/*! 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 15 Multiple-Concept Example 3 revie... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 17: Problem 15

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 m wide. Sound of frequency 607 Hz is coming through the entrance from within the room. The speed of sound is 343 m/s. What is the diffraction angle 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
(a) Approx. 53.7掳; (b) Approx. 23.8掳.

Step by step solution

01

Understand the Phenomenon of Diffraction

Diffraction occurs when a wave encounters an obstacle or a slit. The wave bends around the edges of the obstacle or the slit. In this problem, sound waves are diffracting through a doorway, which acts as a single or double slit depending on how many doors are open.
02

Calculate the Wavelength of the Sound

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

Calculate the Diffraction Angle for One Open Door

When one door is open, the width \( a \) is 0.700 m. The diffraction angle \( \theta \) can be approximated by \[ \sin \theta \approx \frac{\lambda}{a} \]Substitute the values:\[ \sin \theta \approx \frac{0.565}{0.700} \approx 0.807 \]\( \theta \approx \arcsin(0.807) \approx 53.7^\circ \)
04

Calculate the Diffraction Angle for Both Doors Open

When both doors are open, the effective width \( a \) becomes the sum of the width of the two doors, which is 1.400 m. The diffraction angle \( \theta \) now is \[ \sin \theta \approx \frac{\lambda}{a} \]Substitute the values:\[ \sin \theta \approx \frac{0.565}{1.400} \approx 0.404 \]\( \theta \approx \arcsin(0.404) \approx 23.8^\circ \)

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.

Sound Waves
Sound waves are a type of mechanical wave that travels through a medium such as air, water, or solid materials. They are created when an object vibrates, causing the particles in the surrounding medium to vibrate as well. This vibration travels as a wave. Sound waves are longitudinal, meaning that the particle displacement is parallel to the direction of wave propagation. These waves can vary in frequency and amplitude, which affect the pitch and volume of the sound respectively.
  • The frequency of a sound wave determines its pitch. Higher frequencies equate to higher-pitched sounds.
  • The amplitude determines the loudness. Greater amplitude means a louder sound.
  • Sound waves can undergo diffraction, which is the bending of waves around obstacles or through openings.
Diffraction occurs prominently with sound waves because their wavelengths are often of a comparable size to everyday obstacles, such as doors.
Wavelength
Wavelength refers to the physical distance between successive crests of a wave. It is a crucial parameter in wave physics that determines how the wave will interact with its surroundings. The wavelength of a sound can be calculated using the formula:\[ \lambda = \frac{v}{f} \]where:
  • \( \lambda \) is the wavelength,
  • \( v \) is the speed of the wave (for sound in air, about 343 m/s),
  • \( f \) is the frequency of the wave.
In the problem where sound with a frequency of 607 Hz is considered, the wavelength calculates as approximately 0.565 m. This wavelength is instrumental in determining how pronounced the diffraction effects will be when it encounters an opening like a door.
Diffraction Angle
The diffraction angle is the angle at which a wave deviates from its original path as it passes through an aperture or past an obstacle. It gives insight into the extent of bending or spreading of waves. The formula to estimate the diffraction angle for a slit or opening is:\[\sin \theta \approx \frac{\lambda}{a}\]where:
  • \( \theta \) is the diffraction angle,
  • \( \lambda \) is the wavelength,
  • \( a \) is the width of the opening.
The larger the wavelength relative to the width of the opening, the larger the diffraction angle will be. In scenarios where sound waves pass through single or double doors, varying the door configuration affects the diffraction angle, demonstrating how wave behavior changes with structural alterations.
Single Slit
A single slit refers to a narrow opening through which a wave, such as sound, passes, leading to diffraction. The width of this slit plays a critical role in determining how much the wave spreads as it passes through. For a single door open, the door width forms the slit:
  • When a door with a width of 0.700 m is open, the sound waves diffract through this single slit.
  • The diffraction angle becomes larger with a smaller opening and is calculated using the approach described previously.
Understanding single slit diffraction is key to predicting how effectively waves can bend around obstacles or spread out when emerging from confined paths.
Double Slit
The double slit condition arises when two adjacent openings allow waves to pass through, a configuration that changes the diffraction behavior. Opening both doors:
  • The two door widths combine to form a single effective slit of 1.400 m.
  • This wider opening reduces the diffraction angle compared to the single slit.
  • The calculated diffraction angle is smaller, illustrating the principle that wider openings lead to less wave spreading.
The double slit concept is fundamental in optics and wave physics, providing a basis for understanding interference patterns and the impact of aperture width on wave behavior.

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

Two out-of-tune flutes play the same note. One produces a tone that has a frequency of 262 Hz, while the other produces 266 Hz. When a tuning fork is sounded together with the 262-Hz tone, a beat frequency of 1 Hz is produced. When the same tuning fork is sounded together with the 266-Hz tone, a beat frequency of 3 Hz is produced. What is the frequency of the tuning fork?

Divers working in underwater chambers at great depths must deal with the danger of nitrogen narcosis (the 鈥渂ends鈥�), 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鈥檚 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 \(20^{\circ} C\) . One is filled with air, in which the speed of sound is 343 \(m/s\) . The other is filled with helium, in which the speed of sound is \(1.00 \times 10^{3} m /s\) . To see the effect of helium on voice quality, calculate the ratio of the nth natural frequency of the helium-filled tube to the \(nth\) natural frequency of the air-filled tube.

Two wires, each of length 1.2 m, are stretched between two fixed supports. On wire A there is a second-harmonic standing wave whose frequency is 660 Hz. However, the same frequency of 660 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 m apart, and are each playing a 73.0-Hz tone. The speed of sound is 343 m/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?

The A string on a string bass vibrates at a fundamental frequency of 55.0 Hz. If the string鈥檚 tension were increased by a factor of four, what would be the new fundamental frequency?
See all solutions

Recommended explanations on Physics Textbooks

Space Physics

Read Explanation

Fundamentals of Physics

Read Explanation

Engineering Physics

Read Explanation

Famous Physicists

Read Explanation

Electricity

Read Explanation

Magnetism

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.