/*! 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] | 91Ó°ÊÓ

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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
Diffraction angle is approximately \(53.62^\circ\) for one door and \(23.84^\circ\) for both doors open.

Step by step solution

01

Calculate Wavelength of Sound

The first step is to determine the wavelength of the sound using the formula \( \lambda = \frac{v}{f} \), where \( v = 343 \, \text{m/s} \) (speed of sound) and \( f = 607 \, \text{Hz} \) (frequency). Calculating, we get:\[\lambda = \frac{343}{607} \approx 0.565 \text{ m}\]
02

Diffraction with One Door Open

When one door is open, the width \( a \) of the opening is \( 0.700 \, \text{m} \). The diffraction angle \( \theta \) can be found using the formula \( \sin\theta = \frac{\lambda}{a} \). Substituting the values, we have:\[\sin\theta = \frac{0.565}{0.700} \approx 0.807\]Solving for \( \theta \):\[\theta \approx \arcsin(0.807) \approx 53.62^\circ\]
03

Diffraction with Both Doors Open

When both doors are open, the width of the doorway double, so \( a = 2 \times 0.700 = 1.400 \, \text{m} \). Again, use \( \sin\theta = \frac{\lambda}{a} \):\[\sin\theta = \frac{0.565}{1.400} \approx 0.404\]Solving for \( \theta \):\[\theta \approx \arcsin(0.404) \approx 23.84^\circ\]

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Key Concepts

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

Wavelength Calculation
To understand diffraction, we first need to calculate the wavelength of the sound. The wavelength determines how the sound waves will spread as they pass through an opening, such as a doorway. Here, frequency and speed of sound are key. Given the speed of the sound is 343 meters per second, and the frequency is 607 Hz, we use the formula: \[ \lambda = \frac{v}{f} \]where \( \lambda \) is the wavelength, \( v \) is the speed of sound, and \( f \) is the frequency.- **Example Calculation:** - Substitute the known values into the formula: \[\lambda = \frac{343}{607} \approx 0.565 \text{ m}\]This calculation shows the sound wavelength to be approximately 0.565 meters. Recognizing that wavelength is the distance between consecutive peaks of a wave is crucial for understanding how sound interacts with openings of different sizes.
Diffraction Angle
Diffraction is a phenomenon that occurs when waves bend around obstacles or through openings. It's particularly noticeable when the size of the opening is comparable to the wavelength of the sound. The diffraction angle gives us an idea of how much the sound wave spreads after passing through an opening.When a single door (0.700 m wide) is open, we can calculate the diffraction angle \( \theta \) using the equation:\[ \sin\theta = \frac{\lambda}{a} \]where \( a \) is the width of the open door. For one door open:- **Substitute the Values:** \[\sin\theta = \frac{0.565}{0.700} \approx 0.807\] - Solve for \( \theta \): \[\theta \approx \arcsin(0.807) \approx 53.62^\circ\]For both doors open (1.400 m wide), using the same formula:- **Substitute the Values:** \[\sin\theta = \frac{0.565}{1.400} \approx 0.404\] - Solve for \( \theta \): \[\theta \approx \arcsin(0.404) \approx 23.84^\circ\]With both doors open, the diffraction angle decreases significantly, indicating less spreading of the sound.
Physics Problem Solving
Understanding physics problems, like calculating diffraction, involves grasping the core principles at play. This exercise showcases the application of wave concepts in a real-world scenario.1. **Identifying Known Values:** Begin by identifying the given information and crucial constants, such as the speed of sound and frequency.
2. **Formulating the Problem:** Use basic and relevant equations like the wavelength formula \( \lambda = \frac{v}{f} \) and the diffraction angle formula \( \sin\theta = \frac{\lambda}{a} \).
3. **Substitution and Calculation:** Substitute the known values into formulas systematically, ensuring that units match and calculations are correct.
4. **Interpreting Results:** Analyze the resulting values to understand their physical meaning in the context of the problem, checking if the results are sensible given the scenario.Physics problem solving encourages critical thinking and logical progression from basic principles to specific solutions. Each step in the calculation process should build confidence in applying formulas to similar problems.

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Most popular questions from this chapter

A pipe open only at one end has a fundamental frequency of 256 Hz. A second pipe, initially identical to the first pipe, is shortened by cutting off a portion of the open end. Now, when both pipes vibrate at their fundamental frequencies, a beat frequency of \(12 \mathrm{Hz}\) is heard. How many centimeters were cut off the end of the second pipe? The speed of sound is \(343 \mathrm{m} / \mathrm{s}\)

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?

A tube, open at only one end, is cut into two shorter (nonequal) lengths. The piece that is open at both ends has a fundamental frequency of \(425 \mathrm{Hz},\) while the piece open only at one end has a fundamental frequency of \(675 \mathrm{Hz}\). What is the fundamental frequency of the original tube?

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.

A vertical tube is closed at one end and open to air at the other end. The air pressure is \(1.01 \times 10^{5} \mathrm{Pa}\). The tube has a length of \(0.75 \mathrm{m} .\) Mercury (mass density \(\left.=13600 \mathrm{kg} / \mathrm{m}^{3}\right)\) is poured into it to shorten the effective length for standing waves. What is the absolute pressure at the bottom of the mercury column, when the fundamental frequency of the shortened, air-filled tube is equal to the third harmonic of the original tube?

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