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Chapter 35: Problem 2

Two speakers that are 15.0 \(\mathrm{m}\) apart produce in-phase sound waves of frequency 250.0 \(\mathrm{Hz}\) in a room where the speed of sound is 340.0 \(\mathrm{m} / \mathrm{s} .\) A woman starts out at the midpoint between the two speakers. The room's walls and ceiling are covered with absorbers to eliminate reflections, and she listens with only one ear for best precision. (a) What does she hear: constructive or destructive interference? Why? (b) She now walks slowly toward one of the speakers. How far from the center must she walk before she first hears the sound reach a minimum intensity? (c) How far from the center must she walk before she first hears the sound maximally enhanced?

Short Answer

Expert verified
(a) Constructive interference; (b) 0.34 m; (c) 0.68 m.

Step by step solution

01

Determine the Wavelength

First, calculate the wavelength \( \lambda \) of the sound waves using the formula \( \lambda = \frac{v}{f} \), where \( v = 340.0 \, \mathrm{m/s} \) is the speed of sound and \( f = 250.0 \, \mathrm{Hz} \) is the frequency of the sound waves. \( \lambda = \frac{340.0}{250.0} = 1.36 \, \mathrm{m} \).
02

Analyze Starting Position for Interference

At the midpoint, the difference in path lengths from both speakers to the woman is zero, making the path difference \( \Delta L = 0 \). For constructive interference, \( \Delta L = m\lambda \) (where \( m \) is an integer). For \( m = 0 \), there is constructive interference at the midpoint, so she hears maximum intensity.
03

Find the Position for First Minimum Intensity

For destructive interference, the path difference should satisfy \( \Delta L = (m + \frac{1}{2})\lambda \). Starting at the midpoint, the woman moves towards one speaker. Set \( m = 0 \) and find the distance \( d \) from the center where \( \Delta L = \frac{1}{2}\lambda \). The speakers are 15.0 m apart, so the path difference is \( 2d = \frac{1}{2}\times1.36 = 0.68 \, \mathrm{m} \). Solving gives \( d = 0.34 \, \mathrm{m} \).
04

Find the Position for Maximum Enhanced Sound

To find the position for the next maximum (constructive interference), use \( \Delta L = m\lambda \). Let \( m = 1 \) for the first maxima after the midpoint. Solve \( 2d = 1.36 \, \mathrm{m} \), giving \( d = 0.68 \, \mathrm{m} \).

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Key Concepts

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

Constructive Interference
Constructive interference occurs when two or more sound waves overlap in such a way that they reinforce each other, leading to an increase in sound intensity. This happens when the waves are in phase, meaning the peaks and troughs of the waves align perfectly. In simpler terms, constructive interference makes the sound louder.

For the woman at the midpoint between the two speakers, the path difference is zero. Since both sound waves travel the same distance to reach her, they arrive in phase. This perfect alignment results in constructive interference.

When sound waves have a path difference of multiples of their wavelength ( \( \Delta L = m \lambda \), where \( m \) is an integer), constructive interference occurs. At this overlap, the interference maximizes sound intensity. So, when the woman stands at the midpoint, she experiences the sound at its maximum volume due to constructive interference.
Destructive Interference
Destructive interference, on the other hand, is when sound waves meet in such a way that they cancel each other out. This leads to a decrease in sound intensity or even silence in certain conditions. It occurs when the sound waves are out of phase, where one wave's peak meets another wave's trough.

For the woman walking towards one of the speakers, she wants the path to create a condition for destructive interference. This happens when the path difference equals half of the wavelength plus an integer multiple of the wavelength ( \( \Delta L = (m + \frac{1}{2}) \lambda \)). At this point, the waves do not align properly, leading to reduced sound intensity.

In the exercise, setting \( m = 0 \) and creating a path difference of \( \frac{1}{2} \lambda \) achieves the destructive interference. Solving for distance unveils the woman needs to move 0.34 m from the center to hear the first sound minimum. This movement ensures the sound waves clash against each other, lowering the audible sound.
Path Difference
Path difference is key to understanding sound wave interference. It refers to the difference in the distance that two sound waves travel to reach a common point. This difference determines whether the interference will be constructive or destructive. By managing the path difference, one can control sound intensity and quality.

At the midpoint, the path difference is zero, meaning both waves travel identical distances. This leads to constructive interference because the waves are perfectly in phase. But as the woman moves towards one speaker, path difference plays its role. When it's at half a wavelength, the waves become out of phase, leading to destructive interference.

The calculations show that for destructive interference, the path difference must be \( (m + \frac{1}{2}) \lambda \). For constructive interference following the midpoint, the path difference should equal the whole wavelength, like \( m \lambda \). Sounds occurring from slight differences in path lengths can create unique interference patterns, affecting how we perceive sound.

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Most popular questions from this chapter

Light with wavelength 648 \(\mathrm{nm}\) in air is incident perpendicularly from air on a film 8.76\(\mu \mathrm{m}\) thick and with refractive index \(1.35 .\) Part of the light is reflected from the first surface of the film, and part enters the film and is reflected back at the second surface, where the film is again in contact with air. (a) How many waves are contained along the path of this second part of the light in its round trip through the film? (b) What is the phase difference between these two parts of the light as they leave the film?

Radio Interference. Two radio antennas \(A\) and \(B\) radiate in phase. Antenna \(B\) is 120 \(\mathrm{m}\) to the right of antenna \(A .\) Consider point \(Q\) along the extension of the line connecting the antennas, a horizontal distance of 40 \(\mathrm{m}\) to the right of antenna \(B\) . The frequency, and hence the wavelength, of the emitted waves can be varied. (a) What is the longest wavelength for which there will be destructive interference at point \(Q ?\) (b) What is the longest wave-length for which there will be constructive interference at point \(Q ?\)

A source \(S\) of monochromatic light and a detector \(D\) are both located in air a distance \(h\) above a horizontal plane sheet of glass and are separated by a horizontal distance \(x .\) Waves reaching \(D\) directly from \(S\) interfere with waves that reflect off the glass. The distance \(x\) is small compared to \(h\) so that the reflection is at close to normal incidence. (a) Show that the condition for constructive interference is \(\sqrt{x^{2}+4 h^{2}}-x=\left(m+\frac{1}{2}\right) \lambda,\) and the condition for destructive interference is \(\sqrt{x^{2}+4 h^{2}}-x=m \lambda\) (Hint: Take into account the phase change on reflection.) (b) Let \(h=24 \mathrm{cm}\) and \(x=14 \mathrm{cm} .\) What is the longest wavelength for which there will be constructive interference?

What is the thinnest soap film (excluding the case of zero thickness) that appears black when illuminated with light with wavelength 480 \(\mathrm{nm} ?\) The index of refraction of the film is \(1.33,\) and there is air on both sides of the film.

A uniform thin film of material of refractive index 1.40 coats a glass plate of refractive index \(1.55 .\) This film has the proper thickness to cancel normally incident light of wavelength 525 nm that strikes the film surface from air, but it is somewhat greater than the minimum thickness to achieve this cancellation. As time goes by, the film wears away at a steady rate of 4.20 \(\mathrm{nm}\) per year. What is the minimum number of years before the reflected light of this wavelength is now enhanced instead of cancelled?
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