/*! 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 12 Sound exits a diffraction horn l... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 17: Problem 12

Sound exits a diffraction horn loudspeaker through a rectangular opening like a small doorway. Such a loudspeaker is mounted outside on a pole. In winter, when the temperature is \(273 \mathrm{~K}\), the diffraction angle \(\theta\) has a value of \(15.0^{\circ} .\) What is the diffraction angle for the same sound on a summer day when the temperature is \(311 \mathrm{~K}\) ?

Short Answer

Expert verified
The diffraction angle on a summer day at 311 K is approximately 13.56掳.

Step by step solution

01

Understand the Scenario

We need to compute the change in the diffraction angle of sound as the temperature changes. According to the problem, the sound exits a loudspeaker through a rectangular opening like a small doorway. The diffraction angle is the angle at which sound spreads out as it exits this opening.
02

Consider Diffraction of Sound

The diffraction angle (\( \theta \)) can be evaluated using the formula for diffraction, which relates temperature to sound velocity. The formula is derived from the relationship between the speed of sound and temperature, namely: \[ v = \sqrt{ \frac{\gamma \cdot R \cdot T}{M} } \] where \( v \) is the speed of sound, \( T \) is the absolute temperature, \( \gamma \) is the adiabatic index, \( R \) is the ideal gas constant, and \( M \) is the molar mass of air.
03

Proportionality Between Temperature and Velocity

Since temperature affects the velocity of sound, we know: \[ v_{273} \propto \sqrt{273} \] and \[ v_{311} \propto \sqrt{311} \]. That is, the velocity of sound is proportional to the square root of the absolute temperature. Since the diffraction angle depends on the inverse of velocity, we have \( \theta_{311} \propto \frac{1}{v_{311}} \) and \( \theta_{273} \propto \frac{1}{v_{273}} \).
04

Calculate the Change in Diffraction Angle

Use the inversely proportional relationship between diffraction angle and sound speed: \[ \frac{\theta_{311}}{\theta_{273}} = \frac{v_{273}}{v_{311}} \]. Simplifying this expression using the proportionality to the square root of temperature, we have: \[ \theta_{311} = \theta_{273} \times \sqrt{\frac{273}{311}} \]. Substituting \( \theta_{273} = 15.0^{\circ} \), we find \( \theta_{311} \approx 15.0^{\circ} \times \sqrt{\frac{273}{311}} \).
05

Compute Final Answer

Calculate: \[ \theta_{311} \approx 15.0 \times \sqrt{\frac{273}{311}} \approx 13.56^{\circ} \]. Therefore, on a summer day at a temperature of 311 K, the diffraction angle is approximately 13.56 degrees.

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 91影视!

Key Concepts

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

Diffraction Angle
The diffraction angle is a fascinating concept in the field of wave physics. When you speak through a megaphone or hear sound from a loudspeaker, you're witnessing diffraction. Diffraction is the bending and spreading of waves when they pass through an opening or obstacle. For sound, a diffraction horn speaker uses a rectangular opening to project sound efficiently. The diffraction angle, denoted by \(\theta\), measures how much the sound wave spreads out as it exits the speaker.The diffraction angle is dependent on various factors such as the wavelength of the sound and the size of the opening. If the opening is large compared to the wavelength, the angle is small. In contrast, if the opening is small, the angle increases. This concept explains why sound can be heard around corners or obstacles鈥攊t's due to diffraction!Understanding diffraction helps in designing better sound systems, allowing us to have clear and directed sound in venues like concert halls.
Temperature Effect on Sound
Sound doesn't travel in a vacuum; it needs a medium like air. And, an interesting thing happens with sound in different temperatures! The speed of sound waves is influenced by the temperature of the medium. As temperature increases, molecules move faster, making it easier for sound waves to travel through the air. As a result, warmer air increases the speed of sound. On a colder day, you might notice sounds are not as loud or clear from afar. That's because the speed of sound is slower due to less energetic molecule movement. Consequently, the diffraction angle changes with temperature because speed affects how tightly or loosely the sound wave fronts are spread out as they exit the speaker. It's all about the dance and energy of air molecules, directing sound in varying ways depending on the warmth of the day!
Speed of Sound
The speed of sound is a measure of how quickly sound waves travel through a medium, like air. It's expressed in meters per second (m/s). One of the fascinating aspects of sound speed is its dependence on temperature. The speed of sound in air is calculated using the formula: \[ v = \sqrt{ \frac{\gamma \cdot R \cdot T}{M} } \]Here, \(v\) is the speed of sound, \(T\) is the absolute temperature in Kelvin, \(\gamma\) is the adiabatic index (which describes how air molecules exchange heat), \(R\) is the ideal gas constant, and \(M\) is the molar mass of air.Temperature plays a crucial role here. As you plug in a higher temperature value into the formula, the speed of sound increases. This stems from the molecual physics where warmer molecules transfer sound vibrations more efficiently.Knowing the speed of sound is essential for applications like acoustics, meteorology, and even in designing musical instruments. It's a fundamental characteristic that defines how we experience sound in our daily lives.

