/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 16 The following two lists give the... [FREE SOLUTION] | 91影视

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Chapter 17: Problem 16

The following two lists give the diameters and sound frequencies for three loudspeakers. Pair each diameter with a frequency, so that the diffraction angle is the same for each of the speakers, and then find the common diffraction angle. Take the speed of sound to be \(343 \mathrm{m} / \mathrm{s}\). $$\begin{array}{c}\hline \text { Diameter, } D \\\\\hline 0.050 \mathrm{m} \\\0.10 \mathrm{m} \\\0.15 \mathrm{m} \\\\\hline \end{array}$$ $$ \begin{array}{c} \hline \text { Frequency, } f \\ \hline 6.0 \mathrm{kHz} \\ 4.0 \mathrm{kHz} \\ 12.0 \mathrm{kHz} \\ \hline \end{array} $$

Short Answer

Expert verified
Diameters 0.050 m, 0.10 m, 0.15 m pair with frequencies 12.0 kHz, 6.0 kHz, 4.0 kHz for a common diffraction angle of approximately 34.87掳.

Step by step solution

01

Understand the problem

We need to match each loudspeaker diameter with a frequency such that they all have the same diffraction angle. Then, find the common diffraction angle. The formula for the diffraction angle \( \theta \) is given by \( \sin \theta = \frac{\lambda}{D} \), where \( \lambda \) is the wavelength, and \( \lambda = \frac{v}{f} \) with \( v = 343 \, \text{m/s} \).
02

Calculate Wavelengths

Calculate the wavelength \( \lambda \) for each frequency using the formula \( \lambda = \frac{v}{f} \).For 6.0 kHz: \( \lambda = \frac{343}{6000} = 0.05717 \, \text{m} \).For 4.0 kHz: \( \lambda = \frac{343}{4000} = 0.08575 \, \text{m} \).For 12.0 kHz: \( \lambda = \frac{343}{12000} = 0.02858 \, \text{m} \).
03

Apply the Same Diffraction Angle Formula

Use \( \sin \theta = \frac{\lambda}{D} \) for each pair of \( (D, \lambda) \). For each diameter, calculate \( \sin \theta \) with every calculated wavelength and find pairs with the same value.
04

Match Diameters with Frequencies

Evaluate each diameter-frequency pair:- For \( D = 0.050 \, \text{m} \): - \( \sin \theta = \frac{0.05717}{0.050} = 1.1434 \) (not possible, must be <= 1) - \( \sin \theta = \frac{0.08575}{0.050} = 1.7150 \) (not possible) - \( \sin \theta = \frac{0.02858}{0.050} = 0.5716 \). - For \( D = 0.10 \, \text{m} \): - \( \sin \theta = \frac{0.05717}{0.10} = 0.5717 \). - \( \sin \theta = \frac{0.08575}{0.10} = 0.8575 \) (not possible, choose closest possible for matching pairs) - \( \sin \theta = \frac{0.02858}{0.10} = 0.2858 \). - For \( D = 0.15 \, \text{m} \): - \( \sin \theta = \frac{0.05717}{0.15} = 0.3811 \). - \( \sin \theta = \frac{0.08575}{0.15} = 0.5717 \). - \( \sin \theta = \frac{0.02858}{0.15} = 0.1905 \).Pairs with closest diffraction angle: \( (D=0.050, f=12.0 \, \text{kHz}), (D=0.10, f=6.0 \, \text{kHz}), (D=0.15, f=4.0 \, \text{kHz}) \)
05

Calculate Common Diffraction Angle

With the selected pairs, calculate the common diffraction angle. Use \( \sin \theta = 0.5716 \), which is common to our matched pairs.Calculate \( \theta \) using \( \theta = \arcsin(0.5716) \). Compute to find \( \theta \approx 34.87^\circ \).

