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Chapter 16: Problem 33

Two loudspeakers, \(A\) and \(B(\mathrm{Fig} .16 .40),\) are driven by the same amplifier and emit sinusoidal waves in phase. Speaker \(B\) is 2.00 \(\mathrm{m}\) to the right of speaker \(A\) . Consider point \(Q\) along the exten- sion of the line connecting the speakers, 1.00 \(\mathrm{m}\) to the right of speaker \(B .\) Both speakers emit sound waves that travel directly from the speaker to point \(Q .\) (a) What is the lowest frequency for which constructive interference occurs at point \(Q ?\) (b) What is the lowest frequency for which destructive interference occurs at point \(Q ?\)

Short Answer

Expert verified
Lowest frequency for constructive interference: 343 Hz. For destructive interference: 171.5 Hz.

Step by step solution

01

Understanding the Scenario

Two loudspeakers, \(A\) and \(B\), are placed in a line, \(2.00\) meters apart with \(B\) to the right of \(A\). Point \(Q\) is \(3.00\) meters to the right of \(A\) since \(Q\) is \(1.00\) meter to the right of \(B\). The problem involves finding the frequencies for constructive and destructive interference at point \(Q\).
02

Determining Path Differences

The distance from \(A\) to \(Q\) is \(3.00\) meters, and from \(B\) to \(Q\) is \(1.00\) meter. The path difference, \(\Delta d\), is the difference in distances traveled by the waves from each speaker to point \(Q\): \(\Delta d = 2.00 - 1.00 = 1.00\) meter.
03

Find Wavelengths for Constructive Interference

Constructive interference occurs when the path difference \(\Delta d\) is an integer multiple of the wavelength \(\lambda\), \(\Delta d = m\lambda\), where \(m\) is an integer. For the lowest frequency (fundamental frequency), we set \(m = 1\), giving \(\lambda = \Delta d = 1.00\) meter.
04

Calculate Lowest Frequency for Constructive Interference

The speed of sound, \(v\), is approximately \(343\) m/s in air. The frequency \(f\) is related to the wavelength by \(f = \frac{v}{\lambda}\). Therefore, the lowest frequency for constructive interference is \(f = \frac{343}{1} = 343\) Hz.
05

Find Wavelengths for Destructive Interference

Destructive interference occurs when the path difference \(\Delta d\) is a half-integer multiple of the wavelength, \(\Delta d = \left(m + \frac{1}{2}\right)\lambda\). For the lowest frequency, let \(m = 0\), giving \(\lambda = 2 \times \Delta d = 2 \times 1.00 = 2.00\) meters.
06

Calculate Lowest Frequency for Destructive Interference

Using the relationship \(f = \frac{v}{\lambda}\), the lowest frequency for destructive interference is \(f = \frac{343}{2} = 171.5 \) Hz.

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Key Concepts

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

Constructive Interference
When two sound waves meet, the phenomenon of constructive interference can lead to a louder sound at specific points due to the combined effect of the waves. This occurs when the peaks (or troughs) of one wave align with the peaks (or troughs) of another, resulting in a larger amplitude wave. This process effectively amplifies the sound at those points.

For constructive interference to take place, the path difference between the waves鈥攎eaning the distance each wave travels to a common point, like point \(Q\) in our initial scenario鈥攎ust be an integer multiple of the wavelength \(\lambda\). This can be expressed mathematically as:
  • \( \Delta d = m\lambda \)
where \(\Delta d\) is the path difference and \(m\) is any integer (e.g., \(0, 1, 2, 3...\)). This condition ensures that the waves are "in phase" at the point of intersection, thus creating the loudest possible version of the sound.
By using the speed of sound and the calculated wavelength, we found that the lowest frequency that causes constructive interference at point \(Q\) is 343 Hz, as derived from the formula \( f = \frac{v}{\lambda} \).
Destructive Interference
Destructive interference represents the phenomenon where sound waves cancel each other out. This occurs when the peaks of one wave align with the troughs of another. Such alignment leads to the reduction or even complete cancellation of the wave's amplitude, resulting in a quieter sound at that point.

For destructive interference to happen, the path difference must be a half-integer multiple of the wavelength. This can be represented as:
  • \( \Delta d = \left(m + \frac{1}{2}\right)\lambda \)
where \(\Delta d\) is the path difference and \(m\) is any integer. This condition ensures that the waves are "out of phase," leading to the cancellation of sound.
By using the speed of sound and the calculated wavelength in our scenario, we determined that the lowest frequency resulting in destructive interference at point \(Q\) is 171.5 Hz. This frequency is calculated based on the wavelength required to meet the condition of destructive interference, namely \( \lambda = 2 \times \Delta d \).
Sound Waves
Sound waves are an essential part of understanding wave interference. They are longitudinal waves that travel through air (and other media), causing particles in the medium to vibrate parallel to the direction of wave travel. These waves can be characterized by their frequency, wavelength, and speed.

Key attributes of sound waves include:
  • Frequency: This measures how many cycles (oscillations) a sound wave completes in a second, typically measured in Hertz (Hz).
  • Wavelength: This is the distance between successive points of a wave, such as from peak to peak or trough to trough.
  • Speed: Sound travels at approximately 343 m/s in air, though this can vary based on conditions like temperature and pressure.
In the context of interference, these attributes define how waves will behave when they meet. The principles of constructive and destructive interference depend heavily on the relationships between the frequency and wavelength of interacting sound waves.
By comprehending these elements, one can predict and understand how different frequencies will interact to produce varying levels of sound at any point in their path.

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Most popular questions from this chapter

On the planet Arrakis a male ornithoid is flying toward his mate at 25.0 \(\mathrm{m} / \mathrm{s}\) while singing at a frequency of 1200 \(\mathrm{Hz}\) . If the stationary female hears a tone of 1240 \(\mathrm{Hz}\) , what is the speed of sound in the atmosphere of Arrakis?

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 \(\mathrm{Hz}\) Point \(P\) is 12.0 \(\mathrm{m}\) from \(A\) and 13.4 \(\mathrm{m}\) from \(B .\) Is the interference at \(P\) constructive or destructive? Give the reasoning behind your answer.

Two identical loudspcakers are located at points \(A\) and \(B\) , 2.00 \(\mathrm{m}\) apart. The loudspeakers are driven by the same amplifier and produce sound waves with a frequency of 784 \(\mathrm{Hz}\) . Take the speed of sound in air to be 344 \(\mathrm{m} / \mathrm{s}\) . A small microphone is moved out from point \(B\) along a line perpendicular to the line connecting \(A\) and \(B(\text { line } B C \text { in Fig. } 16.44)\) (a) At what distances from \(B\) will there be destructive interference? (b) At what distances from \(B\) will there be constructive interference? (c) If the frequency is made low enough, there will be no positions along the line \(B C\) at which destructive interference occurs. How low must the frequency be for this to be the case?

The Sacramento City Council recently adopted a law to reduce the allowed sound intensity level of the much despised leaf blowers from their current level of about 95 \(\mathrm{dB}\) to 70 \(\mathrm{dB}\) . With the new law, what is the ratio of the new allowed intensity to the previously allowed intensity?

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 \(f_{\text { keet }}=f_{0}(\Delta F / 2 F)\) . (b) Two identical violin strings, when in tune and stretched with the same tension, have a fundamental frequency of 440.0 \(\mathrm{Hz}\) . 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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