/*! 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 18 Sound (speed \(=343 \mathrm{m} /... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Sound (speed \(=343 \mathrm{m} / \mathrm{s})\) exits a diffraction horn loudspeaker through a rectangular opening like a small doorway. A person is sitting at an angle \(\alpha\) off to the side of a diffraction horn that has a width \(D\) of \(0.060 \mathrm{m} .\) This individual does not hear a sound wave that has a frequency of \(8100 \mathrm{Hz} .\) When she is sitting at an angle \(\alpha / 2,\) the frequency that she does not hear is different. What is this frequency?

Short Answer

Expert verified
The new frequency she does not hear is approximately 6482 Hz.

Step by step solution

01

Understand Diffraction Criteria

The problem is about sound wave diffraction, which implies using the formula \( D \sin(\alpha) = m \lambda \) for maximum or minimum diffraction. Here, \( \lambda \) is the wavelength, \( D \) the width of the opening, \( \alpha \) the angle, and \( m \) is an integer related to the diffraction order.
02

Calculate Wavelength for Initial Frequency

The speed of sound \( v \) is given as 343 m/s, and the initial frequency is 8100 Hz. Calculate the wavelength using the formula \( \lambda = \frac{v}{f} = \frac{343}{8100} \approx 0.0423457 \mathrm{m} \).
03

Use Wavelength to Find Angle Condition

Using the diffraction condition for no sound \( D \sin(\alpha) = m \lambda \), substitute the known values: \( 0.060 \sin(\alpha) = m \times 0.0423457 \). If we assume the first diffraction condition (first zero), then \( m = 1 \). Calculate \( \sin(\alpha) = \frac{m \lambda}{D} = \frac{0.0423457}{0.060} \approx 0.705762 \).
04

Calculate New Frequency Condition at \( \alpha/2 \)

For \( \alpha/2 \), the diffraction condition becomes \( D \sin(\alpha/2) = m' \lambda' \). Using the previously found \( \sin(\alpha) \), find \( \sin(\alpha/2) = \sqrt{\frac{1+\sin(\alpha)}{2}} = \sqrt{\frac{1+0.705762}{2}} \approx 0.881917 \).
05

Solve for New Frequency

Substitute into the diffraction condition for the new frequency: \( 0.060 \times 0.881917 = \lambda' \). Since \( \lambda' = \frac{v}{f'} = \frac{343}{f'} \), solve for \( f' \): \( f' \approx \frac{343}{0.052915} \approx 6482.39 \mathrm{Hz} \).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Diffraction Horn
A diffraction horn is a special type of loudspeaker designed to control how sound spreads. This control is achieved through a specific shape and structure, similar to a small doorway, which helps guide the sound waves.
The horn affects the spread or 'diffraction' of sound due to its physical dimensions, particularly its width. This width, denoted as \( D \), interacts with the sound waves, causing them to bend around corners and spread out.
For the diffraction to be significant, the wavelength of the sound should be comparable to this width. This design is crucial for aiming sound within a specific space or for projecting it at a particular distance.
Frequency
Frequency refers to how many times a sound wave oscillates in one second, measured in Hertz (Hz). A higher frequency means a higher pitch. The problem at hand involves an initial sound frequency of 8100 Hz. Under this condition, specific waves can't be heard at particular angles, indicating points of diffraction where the sound wave phase changes or cancels out.
The difference in frequency noticed when changing the listening angle to \( \alpha/2 \) demonstrates how frequency is crucial in determining how and where the sound becomes inaudible due to diffraction effects. Adjusting frequency alters the wavelength, affecting where standing waves occur and thus where sound might not be perceived.
Wavelength
Wavelength, denoted \( \lambda \), is the distance over which a sound wave's shape repeats. It is intimately tied to the speed of sound and its frequency through the equation \( \lambda = \frac{v}{f} \), where \( v \) is the speed of sound in air (343 m/s in this exercise).
For a sound with a frequency of 8100 Hz, the wavelength is calculated as approximately 0.04235 meters. This measurement is essential because it determines the effectiveness of diffraction given the opening width \( D \). If the wavelength is larger or of the same order as the horn width, significant diffraction occurs, affecting what sounds are perceivable.
At the \( \alpha/2 \) condition, a new wavelength corresponding to a frequency of approximately 6482 Hz occurs, showing the interplay between frequency and what the observer experiences.
Angle Condition
The angle \( \alpha \) represents the position relative to the diffraction horn's line of sound travel where sound is either heard or absent. Mathematically, it ties into the diffraction condition \( D \sin(\alpha) = m \lambda \). Here, \( m \) represents diffraction order, typically starting at 1 for the first occurrence of a 'zero' or minimal sound condition.
By calculating the sine of different angle setups (\( \alpha \) or \( \alpha/2 \)), we can predict the conditions under which the sound wave phases interfere destructively, resulting in minimized or inaudible sound.
In practice, this concept helps in adjusting sound systems to ensure optimal sound delivery within a defined space, optimizing angles for best acoustic outcomes.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

One method for measuring the speed of sound uses standing waves. A cylindrical tube is open at both ends, and one end admits sound from a tuning fork. A movable plunger is inserted into the other end at a distance \(L\) from the end of the tube where the tuning fork is. For a fixed frequency, the plunger is moved until the smallest value of \(L\) is measured that allows a standing wave to be formed. Suppose that the tuning fork produces a \(485-\mathrm{Hz}\) tone, and that the smallest value observed for \(L\) is \(0.264 \mathrm{m} .\) What is the speed of sound in the gas in the tube?

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 pianos each sound the same note simultaneously, but they are both out of tune. On a day when the speed of sound is \(343 \mathrm{m} / \mathrm{s},\) piano A produces a wavelength of \(0.769 \mathrm{m},\) while piano \(\mathrm{B}\) produces a wavelength of \(0.776 \mathrm{m}\). How much time separates successive beats?

Two waves are traveling in opposite directions on the same string. The displacements caused by the individual waves are given by \(y_{1}=(24.0 \mathrm{mm}) \sin (9.00 \pi t-1.25 \pi x)\) and \(y_{2}=(35.0 \mathrm{mm}) \sin (2.88 \pi t+\) \(0.400 \pi x\) ). Note that the phase angles \((9.00 \pi t-1.25 \pi x)\) and \((2.88 \pi t+\) \(0.400 \pi x\) ) are in radians, \(t\) is in seconds, and \(x\) is in meters. At \(t=4.00 \mathrm{s}\) what is the net displacement (in \(\mathrm{mm}\) ) of the string at (a) \(x=2.16 \mathrm{m}\) and (b) \(x=2.56 \mathrm{m} ?\) Be sure to include the algebraic sign \((+\) or \(-)\) with your answers.

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?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.