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Multiple-Concept Example 3 reviews the concepts that are important in this problem. The entrance to a large lecture room consists of two side-by-side doors, one hinged on the left and the other hinged on the right. Each door is \(0.700 \mathrm{~m}\) wide. Sound of frequency \(607 \mathrm{~Hz}\) is coming through the entrance from within the room. The speed of sound is 343 \(\mathrm{m} / \mathrm{s}\). What is the diffraction angle \(\theta\) of the sound after it passes through the doorway when (a) one door is open and (b) both doors are open?

Short Answer

Expert verified
For one open door, \(\theta \approx 54.7^\circ\); for both open doors, \(\theta \approx 23.8^\circ\).

Step by step solution

01

Understanding Diffraction

Diffraction occurs when a wave encounters an obstacle or a slit that is comparable in size to its wavelength. It results in the spreading of the wave.
02

Calculating Wavelength

The wavelength \( \lambda \) of sound can be calculated using the formula \( \lambda = \frac{v}{f} \), where \( v = 343 \, \text{m/s} \) is the speed of sound and \( f = 607 \, \text{Hz} \) is the frequency of sound. Substituting the values, \( \lambda = \frac{343}{607} \approx 0.565 \, \text{m} \).
03

Single Door Opening

When one door is open, the width \( a \) is \( 0.700 \, \text{m} \). The diffraction angle \( \theta \) can be found using the formula \( \sin \theta = \frac{\lambda}{a} \). Thus, \( \sin \theta = \frac{0.565}{0.700} \approx 0.807 \).
04

Calculate Diffraction Angle for One Door

Find \( \theta \) by taking the arcsin of the result: \( \theta = \sin^{-1}(0.807) \approx 54.7^\circ \).
05

Both Doors Open

When both doors are open, the width \( a \) is doubled, \( a = 2 \times 0.700 = 1.400 \, \text{m} \). Use the same formula \( \sin \theta = \frac{\lambda}{a} \). So \( \sin \theta = \frac{0.565}{1.400} \approx 0.404 \).
06

Calculate Diffraction Angle for Both Doors

Again find \( \theta \) by taking the arcsin: \( \theta = \sin^{-1}(0.404) \approx 23.8^\circ \).

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Key Concepts

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

Sound Wave Propagation
Sound wave propagation refers to the movement of sound waves through a medium. Unlike light waves, which can travel through a vacuum, sound requires a physical medium like air, water, or solids to travel.
The propagation of sound includes a series of compressions and rarefactions, which are essentially high and low-pressure zones moving through the medium.
This can be described as follows:
  • Compression: Where particles in the medium are close together.
  • Rarefaction: Where particles are spread apart.
Sound travels at different speeds depending on the medium. In air, at room temperature, the speed of sound is approximately 343 m/s. Temperature, humidity, and pressure can affect this speed. Sound wave propagation is crucial in understanding phenomena like diffraction, where waves bend around obstacles or spread when passing through openings.
Wave Physics
Wave physics involves the study of waves and their behavior. Waves are all around us and include not just sound, but also light and water waves.
A few key properties of waves are:
  • Amplitude: The height of a wave, which determines its loudness.
  • Frequency: Number of waves passing a point in a second, measured in Hertz (Hz). Higher frequencies produce higher-pitched sounds.
  • Wavelength: The distance between two successive compressions or rarefactions in a wave.
  • Speed: Determined by both the medium through which the wave is traveling and the wave's frequency and wavelength.
Understanding these properties is essential for solving problems related to diffraction and sound waves. Waves interact with obstacles (like doors) to produce effects such as reflection, refraction, and diffraction. Through these interactions, waves demonstrate their properties, adapting their propagation path or wavelength depending on the scenario.
Acoustics
Acoustics is the study of how sound behaves in different environments. This area of study is critical for designing spaces like auditoriums, concert halls, and lecture rooms, ensuring the sound reaches all areas uniformly.
In acoustics, several factors influence sound behavior, such as:
  • Reflection: When sound bounces off surfaces.
  • Absorption: When materials take in sound, reducing echo and reverberation.
  • Diffraction: Sound waves bending around obstacles, filling spaces even when direct "line of sight" is not possible.
Acoustic design uses these principles to enhance sound quality. In the example provided, understanding how sound diffracts through a doorway allows us to predict the sound coverage in a lecture room setting. It's all about ensuring clarity of communication, making sure every listener has a similar auditory experience.
Wavelength Calculation
Calculating the wavelength of a sound wave is essential to understanding its diffraction properties. Wavelength is calculated using the formula:\[\lambda = \frac{v}{f}\]where:
  • \(\lambda\) is the wavelength.
  • \(v\) is the speed of sound, which is 343 m/s in air at room temperature.
  • \(f\) is the frequency of the sound wave.
In the provided example, a sound frequency of 607 Hz was used to calculate the wavelength. Substituting the given values results in:\[\lambda = \frac{343}{607} \approx 0.565 \, \text{m}\]Knowing the wavelength helps in evaluating how sound will diffract when it encounters barriers or openings, such as doors. The wavelength also determines if the obstacle is comparable in size to the wavelength itself, which is crucial in predicting diffraction patterns and angles.

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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 high-pitched 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 \(n^{\text {th }}\) natural frequency of the helium- filled tube to the \(n^{\text {th }}\) natural frequency of the air-filled tube.

An organ pipe is open at both ends. It is producing sound at its third harmonic, the frequency of which is \(262 \mathrm{~Hz}\). The speed of sound is \(343 \mathrm{~m} / \mathrm{s}\). What is the length of the pipe?

Two tuning forks \(X\) and \(Y\) have different frequencies and produce an \(8-H z\) beat frequency when sounded together. When \(X\) is sounded along with a \(392-\mathrm{Hz}\) tone, a \(3-\mathrm{Hz}\) beat frequency is detected. When \(Y\) is sounded along with the \(392-\mathrm{Hz}\) tone, a \(5-\mathrm{Hz}\) beat frequency is heard. What are the frequencies \(f_{X}\) and \(f_{Y}\) when \((a) f_{X}\) is greater than \(f_{Y}\) and (b) \(f_{X}\) is less than \(f_{Y}\) ?

Standing waves are set up on two strings fixed at each end, as shown in the drawing. The tensions in the strings are the same, and each string has the same mass per unit length. However, one string is longer than the other, (a) Do the waves on the longer string have a larger speed, a smaller speed, or the same speed as those on the shorter string? Justify your answer. (b) Will the longer string vibrate at a higher frequency, a lower frequency, or the same frequency as the shorter string? Provide a reason for your answer. (c) Will the beat frequency produced by the two standing waves increase or decrease if the longer string is increased in length? Why? The two strings in the drawing have the same tension and mass per unit length, but they differ in length by \(0.57 \mathrm{~cm} .\) The waves on the shorter string propagate with a speed of \(41.8 \mathrm{~m} / \mathrm{s},\) and the fundamental frequency of the shorter string is \(225 \mathrm{~Hz}\). Determine the beat frequency produced by the two standing waves.

A tube is open only at one end. A certain harmonic produced by the tube has a frequency of \(450 \mathrm{~Hz}\). The next higher harmonic has a frequency of \(750 \mathrm{~Hz}\). The speed of sound in air is \(343 \mathrm{~m} / \mathrm{s}\). (a) What is the integer \(n\) that describes the harmonic whose frequency is \(450 \mathrm{~Hz} ?\) (b) What is the length of the tube?

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