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Chapter 17: Problem 2

Two speakers, one directly behind the other, are each generating a 245-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

Understand Destructive Interference Conditions

Destructive interference occurs when two identical waves are out of phase by half a wavelength (or an odd multiple of half wavelengths). This means for destructive interference, the path difference must be \( \frac{\lambda}{2}, \frac{3\lambda}{2}, \frac{5\lambda}{2}, ... \)
02

Calculate the Wavelength

Use the formula \( \lambda = \frac{v}{f} \) to find the wavelength, where \( v \) is the speed of sound (343 m/s) and \( f \) is the frequency (245 Hz). \[ \lambda = \frac{343 \, \text{m/s}}{245 \, \text{Hz}} \approx 1.4 \, \text{m} \]
03

Find the Smallest Separation for Destructive Interference

To find the smallest separation distance for destructive interference, use the condition that the path difference should be \( \frac{\lambda}{2} \). With \( \lambda = 1.4 \), the smallest path difference is: \[ \frac{\lambda}{2} = \frac{1.4}{2} = 0.7 \, \text{m} \] This is the smallest separation distance at which destructive interference occurs.

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Key Concepts

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

Sound Wave Frequency
Sound waves are waves of pressure that travel through air. Their characteristics can be described by properties like frequency and wavelength. Frequency refers to how many times the wave oscillates per second and is measured in Hertz (Hz). In this exercise, the sound wave frequency is 245 Hz. This means the wave undergoes 245 cycles every second.
Understanding frequency is crucial because it is directly related to both the pitch of the sound we hear and the wavelength of the wave.
It's also a key factor in determining the phenomena of interference, where waves can amplify or cancel each other out based on their phase and frequency.
Wavelength Calculation
To calculate the wavelength of a sound wave, use the formula \[\lambda = \frac{v}{f}\]where \( \lambda \) is the wavelength, \( v \) is the speed of sound, and \( f \) is the frequency. In this case, the speed of sound is given as 343 m/s, and the frequency of the wave is 245 Hz. Plugging these values in, you get:
  • \( \lambda = \frac{343 \, \text{m/s}}{245 \, \text{Hz}} \approx 1.4 \, \text{m} \)
This calculation tells us that one full wavelength of the 245 Hz sound wave is approximately 1.4 meters long. Knowing the wavelength allows us to predict how sound will behave in different conditions, such as when it meets another wave.
Wave Phase Difference
Phase difference refers to the relative offset of two waves that are otherwise similar in frequency and wavelength. When waves meet, they can interfere with each other, either destructively or constructively. Destructive interference occurs when the waves are out of phase by half a wavelength, which effectively cancels them out. This happens when the path difference is an odd multiple of half wavelengths:
  • \( \frac{\lambda}{2}, \frac{3\lambda}{2}, \frac{5\lambda}{2}, \ldots \)

For this problem, to achieve destructive interference, the smallest separation distance is found by computing \( \frac{\lambda}{2} \). With a wavelength of 1.4 m, the smallest path difference for destructive interference is \[\frac{1.4}{2} = 0.7 \, \text{m}\]This distance represents the closest spacing at which the sound waves will be completely out of phase, canceling each other at the listener's position.

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Most popular questions from this chapter

Sound (speed \(=343 \mathrm{m} / \mathrm{s})\) exits a diffraction horn loudspeaker through a rectangular opening like a small doorway. A person is sitting at an angle \(\alpha\) off to the side of a diffraction horn that has a width \(D\) of \(0.060 \mathrm{m} .\) This individual does not hear a sound wave that has a frequency of \(8100 \mathrm{Hz} .\) When she is sitting at an angle \(\alpha / 2,\) the frequency that she does not hear is different. What is this frequency?

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 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 \(343 \mathrm{m} / \mathrm{s}\)

Multiple-Concept Example 4 deals with the same concepts as this problem. A 41-cm length of wire has a mass of 6.0 g. It is stretched between two fixed supports and is under a tension of \(160 \mathrm{N}\). What is the fundamental frequency of this wire?

A tube with a cap on one end, but open at the other end, has a fundamental frequency of \(130.8 \mathrm{Hz}\). The speed of sound is \(343 \mathrm{m} / \mathrm{s}\) (a) If the cap is removed, what is the new fundamental frequency of the tube? (b) How long is the tube?
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