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Key Concepts

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

Wavelength Calculation
To understand wavelength calculation, we first need to explore its relation to sound frequencies and the speed of sound. The wavelength of a sound wave is the distance between successive crests of the wave. It is calculated using the formula \( \lambda = \frac{v}{f} \), where \( \lambda \) is the wavelength, \( v \) is the speed of sound, and \( f \) is the frequency of the sound wave. For instance, if we have a frequency of 6 kHz (or 6000 Hz), and the speed of sound is given as 343 m/s, the wavelength can be calculated as follows:
\[ \lambda = \frac{343 \, \text{m/s}}{6000 \, \text{Hz}} = 0.05717 \, \text{m} \]
This calculation provides the wavelength, which is crucial for understanding how sound waves travel and interact with different objects, like loudspeakers.
Sound Frequency
Sound frequency plays a pivotal role in the behavior of sound waves. Frequency refers to the number of sound wave cycles per second, measured in hertz (Hz). High-frequency sounds, such as 12 kHz, have shorter wavelengths compared to lower frequencies like 4 kHz. This difference impacts how sound is perceived and how it interacts with environments. For example, higher frequencies tend to be more easily absorbed by materials and may not travel as far as lower frequencies.
  • Low Frequency: Ex. 4 kHz results in a longer wavelength \( \lambda = 0.08575 \, \text{m} \).
  • Mid Frequency: Ex. 6 kHz results in \( \lambda = 0.05717 \, \text{m} \).
  • High Frequency: Ex. 12 kHz results in \( \lambda = 0.02858 \, \text{m} \).
The choice of frequency affects the diffraction angle, which is significant in designing and utilizing loudspeakers for optimal sound distribution.
Loudspeaker Physics
Exploring loudspeaker physics, particularly the relationship between the diameter of the loudspeaker and the sound frequency, helps us understand sound behavior. The diameter of a loudspeaker affects the diffraction of sound waves. Diffraction refers to the bending of waves around the edges of an obstacle or opening.
In this context, the diffraction angle \( \theta \) is determined by the formula \( \sin \theta = \frac{\lambda}{D} \), where \( D \) is the diameter of the loudspeaker. This formula highlights how various diameters and wavelengths result in different diffraction angles. For a given wavelength, a larger diameter will produce a smaller diffraction angle, allowing sound to travel further and maintain directionality.
  • For \( D = 0.050 \, \text{m} \) and \( f = 12.0 \, \text{kHz} \), \( \sin \theta = 0.5716 \).
  • For \( D = 0.10 \, \text{m} \) and \( f = 6.0 \, \text{kHz} \), \( \sin \theta = 0.5716 \).
  • For \( D = 0.15 \, \text{m} \) and \( f = 4.0 \, \text{kHz} \), \( \sin \theta = 0.5717 \).
This understanding is essential for creating sound systems that optimize performance and audience experience, particularly in venues where sound coverage is crucial.

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Most popular questions from this chapter

Divers working in underwater chambers at great depths must deal with the danger of nitrogen narcosis (the "bends"), in which nitrogen dissolves into the blood at toxic levels. One way to avoid this danger is for divers to breathe a mixture containing only helium and oxygen. Helium, however, has the effect of giving the voice a highpitched quality, like that of Donald Duck's voice. To see why this occurs, assume for simplicity that the voice is generated by the vocal cords vibrating above a gas-filled cylindrical tube that is open only at one end. The quality of the voice depends on the harmonic frequencies generated by the tube; larger frequencies lead to higher-pitched voices. Consider two such tubes at \(20^{\circ} \mathrm{C}\). One is filled with air, in which the speed of sound is \(343 \mathrm{m} / \mathrm{s} .\) The other is filled with helium, in which the speed of sound is \(1.00 \times 10^{3} \mathrm{m} / \mathrm{s} .\) To see the effect of helium on voice quality, calculate the ratio of the \(n\) th natural frequency of the helium-filled tube to the \(n\) th natural frequency of the air-filled tube.

Two ultrasonic sound waves combine and form a beat frequency that is in the range of human hearing for a healthy young person. The frequency of one of the ultrasonic waves is \(70 \mathrm{kHz}\). What are (a) the smallest possible and (b) the largest possible value for the frequency of the other ultrasonic wave?

A person hums into the top of a well and finds that standing waves are established at frequencies of \(42,70.0,\) and \(98 \mathrm{Hz}\). The frequency of \(42 \mathrm{Hz}\) is not necessarily the fundamental frequency. The speed of sound is \(343 \mathrm{m} / \mathrm{s}\). How deep is the well?

Two wires, each of length \(1.2 \mathrm{m},\) are stretched between two fixed supports. On wire \(A\) there is a second-harmonic standing wave whose frequency is \(660 \mathrm{Hz}\). However, the same frequency of \(660 \mathrm{Hz}\) is the third harmonic on wire B. Find the speed at which the individual waves travel on each wire.

Speakers A and B are vibrating in phase. They are directly facing each other, are \(7.80 \mathrm{m}\) apart, and are each playing a \(73.0-\mathrm{Hz}\) tone. The speed of sound is \(343 \mathrm{m} / \mathrm{s}\). On the line between the speakers there are three points where constructive interference occurs. What are the distances of these three points from speaker A?
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