \]
05

Understanding Pressure Amplitude

The pressure amplitude \( p_m \) is related to the displacement amplitude and can be calculated using the formula:\[ p_m = \rho v \omega s_m \]where \( \omega = 2\pi f \) is the angular frequency.
06

Calculating Pressure Amplitude

Substitute the known values into the formula:\[ p_m = 1.21 \times 343 \times (2\pi \times 2000) \times 5.41 \times 10^{-6} \]Calculate to find \( p_m \):\[ p_m = 1.21 \times 343 \times 12566.37 \times 5.41 \times 10^{-6} \approx 29.3 \text{ N/m}^2 \]

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Key Concepts

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

Understanding the Inverse Square Law
The inverse square law is a fundamental principle in acoustics and other wave-related phenomena. It describes how the intensity of sound decreases as the distance from the sound source increases.
The formula for the inverse square law is:
  • \( I_2 = I_1 \left( \frac{r_1}{r_2} \right)^2 \)
Here, \( I_1 \) represents the initial intensity at a distance \( r_1 \), and \( I_2 \) is the intensity at a farther distance \( r_2 \).
The law states that as you move away from the sound source, the intensity diminishes proportionally to the square of the distance. This is due to the spreading out of energy over a larger area. For example, doubling the distance from the source results in the sound intensity being reduced to a quarter of its original value.
Exploring Displacement Amplitude
Displacement amplitude refers to the maximum displacement of particles in the medium through which the sound travels. It is directly related to how powerful the sound feels.
The displacement amplitude \( s_m \) is calculated using:
  • \( s_m = \sqrt{\frac{2I}{\rho v (2\pi f)^2}} \)
In this equation:
  • \( I \) is the intensity of sound
  • \( \rho \) is the density of the medium (air)
  • \( v \) is the speed of sound in the medium
  • \( f \) is the frequency
The formula shows us that the displacement amplitude is influenced by the intensity and frequency of the sound wave, as well as the properties of the medium through which it travels.
Delving into Pressure Amplitude
Pressure amplitude is closely linked to displacement amplitude but represents the variation in pressure caused by the sound wave in the medium. It shows the sound wave's ability to compress and expand the medium.
The pressure amplitude \( p_m \) is given by:
  • \( p_m = \rho v \omega s_m \)
Where:
  • \( \rho \) is the medium's density
  • \( v \) is the speed of sound
  • \( \omega = 2\pi f \) is the angular frequency
  • \( s_m \) is the displacement amplitude
Pressure amplitude helps us understand the impact of the sound wave on the medium, giving insight into how strong it is perceived by our ears or sensors.
Introduction to Frequency
Frequency is a key concept in understanding sound, describing how often the sound waves cycle per second. It is measured in Hertz (Hz). For the given speaker system, it is stated as \(2000 \text{ Hz}\).
Frequency influences both the pitch and the intensity of sound. Higher frequencies correspond to higher-pitched sounds, and they may interact with the medium differently than lower frequencies.
Understanding frequency helps in studying the behavior of sound waves and their interaction with different environments. It influences the calculations related to displacement and pressure amplitudes, as it is intrinsic to the sound's physical properties.

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Most popular questions from this chapter

A well with vertical sides and water at the bottom resonates at \(7.00 \mathrm{~Hz}\) and at no lower frequency. The air-filled portion of the well acts as a tube with one closed end (at the bottom) and one open end (at the top). The air in the well has a density of \(1.10 \mathrm{~kg} / \mathrm{m}^{3}\) and a bulk modulus of \(1.33 \times 10^{5} \mathrm{~Pa}\). How far down in the well is the water surface?

Approximately a third of people with normal hearing have ears that continuously emit a low-intensity sound outward through the ear canal. A person with such spontaneous otoacoustic emission is rarely aware of the sound, except perhaps in a noisefree environment, but occasionally the emission is loud enough to be heard by someone else nearby. In one observation, the sound wave had a frequency of \(1665 \mathrm{~Hz}\) and a pressure amplitude of \(1.13 \times 10^{-3} \mathrm{~Pa}\). What were (a) the displacement amplitude and (b) the intensity of the wave emitted by the ear?

A girl is sitting near the open window of a train that is moving at a velocity of \(10.00 \mathrm{~m} / \mathrm{s}\) to the east. The girl's uncle stands near the tracks and watches the train move away. The locomotive whistle emits sound at frequency \(500.0 \mathrm{~Hz}\). The air is still. (a) What frequency does the uncle hear? (b) What frequency does the girl hear? A wind begins to blow from the east at \(10.00\) \(\mathrm{m} / \mathrm{s}\). (c) What frequency does the uncle now hear? (d) What frequency does the girl now hear?

Suppose that the sound level of a conversation is initially at an angry \(70 \mathrm{~dB}\) and then drops to a soothing \(50 \mathrm{~dB}\). Assuming that the frequency of the sound is \(500 \mathrm{~Hz}\), determine the (a) initial and (b) final sound intensities and the (c) initial and (d) final sound wave amplitudes.

A sinusoidal sound wave moves at \(343 \mathrm{~m} / \mathrm{s}\) through air in the positive direction of an \(x\) axis. At one instant during the oscillations, air molecule \(A\) is at its maximum displacement in the negative direction of the axis while air molecule \(B\) is at its equilibrium position. The separation between those molecules is \(15.0 \mathrm{~cm}\), and the molecules between \(A\) and \(B\) have intermediate displacements in the negative direction of the axis. (a) What is the frequency of the sound wave? In a similar arrangement but for a different sinusoidal sound wave, at one instant air molecule \(C\) is at its maximum displacement in the positive direction while molecule \(D\) is at its maximum displacement in the negative direction. The separation between the molecules is again \(15.0 \mathrm{~cm}\), and the molecules between \(C\) and \(D\) have intermediate displacements. (b) What is the frequency of the sound wave?
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