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Chapter 26: Problem 48

Two identical audio speakers connected to the same amplifier produce in-phase sound waves with a single frequency that can be varied between 300 and \(600 \mathrm{~Hz}\). The speed of sound is \(340 \mathrm{~m} / \mathrm{s}\). You find that where you are standing, you hear minimum-intensity sound. (a) Explain why you hear minimum-intensity sound. (b) If one of the speakers is moved \(39.8 \mathrm{~cm}\) toward you, the sound you hear has maximum intensity. What is the frequency of the sound? (c) How much closer to you from the position in part (b) must the speaker be moved to the next position where you hear maximum intensity?

Short Answer

Expert verified
(a) Due to destructive interference. (b) Frequency is 426 Hz. (c) Move an additional 0.399 m.

Step by step solution

01

Understanding Minimum Intensity

Minimum intensity in sound waves occurs when there is destructive interference between the waves produced by the two speakers. This happens when the path difference between the waves from the two speakers is an odd multiple of half wavelengths: \( \Delta L = (n + \frac{1}{2}) \lambda \), where \( n \) is an integer and \( \lambda \) is the wavelength.
02

Maximum Intensity Criterion

Maximum intensity occurs due to constructive interference, where the path difference between the waves from the speakers is a whole number multiple of the wavelength: \( \Delta L = m \lambda \), where \( m \) is an integer. This information is crucial for determining the conditions necessary for maximum sound intensity after moving one of the speakers.
03

Determine Frequency (b)

When one speaker is moved closer by \( 39.8 \text{ cm} \) resulting in maximum intensity, the path difference satisfies \( \Delta L = m \lambda \). Given \( v = 340 \text{ m/s} \), the wavelength is \( \lambda = \frac{v}{f} \). If \( \Delta L = 39.8 \times 10^{-2} \text{ m} = \lambda \), solve for \( f \): \[ f = \frac{v}{\Delta L} = \frac{340}{0.398} \approx 854 \text{ Hz} \]. Since this frequency is not possible given the range, we consider harmonics (e.g., \( n \text{ times} \) fundamental frequency) to find a suitable frequency within the range, like \( f \approx 426 \text{ Hz} \).
04

Next Maximum Intensity (c)

To find out how much closer to move the speaker to the next maximum intensity, consider that the path difference must now be \( (m+1) \lambda \). Therefore, move closer by an additional \( \lambda = \frac{340}{426} \approx 0.798 \text{ m} \). From part (b)'s position, the additional movement should be half a wavelength to maintain constructive interference.

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Key Concepts

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

Destructive Interference
When you hear a minimum-intensity sound from two speakers that are in phase, it means you are experiencing destructive interference. This happens when the sound waves from the speakers meet at a point where their peaks align with the troughs of the other wave, canceling each other out. In technical terms, this occurs when the path difference between the two waves is an odd multiple of half the wavelength, or \( \Delta L = (n + \frac{1}{2}) \lambda \) where \( n \) is an integer and \( \lambda \) is the wavelength.

Imagine the waves as ripples in a pond鈥攊f two sets of ripples meet perfectly out of sync, they can flatten each other out, leading to quietness despite the sources being active.
Constructive Interference
Constructive interference is when you hear sound at maximum intensity, a result of two sound waves meeting in such a way that their peaks and troughs align perfectly, amplifying the sound. This usually happens when the path difference between waves is an integral multiple of the wavelength, expressed as \( \Delta L = m \lambda \) where \( m \) is an integer.

This scenario can be likened to two people pushing a swing at the same time, thereby making it go higher. In terms of sound, when the path difference hits whole wavelength multiples, each crest or trough aligns, increasing the sound's volume.
Wavelength
Wavelength, denoted as \( \lambda \), is essentially the length of one complete wave cycle, from crest to crest or trough to trough. It is a crucial factor in understanding sound wave interference. The formula \( \lambda = \frac{v}{f} \) is used to calculate it, where \( v \) is the speed of sound and \( f \) is the frequency. For the given speakers, \( v \) is 340 m/s and frequencies range between 300 to 600 Hz.

Knowing the wavelength helps you determine how far apart the two points of constructive or destructive interference occur, and is vital for calculating the shifts in sound intensity. Like measuring the distance between buoy tops bobbing in a lake, the wavelength shows how "wide" each wave cycle is.
Frequency Range
In the context of sound, the frequency range refers to the span of frequencies that can be produced or heard in a given setting, measured in Hertz (Hz). For the loudspeakers in our exercise, the frequency can vary between 300 and 600 Hz. The frequency affects the pitch of the sound produced and, alongside wavelength, influences the interference patterns observed.

Low-frequency sounds have longer wavelengths, while high-frequency sounds have shorter ones. This means changes in frequencies result in different interference patterns, altering where and how strongly you might experience constructive or destructive interference. Therefore, understanding the frequency range is crucial for predicting and calculating the conditions for maximum or minimum sound intensity.

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Most popular questions from this chapter

Light of wavelength \(631 \mathrm{nm}\) passes through a diffraction grating having 485 lines \(/ \mathrm{mm}\). (a) What is the total number of bright spots (indicating complete constructive interference) that will occur on a large distant screen? Solve this problem without finding the angles. (Hint: What is the largest that \(\sin \theta\) can be? What does this imply for the largest value of \(m ?\) (b) What is the angle of the bright spot farthest from the center?

Red light of wavelength \(633 \mathrm{nm}\) from a helium-neon laser passes through a slit \(0.350 \mathrm{~mm}\) wide. The diffraction pattern is observed on a screen \(3.00 \mathrm{~m}\) away. Define the width of a bright fringe as the distance between the minima on either side. (a) What is the width of the central bright fringe? (b) What is the width of the first bright fringe on either side of the central one?

A plate of glass \(9.00 \mathrm{~cm}\) long is placed in contact with a second plate and is held at a small angle with it by a metal strip \(0.0800 \mathrm{~mm}\) thick placed under one end. The space between the plates is filled with air. The glass is illuminated from above with light having a wavelength in air of \(656 \mathrm{nm}\). How many interference fringes are observed per centimeter in the reflected light?

Light of wavelength \(585 \mathrm{nm}\) falls on a slit \(0.0666 \mathrm{~mm}\) wide. (a) On a very large distant screen, how many totally dark fringes (indicating complete cancellation) will there be, including both sides of the central bright spot? Solve this problem without calculating all the angles! (Hint: What is the largest that \(\sin \theta\) can be? What does this tell you is the largest that \(m\) can be?) (b) At what angle will the dark fringe that is most distant from the central bright fringe occur?

A laser beam of wavelength \(\lambda=632.8 \mathrm{nm}\) shines at normal incidence on the reflective side of a compact disk. (a) The tracks of tiny pits in which information is coded onto the \(\mathrm{CD}\) are \(1.60 \mu \mathrm{m}\) apart. For what angles of reflection (measured from the normal) will the intensity of light be maximum? (b) On a DVD, the tracks are only \(0.740 \mu \mathrm{m}\) apart. Repeat the calculation of part (a) for the DVD.
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