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Chapter 35: Problem 49

Two speakers \(A\) and \(B\) are 3.50 \(\mathrm{m}\) apart, and each one is emitting a frequency of 444 \(\mathrm{Hz}\) . However, because of signal delays in the cables, speaker \(A\) is one-fourth of a period ahead of speaker \(B\) . For points far from the speakers, find all the angles relative to the centerline (Fig. \(\mathrm{P} 35.49\) ) at which the sound from these speakers cancels. Include angles on both sides of the centerline. The speed of sound is 340 \(\mathrm{m} / \mathrm{s}\) .

Short Answer

Expert verified
Angles for destructive interference occur at \( \sin^{-1}(\pm 0.109 \times m) \) for odd \( m \).

Step by step solution

01

Calculate Wavelength

The wavelength \( \lambda \) of the sound can be found using the formula \( \lambda = \frac{v}{f} \), where \( v \) is the speed of sound and \( f \) is the frequency. Substituting the values given, \( v = 340 \, \text{m/s} \) and \( f = 444 \, \text{Hz} \), we get: \[ \lambda = \frac{340}{444} \approx 0.766 \text{ m} \]
02

Calculate Phase Difference

Speaker \( A \) is ahead of speaker \( B \) by one-fourth of a period, translating into a phase difference of \( \phi = \frac{1}{4}(2\pi) = \frac{\pi}{2} \).
03

Conditions for Destructive Interference

For destructive interference (sound cancellation) between the two speakers, the condition \( \Delta \phi + k(2\pi) = n\pi \) must be met, where \( \Delta \phi \) is the phase difference and \( n \) is an odd integer. Substituting \( \Delta \phi = \frac{\pi}{2} \) into the condition, we get: \( \frac{\pi}{2} + k(2\pi) = (2m+1)\pi \), which simplifies to: \[ k = \frac{2m+1}{4} \] where \( m \) is an integer.
04

Calculate Path Difference for Destructive Interference

Use the relationship for path difference in terms of angles: \( d \sin(\theta) = (k - \frac{1}{4})\lambda \). Substituting \( k = \frac{2m+1}{4} \) into the equation, the equation becomes: \[ d \sin(\theta) = \left( \frac{2m+1}{4} - \frac{1}{4} \right) \lambda = \frac{m \lambda}{2} \] with \( d = 3.50 \, \text{m} \).
05

Solve for Angles Relative to the Centerline

Solving for \( \theta \): \[ \sin(\theta) = \frac{m \lambda}{2d} = \frac{m}{2} \times \frac{0.766}{3.50} \approx \frac{0.109m}{2} \] For it to be valid, \( |\sin(\theta)| \leq 1 \), which implies \( m = \pm1, \pm3, \pm5, \ldots \). Thus, calculate \( \theta \) for these \( m \) values:

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Key Concepts

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

Phase Difference
When two wave sources, like speakers, produce sound at slightly different times, there is a difference in the phase of the waves they generate. This phase difference influences how sound waves interact with one another. In the original exercise, speaker A is a quarter of a period ahead of speaker B. This means there is a phase difference of \( \Delta \phi = \frac{\pi}{2} \). The phase difference is critical to determining how the waves will combine, as it determines whether they will enhance each other (constructive interference) or cancel each other out (destructive interference). Understanding the phase difference helps predict sound wave behavior, especially when the paths of the waves differ.
Destructive Interference
Destructive interference occurs when two sound waves combine to produce a wave of reduced or zero amplitude. This happens when the phase difference between the waves leads to a perfectly out-of-phase alignment, causing them to cancel each other out.In this exercise, we achieve destructive interference by setting the phase condition \( \Delta \phi + k(2\pi) = n\pi \), where \( n \) is an odd integer. This expression ensures that the sound waves generated by the two speakers meet in such a way that their peaks align with the troughs of the other, resulting in sound cancellation. This principle is particularly useful in applications requiring noise reduction, like in noise-canceling headphones.
Path Difference
The path difference is the physical difference in distance that two waves travel to reach the same point. This plays a vital role in determining interference patterns. For destructive interference, this path difference should satisfy certain conditions derived from the wave properties and their angles of travel.In the context of the given problem, the path difference involves spacing \( d \) between the speakers and angle \( \theta \) relative to the centerline. The equation \( d\sin(\theta) = \frac{m\lambda}{2} \) relates the path difference to the wavelength and ensures destructive interference for specific angles. This equation helps identify positions where the amplitudes of waves from both speakers will destructively interfere and cancel each other.
Wave Interference
Wave interference is the phenomenon where two or more waves superpose to form a resultant wave. This effect can be observed when waves overlap as they travel through a medium like air. There are two primary types of interference: constructive and destructive. This interplay between waves' paths and phases occurs naturally in many contexts, such as light and water waves, but is particularly significant in sound waves as studied in this problem. By understanding wave interference, one can manipulate sound to enhance or cancel specific frequencies, aiding in sound design and audio engineering.
Sound Waves
Sound waves are mechanical waves that require a medium to travel through, such as air, water, or solids. They are longitudinal waves, consisting of compressions and rarefactions moving through the medium.The frequency of the sound waves is key to sound perception鈥攁ffecting pitch. In this exercise, the frequency of 444 Hz falls within the audible range for humans and dictates the wavelength through the relation \( \lambda = \frac{v}{f} \).Understanding sound waves involves considering not only their speed and frequency but how they interact with each other, as explored through concepts like phase and path difference. This comprehension is crucial for many technological applications, including acoustics, speech recognition, and broadcast systems.

