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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 \((\mathrm{b})\) the displacement amplitude and (c) the pressure amplitude?

Short Answer

Expert verified
(a) 0.0392 mW/m虏; (b) 1.87 脳 10鈦烩伕 m; (c) 6.22 脳 10鈦宦� Pa.

Step by step solution

01

Calculate intensity at 30.0 m

We know that intensity \( I \) is inversely proportional to the square of the distance \( r \) from the source when the sound is emitted isotropically. Given the initial intensity, \( I_1 = 0.960 \, \mathrm{mW/m^2} \), at a distance of \( r_1 = 6.10 \, \mathrm{m} \), we use the formula \( \frac{I_1}{I_2} = \left(\frac{r_2}{r_1}\right)^2 \) to find the intensity \( I_2 \) at \( r_2 = 30.0 \, \mathrm{m} \). Substitute the values: \[ \frac{0.960}{I_2} = \left(\frac{30.0}{6.10}\right)^2 \]Solve for \( I_2 \): \[ I_2 = \frac{0.960}{\left(\frac{30.0}{6.10}\right)^2} \approx 0.0392 \, \mathrm{mW/m^2} \].
02

Calculate displacement amplitude at 6.10 m

The displacement amplitude \( s_{max} \) can be found using the formula: \[ I = \frac{1}{2} \rho v \omega^2 s_{max}^2 \]where \( \rho \) is the density of air (approximately \(1.21 \, \mathrm{kg/m^3}\)), \( v \) is the speed of sound (approximately \(343 \, \mathrm{m/s}\)), and \( \omega = 2 \pi f \) is the angular frequency with \( f = 2000 \, \mathrm{Hz} \). Rearrange to solve for \( s_{max} \):\[ s_{max} = \sqrt{\frac{2I}{\rho v \omega^2}} \]Substitute the values: \[ s_{max} = \sqrt{\frac{2 \times 0.960 \times 10^{-3}}{1.21 \times 343 \times (2 \pi \times 2000)^2}} \approx 1.87 \times 10^{-8} \, \mathrm{m} \].
03

Calculate pressure amplitude at 6.10 m

The pressure amplitude \( p_{max} \) can be found using the relationship between displacement and pressure amplitudes: \[ p_{max} = \rho v \omega s_{max} \]Substitute the previously calculated \( s_{max} \), and other known values:\[ p_{max} = 1.21 \times 343 \times 2 \pi \times 2000 \times 1.87 \times 10^{-8} \approx 6.22 \times 10^{-2} \, \mathrm{Pa} \].

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Key Concepts

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

Isotropic Sound Emission
Sound emitting isotropically is like a light bulb radiating light uniformly in all directions. An isotropic sound source is assumed to disperse sound waves evenly throughout space. For this concept, the energy transferred through the medium is evenly distributed over a spherical area that expands as the sound moves away from the source. Hence, the intensity of sound diminishes as the distance from the source increases, because the same amount of energy is spread over a larger area.
  • This can be visualized as the progressive weakening of sound as it travels through a medium鈥攎uch like the decreasing brightness of a light bulb as you walk away from it.
  • Sound radiates outward in an ever-expanding spherical shape, and the intensity is inversely related to the square of the distance from the source.
Displacement Amplitude
The displacement amplitude is about how much particles in the medium move back and forth as a sound wave passes through. It's the maximum change in position these particles experience.
It's important because it connects the movement of the medium with the energy carried by the sound wave. A higher displacement amplitude generally indicates a more energetic sound wave.
  • The larger the amplitude, the stronger the vibration. Thus, louder sounds have higher displacement amplitudes compared to softer sounds.
  • Calculating displacement amplitude can involve parameters like air density, the speed of sound, and frequency of the sound wave.
Pressure Amplitude
Pressure amplitude in sound waves is the maximum change in pressure compared to the ambient atmospheric pressure caused by the wave as it travels through a medium. The pressure amplitude is important because it reflects how intense a sound feels or how loud it appears.
  • As sound waves propagate, they cause rapid fluctuations in pressure. The pressure amplitude measures these changes.
  • Higher pressure amplitudes correspond to louder sounds because air particles are being compressed and rarified to greater extents.
To calculate pressure amplitude, one multiplies the density of the medium, the speed at which the wave travels through it, and the frequency of the wave with the displacement amplitude.
Inverse Square Law of Intensity
The inverse square law of intensity is a fundamental principle that describes how the intensity of sound diminishes with distance in isotropic sound emission. According to this law, the intensity of sound is inversely proportional to the square of the distance from the sound source.
This principle can be written mathematically as:
  • If you move twice as far from the sound source, the intensity drops to one-quarter.
  • In essence, the same energy is being spread over a four times larger area. Therefore, intensity decreases with the square of the increasing distance.
  • It's crucial because it helps predict sound levels at various distances from a sound source, which is very useful in many fields, such as acoustics and audio engineering.

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Most popular questions from this chapter

Two sound waves with an amplitude of \(12 \mathrm{nm}\) and a wave- length of \(35 \mathrm{~cm}\) travel in the same direction through a long tube, with a phase difference of \(\pi / 3\) rad. What are the (a) amplitude and (b) wavelength of the net sound wave produced by their interference? If, instead, the sound waves travel through the tube in opposite directions, what are the (c) amplitude and (d) wavelength of the net wave?

On July \(10,1996,\) a granite block broke away from a wall in Yosemite Valley and, as it began to slide down the wall, was launched into projectile motion. Seismic waves produced by its impact with the ground triggered seismographs as far away as \(200 \mathrm{~km} .\) Later measurements indicated that the block had a mass between \(7.3 \times 10^{7} \mathrm{~kg}\) and \(1.7 \times 10^{8} \mathrm{~kg}\) and that it landed \(500 \mathrm{~m}\) vertically below the launch point and \(30 \mathrm{~m}\) horizontally from it. (The launch angle is not known.) (a) Estimate the block's kinetic energy just before it landed. Consider two types of seismic waves that spread from the impact point - a hemispherical body wave traveled through the ground in an expanding hemisphere and a cylindrical surface wave traveled along the ground in an expanding shallow vertical cylinder (Fig. 17-49). Assume that the impact lasted \(0.50 \mathrm{~s}\), the vertical cylinder had a depth \(d\) of \(5.0 \mathrm{~m},\) and each wave type received \(20 \%\) of the energy the block had just before impact. Neglecting any mechanical energy loss the waves experienced as they traveled, determine the intensities of (b) the body wave and (c) the surface wave when they reached a seismograph \(200 \mathrm{~km}\) away. (d) On the basis of these results, which wave is more easily detected on a distant seismograph?

You are standing at a distance \(D\) from an isotropic point source of sound. You walk \(50.0 \mathrm{~m}\) toward the source and observe that the intensity of the sound has doubled. Calculate the distance \(D\).

A stationary motion detector sends sound waves of frequency \(0.150 \mathrm{MHz}\) toward a truck approaching at a speed of \(45.0 \mathrm{~m} / \mathrm{s}\). What is the frequency of the waves reflected back to the detector?

The pressure in a traveling sound wave is given by the equation $$ \Delta p=(1.50 \mathrm{~Pa}) \sin \pi\left[\left(0.900 \mathrm{~m}^{-1}\right) x-\left(315 \mathrm{~s}^{-1}\right) t\right] $$ Find the (a) pressure amplitude, (b) frequency, (c) wavelength, and (d) speed of the wave.
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