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In noisy factory environments, it's possible to use a loudspeaker to cancel persistent low-frequency machine noise at the position of one worker. The details of practical systems are complex, but we can present a simple example that gives you the idea. Suppose a machine \(5.0 \mathrm{m}\) away from a worker emits a persistent \(80 \mathrm{Hz}\) hum. To cancel the sound at the worker's location with a speaker that exactly duplicates the machine's hum, how far from the worker should the speaker be placed? Assume a sound speed of \(340 \mathrm{m} / \mathrm{s}\).

Short Answer

Expert verified
The loudspeaker should be placed \(7.125 \mathrm{m}\) away from the worker.

Step by step solution

01

Calculate the Wavelength

First, let's calculate the wavelength of the machine's hum. The formula to find the wavelength (λ) of a wave is given by the equation λ = speed/frequency. Here the speed of sound is given as \(340 \mathrm{m/s}\) and the frequency as \(80 \mathrm{Hz}\). So, the wavelength (λ) is calculated as λ = \(340 \mathrm{m/s} / 80 \mathrm{Hz} = 4.25 \mathrm{m}\).
02

Calculate the Distance for Destructive Interference

The next step is to determine the distance the loudspeaker needs to be for the sound wave to arrive halfway (λ/2) out of phase with the machine's sound. As we know that the sound wave needs to be offset by half a wavelength (λ/2), this implies that the sound from the loudspeaker needs to travel a distance of λ/2 longer or shorter than the sound from the machine. So, the loudspeaker needs to be placed \(5.0 \mathrm{m} \pm 4.25 \mathrm{m} / 2\)
03

Determine the Position of the Speaker

Now let's calculate the position of the loudspeaker. Using the formula from Step 2, we have two possible solutions: \(5.0 \mathrm{m} + 4.25 \mathrm{m}/2\) and \(5.0 \mathrm{m} - 4.25 \mathrm{m}/2\). However, the second solution gives a negative result, which is not feasible as the loudspeaker cannot be placed at a negative distance from the worker. So, we discard the second solution. The speaker should therefore be placed \(5.0 \mathrm{m} + 4.25 \mathrm{m}/2 = 7.125 \mathrm{m}\) away from the worker.

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Key Concepts

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

Destructive Interference
In sound interference, especially in practical applications like noise cancellation, understanding destructive interference is crucial. Destructive interference occurs when two sound waves meet in such a way that their amplitudes cancel each other out. For this to happen correctly, the two sound waves need to be out of phase by exactly half a wavelength, or \(\lambda/2\). This means that the highs of one wave align with the lows of the other, effectively reducing the sound to a much lower level or even complete silence.
When using this concept in a noisy environment, like the factory from our example, the loudspeaker emits a sound that is identical to the machine noise but shifted by half a wavelength. The worker then hears nothing much, as these two sound waves cancel each other out at that position. Think of it like "erasing" the sound at a certain spot using the waves themselves. This makes workplaces quieter without needing heavy physical barriers.

To achieve this, the positioning of the interfering wave source (the speaker, in our case) is critical. It must be placed so that its emitted sound travels a path distance that differs by half a wavelength from the noise source to the worker, creating the desired cancellation effect here.
Wavelength Calculation
In any wave phenomenon, calculating the wavelength is the starting point to understanding its behavior and properties. The wavelength (\(\lambda\)) represents the distance over which the wave's shape repeats. This can be determined by using the formula:
  • \(\lambda = \frac{\text{speed of sound}}{\text{frequency}}\)
In the provided example with machine noise, the speed of sound is given as \(340 \text{ m/s}\) and the frequency is \(80 \text{ Hz}\). When substituting these values into the formula, we find that
\(\lambda = \frac{340 \text{ m/s}}{80 \text{ Hz}} = 4.25 \text{ m}\) .
Understanding how to perform this calculation is essential because it determines the path distance difference required for producing destructive interference. The wavelength sets the "scale" needed to apply the principle efficiently enough for noise cancellation work.
Wave Phase Cancellation
The principle of wave phase cancellation is pivotal in situations requiring silence or noise reduction, like cancelling persistent machine hums.
When two sound waves are in phase, this means their peaks and troughs match, resulting in amplified sound. Conversely, when they are out of phase, peaks and troughs from one wave align inversely with those of another, causing cancellation.
  • The term "phase" refers to the position of a point within the wave cycle.
  • Cancelling requires the waves to have opposite phases at the target location; usually exactly half a wavelength apart which is called 180 degrees out of phase.
Thus, when you align the sound waves so that their frequencies and speeds match, but their paths to the listener differ by half a wavelength, they tend to "cancel out." It is the underlying principle employed by noise-cancelling devices to reduce unwanted background sounds efficiently. Understanding this can help in other real-world situations like concert hall design or even enhancing audio equipment performance. It can make a significant difference by applying such cancellation principles.

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Most popular questions from this chapter

A particularly beautiful note reaching your ear from a rare Stradivarius violin has a wavelength of \(39.1 \mathrm{cm} .\) The room is slightly warm, so the speed of sound is \(344 \mathrm{m} / \mathrm{s}\). If the string's linear density is \(0.600 \mathrm{g} / \mathrm{m}\) and the tension is \(150 \mathrm{N},\) how long is the vibrating section of the violin string?

When you voice the vowel sound in "hat," you narrow the opening where your throat opens into the cavity of your mouth so that your vocal tract appears as two connected tubes. The first is in your throat, closed at the vocal cords and open at the back of the mouth. The second is the mouth itself, open at the lips and closed at the back of the mouth-a different condition than for the throat because of the relatively larger size of the cavity. The corresponding formant frequencies are \(800 \mathrm{Hz}\) (for the throat) and \(1500 \mathrm{Hz}\) (for the mouth). What are the lengths of these two cavities? Assume a sound speed of \(350 \mathrm{m} / \mathrm{s}\).

Two loudspeakers directly face each other \(30 \mathrm{m}\) apart, with the left speaker positioned at \(x=0 \mathrm{m}\). The pressure of the sound wave emitted by the left speaker is described by the equation \(\Delta p_{\mathrm{L}}=p_{0} \cos (1.90 x-630 t),\) while that from the right speaker is given by \(\Delta p_{\mathrm{R}}=p_{0} \cos (1.90 x+630 t),\) where \(x\) is measured in \(\mathrm{m}\) and \(t\) in \(\mathrm{s}\). What is the point nearest to the left speaker at which there is a node in the sound wave?

Two strings are adjusted to vibrate at exactly \(200 \mathrm{Hz}\). Then the tension in one string is increased slightly. Afterward, three beats per second are heard when the strings vibrate at the same time. What is the new frequency of the string that was tightened?

Musicians can use beats to tune their instruments. One flute is properly tuned and plays the musical note A at exactly \(440 \mathrm{Hz}\). A second player sounds the same note and hears that her instrument is slightly "flat" (that is, at too low a frequency). Playing at the same time as the first flute, she hears two loud-soft-loud beats per second. What is the frequency of her instrument?

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