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Most popular questions from this chapter

One round face of a 3.25 -m, solid, cylindrical plastic pipe is covered with a thin black coating that completely blocks light. The opposite face is covered with a fluorescent coating that glows when it is struck by light. Two straight, thin, parallel scratches, 0.225 \(\mathrm{mm}\) apart, are made in the center of the black face. When laser light of wavelength 632.8 \(\mathrm{nm}\) shines through the slits perpendicular to the black face, you find that the central bright fringe on the opposite face is 5.82 \(\mathrm{mm}\) wide, measured between the dark fringes that border it on either side. What is the index of refraction of the plastic?

Suppose you illuminate two thin slits by monochromatic coherent light in air and find that they produce their first interference minima at \(\pm 35.20^{\circ}\) on either side of the central bright spot. You then immerse these slits in a transparent liquid and illuminate them with the same light. Now you find that the first minima occur at \(\pm 19.46^{\circ}\) instead. What is the index of refraction of this liquid?

The index of refraction of a glass rod is 1.48 at \(T=20.0^{\circ} \mathrm{C}\) and varies linearly with temperature, with a coefficient of \(2.50 \times 10^{-5} / \mathrm{C}^{\circ} .\) The coefficient of linear expansion of the glass is \(5.00 \times 10^{-6} / \mathrm{C}^{\circ} .\) At \(20.0^{\circ} \mathrm{C}\) the length of the rod is 3.00 \(\mathrm{cm} . \mathrm{A}\) Michelson interferometer has this glass rod in one arm, and the rod is being heated so that its temperature increases at a rate of 5.00 \(\mathrm{C}^{\circ} / \mathrm{min}\) . The light source has wavelength \(\lambda=589 \mathrm{nm},\) and the rod initially is at \(T=20.0^{\circ} \mathrm{C}\) . How many fringes cross the field of view each minute?

Sensitive Eyes. After an eye examination, you put some eyedrops on your sensitive eyes. The cornea (the front part of the eye) has an index of refraction of \(1.38,\) while the eyedrops have a refractive index of \(1.45 .\) After you put in the drops, your friends notice that your eyes look red, because red light of wavelength 600 nm has been reinforced in the reflected light. (a) What is the minimum thickness of the film of eyedrops on your cornea? (b) Will any other wavelengths of visible light be reinforced in the reflected light? Will any be cancelled? (c) Suppose you had contact lenses, so that the eyedrops went on them instead of on your corneas. If the refractive index of the lens material is 1.50 and the layer of eyedrops has the same thickness as in part (a), what wavelengths of visible light will be reinforced? What wavelengths will be cancelled?

Reflective Coatings and Herring. Herring and related fish have a brilliant silvery appearance that camouflages them while they are swimming in a sunlit ocean. The silveriness is due to platelets attached to the surfaces of these fish. Each platelet is made up of several alternating layers of crystalline guanine \((n=1.80)\) and of cytoplasm \((n=1.333,\) the same as water), with a guanine layer on the outside in contact with the surrounding water (Fig. \(\mathrm{P} 35.56\) ). In one typical platelet, the guanine layers are 74 nm thick and the cytoplasm layers are 100 \(\mathrm{nm}\) thick. (a) For light striking the platelet surface at normal incidence, for which vacuum wavelengths of visible light will all of the reflections \(R_{1}\) , \(R_{2}, R_{3}, R_{4},\) and \(R_{5},\) shown in Fig. P35.56, be approximately in phase? If white light is shone on this platelet, what color will be most strongly reflected (see Fig. 32.4\() ?\) The surface of a herring has very many platelets side by side with layers of different thickness, so that all visible wavelengths are reflected. (b) Explain why such a "stack" of layers is more reflective than a single layer of guanine with cytoplasm underneath. (A stack of five guanine layers separated by cytoplasm layers reflects more than 80\(\%\) of incident light at the wavelength for which it is "tuned.") (c) The color that is most strongly reflected from a platelet depends on the angle at which it is viewed. Explain why this should be so. (You can see these changes in color by examining a herring from different angles. Most of the platelets on these fish are oriented in the same way, so that they are vertical when the fish is swimming.)
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