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

The approach to solving this problem is similar to that taken in Multiple- Concept Example 4 . On a cello, the string with the largest linear density \(\left(1.56 \times 10^{-2} \mathrm{~kg} / \mathrm{m}\right)\) is the C string. This string produces a fundamental frequency of \(65.4 \mathrm{~Hz}\) and has a length of \(0.800 \mathrm{~m}\) between the two fixed ends. Find the tension in the string.

Suppose the strings on a violin are stretched with the same tension and each has the same length between its two fixed ends. The musical notes and corresponding fundamental frequencies of two of these strings are \(\mathrm{G}(196.0 \mathrm{~Hz})\) and \(\mathrm{E}(659.3 \mathrm{~Hz}) .\) The linear density of the E string is \(3.47 \times 10^{-4} \mathrm{~kg} / \mathrm{m}\). What is the linear density of the \(\mathrm{G}\) string?

A copper block is suspended in air from a wire in Part 1 of the drawing. A container of mercury is then raised up around the block as in Part \(2 .\) (a) The fundamental frequency of the wire is given by Equation 17.3 with \(n=1: f_{1}=v /(2 L)\). How is the speed \(v\) at which individual waves travel on the wire related to the tension in the wire? (b) Is the tension in the wire in Part 2 less than, greater than, or equal to the tension in Part \(1 ?\) (c) Is the fundamental frequency of the wire in Part 2 less than, greater than, or equal to the fundamental frequency in Part \(1 ?\) Justify each of your answers. In Part 2 of the drawing some \((50.0 \%)\) of the block's volume is submerged in the mercury. The density of copper is \(8890 \mathrm{~kg} / \mathrm{m}^{3},\) and the density of mercury is 13600 \(\mathrm{kg} / \mathrm{m}^{3}\). Find the ratio of the fundamental frequency of the wire in Part 2 to the fundamental frequency in Part \(1 .\) Check to see that your answer is consistent with your answers to the Concept Questions.

A string of length \(0.28 \mathrm{~m}\) is fixed at both ends. The string is plucked and a standing wave is set up that is vibrating at its second harmonic. The traveling waves that make up the standing wave have a speed of \(140 \mathrm{~m} / \mathrm{s}\). What is the frequency of vibration?

The note that is three octaves above middle \(C\) is supposed to have a fundamental frequency of \(2093 \mathrm{~Hz}\). On a certain piano the steel wire that produces this note has a cross-sectional area of \(7.85 \times 10^{-7} \mathrm{~m}^{2}\). The wire is stretched between two pegs. When the piano is tuned properly to produce the correct frequency at \(25.0{ }^{\circ} \mathrm{C},\) the wire is under a tension of \(818.0 \mathrm{~N}\). Suppose the temperature drops to \(20.0{ }^{\circ} \mathrm{C} .\) In addition, as an approximation, assume that the wire is kept from contracting as the temperature drops. Consequently, the tension in the wire changes. What beat frequency is produced when this piano and another instrument (properly tuned) sound the note simultaneously?
See all solutions

Recommended explanations on Physics Textbooks

Waves Physics

Read Explanation

Geometrical and Physical Optics

Read Explanation

Measurements

Read Explanation

Radiation

Read Explanation

Magnetism and Electromagnetic Induction

Read Explanation

Physical Quantities And Units

Read Explanation
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.