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Two loudspeakers, \(A\) and \(B,\) are driven by the same amplifier and emit sinusoidal waves in phase. The frequency of the waves emitted by each speaker is 860 \(\mathrm{Hz}\) Point \(P\) is 12.0 \(\mathrm{m}\) from \(A\) and 13.4 \(\mathrm{m}\) from \(B .\) Is the interference at \(P\) constructive or destructive? Give the reasoning behind your answer.

Short Answer

Expert verified
The interference at point \( P \) is destructive because the path difference is not an integer multiple of the wavelength.

Step by step solution

01

Determine the wavelength

First, calculate the wavelength of the sound waves using the formula \( \lambda = \frac{v}{f} \), where \( v \) is the speed of sound in air (approximately 343 m/s) and \( f \) is the frequency (860 Hz). Thus, \( \lambda = \frac{343 \, \text{m/s}}{860 \, \text{Hz}} \approx 0.398 \text{ m} \).
02

Calculate the path difference

Find the difference in distances from point \( P \) to each speaker: \( \, d_B - d_A = 13.4 \, \text{m} - 12.0 \, \text{m} = 1.4 \, \text{m} \).
03

Compare path difference to wavelength

Assess whether the path difference (1.4 m) is an integer multiple of the wavelength (0.398 m). Calculate: \( \, \frac{1.4 \, \text{m}}{0.398 \text{ m}} \approx 3.52 \). This value is not an integer.
04

Determine interference type

The path difference is not an integer multiple of the wavelength. Constructive interference occurs when the path difference is a whole number multiple of the wavelength. Since the calculated ratio is not an integer, the interference at point \( P \) is destructive.

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Key Concepts

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

Constructive Interference
When two sound waves meet, they can combine in a way that increases the overall amplitude of the wave. This phenomenon is known as constructive interference. It occurs when the path difference between waves, reaching a point like P in the problem, is an integer multiple of the wavelength of the waves. In simpler terms, if the waves are aligned such that their peaks and troughs coincide perfectly, they amplify each other.

This can be mathematically expressed as the path difference \( \Delta d = n\lambda \), where \( n \) is an integer (0, 1, 2, 3,...). When this condition is met, it results in a louder sound at the point of intersection because the sound waves are perfectly in phase. Imagine two people clapping in perfect synchrony; the resulting sound is much louder.

For constructive interference to occur in the example exercise, the path difference would need to be a whole number multiple of the wavelength, which was calculated as about 0.398 m. Since it was not, constructive interference did not occur.
Destructive Interference
Destructive interference takes place when waves meet in such a way that they cancel each other out, reducing the overall amplitude of the sound. This happens when the waves are out of phase, or their peaks align with the other's troughs.

Mathematically, destructive interference occurs when the path difference is a half-integer multiple of the wavelength, expressed as \( \Delta d = (n + 0.5)\lambda \), where \( n \) is an integer. This means the sound waves are 180 degrees out of phase.

Using the exercise's example, the path difference was found to be 1.4 m. When divided by the wavelength (approximately 0.398 m), the result was not a whole number. Instead, this leads to a partial overlap that thoroughly describes destructive interference. The waves meet at point P and diminish each other, meaning the sound at point P is quieter.
Sound Waves
Sound waves are vibrations that travel through a medium like air or water. They are often depicted as sinusoidal in nature, which means they have peaks (high points) and troughs (low points) like a smooth, repeating wave.

These waves have important properties including frequency, measured in Hertz (Hz), and wavelength, which is the distance between consecutive peaks. In this exercise, sound waves were emitted at a frequency of 860 Hz, common for audible human hearing, and a calculated wavelength of approximately 0.398 m.

The speed of sound in air is a critical factor. Generally estimated to be around 343 m/s in dry air at 20 degrees Celsius, it varies with temperature and humidity. By understanding these properties, we can analyze interference patterns, much like the situation with speakers A and B in the example. As these mechanical waves travel, their interference can either amplify or cancel the sound based on their phase relationship at any given point.
Path Difference
The term "path difference" refers to the difference in distance that two waves travel to reach the same point. In acoustic scenarios, like the one given, it's the extra distance sound from one source travels compared to another before they interact.

The path difference is crucial in determining the type of interference—constructive or destructive—that occurs at a point. In the given problem, the path difference was calculated as 1.4 m, the contrast between distances from speakers A and B to point P. Whether the path difference aligns with a whole or half-integer multiple of the wavelength defines the nature of the interference.

To visualize, imagine two swimmers racing from the point they start to a common finishing point but along paths of slightly different lengths. If they reach the finish simultaneously due to precise timing, they exemplify constructive interference. However, if one swimmer finishes before the other, they metaphorically represent destructive interference as their efforts do not combine effectively at the finish line. This analogy helps in understanding how path difference impacts sound wave interference.

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Most popular questions from this chapter

On the planet Arrakis a male ornithoid is flying toward his mate at 25.0 \(\mathrm{m} / \mathrm{s}\) while singing at a frequency of 1200 \(\mathrm{Hz}\) . If the stationary female hears a tone of 1240 \(\mathrm{Hz}\) , what is the speed of sound in the atmosphere of Arrakis?

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Example 16.1 (Section 16.1\()\) showed that for sound waves in air with frequency 1000 \(\mathrm{Hz}\) , a displacement amplitude of\( 1.2 \times 10^{-8} \mathrm{m}\) produces a pressure amplitude of \(3.0 \times 10^{-2} \mathrm{Pa}\) . Water at \(20^{\circ} \mathrm{C}\) has a bulk modulus of \(2.2 \times 10^{9} \mathrm{Pa}\) , and the speed of sound in water at this temperature is 1480 \(\mathrm{m} / \mathrm{s}\) . For \(1000-\mathrm{Hz}\) sound waves in \(20^{\circ} \mathrm{C}\) water, what displacement amplitude is produced if the pressure amplitude is \(3.0 \times 10^{-2}\) Pa? Explain why your answer is much less than \(1.2 \times 10^{-8} \mathrm{Pa}\) .

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