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Chapter 16: Problem 37

Two loudspeakers, \(A\) and \(B,\) are driven by the same amplifier and emit sinusoidal waves in phase. The frequency of the waves emitted by each speaker is 860 Hz. Point \(P\) is 12.0 m from \(A\) and 13.4 \(\mathrm{m}\) from \(B\) . Is the interference at \(P\) constructive or destructive? Give the reasoning behind your answer.

Short Answer

Expert verified
The interference at P is destructive because the path difference is an odd half-wavelength multiple.

Step by step solution

01

Identify Wavelength

The speed of sound in air is approximately 343 m/s. To find the wavelength of the sound waves, use the formula \( \lambda = \frac{v}{f} \), where \( v \) is the speed of sound and \( f \) is the frequency. \( \lambda = \frac{343}{860} \approx 0.399 \mathrm{\ m} \).
02

Calculate Path Difference

The path difference between waves from the two speakers is the difference in the distances from point \(P\) to each speaker. Calculate this as: \( \text{Path Difference} = |13.4 - 12.0| = 1.4 \mathrm{\ m} \).
03

Determine Constructive or Destructive Interference

For constructive interference, the path difference should be an integer multiple of the wavelength (\( m\lambda \)), and for destructive interference, it should be an odd multiple of half-wavelengths (\((m + \frac{1}{2})\lambda\)). Divide the path difference by the wavelength: \( \frac{1.4}{0.399} \approx 3.51\). This is not an integer, but close to 3.5, suggesting it is an odd multiple of half-wavelengths and thus destructive interference.

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Key Concepts

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

Sound Waves
Sound waves are a type of mechanical wave that travels through a medium, typically air. Unlike electromagnetic waves, they require a medium to propagate. Sound waves are generated by vibrating objects which create pressure variations in the medium. As these pressure variations travel through the medium, they are perceived as sound by our ears.

Key characteristics of sound waves include:
  • Wavelength: This is the distance between two consecutive points that are in phase, like two adjacent crests or troughs.
  • Frequency: This refers to the number of wave cycles that pass a point per unit of time, measured in hertz (Hz).
  • Speed: In air, sound travels at approximately 343 meters per second, although this can vary with temperature and humidity.
Sound waves can be transformed by various conditions, such as reflecting off surfaces or bending around obstacles, which is part of the phenomenon known as wave interference.
Constructive Interference
Constructive interference occurs when two or more waves meet in such a way that their crests overlap, resulting in a new wave of greater amplitude. This happens when the waves are in phase, meaning their peaks align perfectly.

Here鈥檚 how constructive interference works:
  • The path difference between interacting waves results in a whole number of wavelengths, denoted as \(m\lambda\), where \(m\) is an integer.
  • In such cases, the amplitude of the resulting wave is the sum of the individual wave amplitudes, leading to louder or more intense sounds.
  • Constructive interference is commonly seen in scenarios where sound, light, or water waves add together.
This principle is essential in various applications, such as sound engineering, where phase alignment is used to amplify audio signals.
Destructive Interference
Destructive interference occurs when two or more waves meet and their crests and troughs are out of phase, effectively canceling each other out. This results in a wave of reduced or zero amplitude.

Here鈥檚 how you can identify destructive interference:
  • The path difference between waves is an odd multiple of a half-wavelength, expressed as \((m + \frac{1}{2})\lambda\), where \(m\) is an integer.
  • When waves are exactly half a wavelength apart, their crests meet troughs, causing cancellation and a pronounced reduction in sound intensity.
  • This phenomenon is critical in various fields, such as acoustics, where it helps in noise cancellation technologies.
In the exercise example, the interference at point \(P\) is destructive because the path difference calculated resulted in an odd multiple of half-wavelengths, leading to cancellation of sound at that specific location.

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Most popular questions from this chapter

Standing sound waves are produced in a pipe that is 1.20 \(\mathrm{m}\) long. For the fundamental and first two overtones, determine the locations along the pipe (measured from the left end) of the dis- placement nodes and the pressure nodes if (a) the pipe is open at both ends and (b) the pipe is closed at the left end and open at the right end.

A car alarm is emitting sound waves of frequency 520 \(\mathrm{Hz}\) You are on a motorcycle, traveling directly away from the car. How fast must you be traveling if you detect a frequency of 490 \(\mathrm{Hz}\) ?

Ep A New Musical Instrument. You have designed a new musical instrument of very simple construction. Your design consists of a metal tube with length \(L\) and diameter \(L / 10 .\) You have stretched a string of mass per unit length \(\mu\) across the open end of the tube. The other end of the tube is closed. To produce the musical effect you're looking for, you want the frequency of the third-harmonic standing wave on the string to be the same as the fundamental frequency for sound waves in the air column in the tube. The speed of sound waves in this air column is \(v_{\mathrm{s}}\) . (a) What must be the tension of the string to produce the desired effect? (b) What happens to the sound produced by the instrument if the tension is changed to twice the value calculated in part (a)? (c) For the tension calculated in part (a), what other harmonics of the string, if any, are in resonance with standing waves in the air column?

CP You have a stopped pipe of adjustable length close to a taut \(85.0-\mathrm{cm}, 7.25-\mathrm{g}\) wire under a tension of 4110 \(\mathrm{N}\) . You want to adjust the length of the pipe so that, when it produces sound at its fundamental frequency, this sound causes the wire to vibrate in its second overtone with very large amplitude. How long should the pipe be?

CP Two identical taut strings under the same tension \(F\) produce a note of the same fundamental frequency \(f_{0}\) . The tension in one of them is now increased by a very small amount \(\Delta F .\) (a) If they are played together in their fundamental, show that the frequency of the beat produced is \(f_{\text { beat }}=f_{0}(\Delta F / 2 F)\) . (b) Two identical violin strings, when in tune and stretched with the same tension, have a fundamental frequency of 440.0 Hz. One of the strings is retuned by increasing its tension. When this is done, 1.5 beats per second are heard when both strings are plucked simultaneously at their centers. By what percentage was the string tension changed?
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