Problem 10 - Physics

Question

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?

Step-by-Step Solution

Answer

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

Step 1: 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.

Step 2: 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, \( \lambda = \frac{343}{607} \approx 0.565 \, \text{m} \).

Step 3: 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, \( \sin \theta = \frac{0.565}{0.700} \approx 0.807 \).

Step 4: Calculate Diffraction Angle for One Door

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

Step 5: 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 \( \sin \theta = \frac{0.565}{1.400} \approx 0.404 \).

Step 6: Calculate Diffraction Angle for Both Doors

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

Key Concepts

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.