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Chapter 17: Problem 2

Two speakers, one directly behind the other, are each generating a \(245-\mathrm{Hz}\) sound wave. What is the smallest separation distance between the speakers that will produce destructive interference at a listener standing in front of them? The speed of sound is \(343\) \(\mathrm{m} / \mathrm{s}\).

Short Answer

Expert verified
The smallest separation distance is 0.7 meters.

Step by step solution

01

Understanding Destructive Interference

Destructive interference occurs when waves are out of phase by an odd multiple of half wavelengths, specifically \( (2n + 1)\frac{\lambda}{2} \), where \( n \) is an integer. For the smallest separation, we take \( n = 0 \).
02

Calculate Wavelength

First, calculate the wavelength \( \lambda \) of the sound wave using the formula \( \lambda = \frac{v}{f} \), where \( v = 343 \ \mathrm{m/s} \) is the speed of sound, and \( f = 245 \ \mathrm{Hz} \) is the frequency. So, \( \lambda = \frac{343}{245} \approx 1.4 \ \mathrm{m} \).
03

Find the Smallest Separation for Destructive Interference

Using the formula for destructive interference distance \( d = (2n + 1)\frac{\lambda}{2} \) with \( n = 0 \), we get \( d = \frac{1.4}{2} = 0.7 \ \mathrm{m} \).

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Key Concepts

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

Wave Phase
Understanding the phase of a wave is crucial when exploring concepts like interference. Wave phase refers to the position of a point in time on a waveform cycle. Two key terms often associated with wave phase are "in phase" and "out of phase." When two sound waves are "in phase," their peaks and troughs align perfectly, reinforcing each other. "Out of phase," however, describes waves that have peaks aligning with the troughs of another wave, leading to interference.

Destructive interference occurs when two waves are perfectly out of phase. In this scenario, the wave amplitudes cancel each other out, reducing the overall sound. This situation is used in the problem to determine the separation distance between speakers that causes this cancelation. Understanding this alignment is crucial for both theoretical and practical applications in acoustics and physics.
Wavelength Calculation
In physics, calculating the wavelength of a wave is an essential step in analyzing wave behavior. The wavelength (\( \lambda \)) can be calculated using the formula:
  • \( \lambda = \frac{v}{f} \)
where \( v \) is the speed of sound, and \( f \) is the frequency of the wave.

For the given exercise, the speed of sound was \(343 \ \mathrm{m/s} \), and the frequency was \(245 \ \mathrm{Hz} \). By plugging these values into the formula, we find that the wavelength is \( \lambda = \frac{343}{245} \approx 1.4 \ \mathrm{m} \).

This calculation is crucial because it directly influences how sound waves interact, such as determining the distances required for interference patterns. Accurately calculating wavelengths allows us to predict and understand phenomena like destructive interference.
Sound Wave Interference
Sound wave interference relates to how sound waves interact with each other when they meet. Specifically, destructive interference, which is central to the exercise, occurs when waves meet and cancel each other out, creating areas of silence or lower amplitude sound.

To produce destructive interference, waves must arrive at the same point in opposite phase. This usually happens when the path difference between the two waves is an odd multiple of half wavelengths:
  • \((2n + 1)\frac{\lambda}{2}\)
where \( n \) is an integer.

In the given problem, the smallest separation distance necessary to create destructive interference was calculated using \( n = 0 \), which simplified the equation to \( d = \frac{\lambda}{2} \). Therefore, with a wavelength of approximately \(1.4 \ \mathrm{m} \), the smallest distance needed was \(0.7 \ \mathrm{m} \). Understanding sound wave interference allows for precise control over sound environments, critical in fields like acoustics, audio technology, and engineering.

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Most popular questions from this chapter

An organ pipe is open at both ends. It is producing sound at its third harmonic, the frequency of which is \(262\) \(\mathrm{Hz}\). The speed of sound is \(343\) \(\mathrm{m} / \mathrm{s}\). What is the length of the pipe?

One method for measuring the speed of sound uses standing waves. A cylindrical tube is open at both ends, and one end admits sound from a tuning fork. A movable plunger is inserted into the other end at a distance \(L\) from the end of the tube where the tuning fork is. For a fixed frequency, the plunger is moved until the smallest value of \(L\) is measured that allows a standing wave to be formed. Suppose that the tuning fork produces a \(485-\mathrm{Hz}\) tone, and that the smallest value observed for \(L\) is \(0.264 \mathrm{m} .\) What is the speed of sound in the gas in the tube?

Review Conceptual Example 5 before attempting this problem. As the drawing shows, the length of a guitar string is \(0.628\) \(\mathrm{m} .\) The frets are numbered for convenience. A performer can play a musical scale on a single string because the spacing between the frets is designed according to the following rule: When the string is pushed against any fret \(j,\) the fundamental frequency of the shortened string is larger by a factor of the twelfth root of two \((\sqrt[12]{2})\) than it is when the string is pushed against the fret \(j-1 .\) Assuming that the tension in the string is the same for any note, find the spacing (a) between fret 1 and fret 0 and (b) between fret 7 and fret \(6 .\)

The fundamental frequencies of two air columns are the same. Column A is open at both ends, while column B is open at only one end. The length of column A is 0.70 m. What is the length of column B?

To review the concepts that play roles in this problem, consult Multiple- Concept Example 4. Sometimes, when the wind blows across a long wire, a low- frequency "moaning" sound is produced. This sound arises because a standing wave is set up on the wire, like a standing wave on a guitar string. Assume that a wire (linear density \(=0.0140\) \(\mathrm{kg} / \mathrm{m}\) ) sustains a tension of 323 N because the wire is stretched between two poles that are \(7.60\) \(\mathrm{m}\) apart. The lowest frequency that an average, healthy human ear can detect is 20.0 Hz. What is the lowest harmonic number \(n\) that could be responsible for the "moaning" sound?
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