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Chapter 17: Problem 36

The range of human hearing is roughly from twenty hertz to twenty kilohertz. Based on these limits and a value of \(343 \mathrm{~m} / \mathrm{s}\) for the speed of sound, what are the lengths of the longest and shortest pipes (open at both ends and producing sound at their fundamental frequencies) that you expect to find in a pipe organ?

Short Answer

Expert verified
Longest pipe: 8.575 m; Shortest pipe: 0.008575 m.

Step by step solution

01

Identify the Fundamental Frequency Formula

The fundamental frequency of a pipe open at both ends is calculated using the formula: \( f = \frac{v}{2L} \), where \( f \) is the frequency, \( v \) is the speed of sound, and \( L \) is the length of the pipe. We are given \( v = 343 \text{ m/s} \).
02

Find the Longest Pipe Length

For the longest pipe, we use the lowest frequency, which is \( 20 \text{ Hz} \). Rearranging the formula to solve for \( L \), we have: \( L = \frac{v}{2f} = \frac{343}{2 \times 20} = \frac{343}{40} = 8.575 \text{ m} \).
03

Find the Shortest Pipe Length

For the shortest pipe, we use the highest frequency, which is \( 20,000 \text{ Hz} \). Using the formula \( L = \frac{v}{2f} \), we get: \( L = \frac{343}{2 \times 20,000} = \frac{343}{40,000} = 0.008575 \text{ m} \).
04

Conclusion

The lengths of the longest and shortest pipes in a pipe organ, producing sound at their fundamental frequencies, are approximately \( 8.575 \text{ m} \) and \( 0.008575 \text{ m} \), respectively.

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Key Concepts

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

Understanding Fundamental Frequency in Pipe Organs
Pipe organs produce sound based on the principle of resonance, where each pipe produces a specific note or frequency known as its fundamental frequency. To determine this frequency, we use the formula \( f = \frac{v}{2L} \). Here, \( f \) represents the frequency, \( v \) the speed of sound in the air, and \( L \) the length of the pipe. This calculation helps in identifying the note that the pipe will produce. For organs, the fundamental frequency is critical in achieving the correct pitch. When a pipe organ is tuned, the lengths of the pipes are adjusted so that each one resonates at its desired frequency. This fundamental concept not only sets the base note but also influences overtones that result in the rich, layered sound characteristic of pipe organs.
How Speed of Sound Affects Pipe Organ Acoustics
The speed of sound is essential in the acoustics of pipe organs, as it directly influences the frequencies produced by the pipes. Sound travels at approximately 343 meters per second in air at room temperature. This constant determines how fast sound waves move through the air and affects how quickly a listener perceives the sound. In the context of a pipe organ, the speed of sound is used along with the length of each pipe to calculate the fundamental frequency. A higher speed of sound would result in higher frequencies for the same pipe length, and vice versa. It's essential for organ builders to understand this property so that they can precisely calibrate the lengths of the pipes to produce accurate notes across a wide range of frequencies.
The Relationship Between Pipe Length and Human Hearing Range
The range of human hearing spans from 20 Hz to 20,000 Hz. Within this range, a pipe organ must produce sounds that are both audible and distinct to the human ear. The length of the pipes in a pipe organ plays a critical role in this hearing range. Longer pipes produce lower frequencies, which correspond to deeper sounds such as bass notes. Conversely, shorter pipes produce higher frequencies, producing treble notes. For example, a pipe vibrating at 20 Hz, which is the lower limit of human hearing, would need a length of approximately 8.575 meters, while a pipe at 20,000 Hz, near the upper limit, would only require about 0.008575 meters. Understanding how pipe length and frequency interact within the human hearing range helps designers create organs with a wide array of notes that are clear and resonant. This knowledge ensures that the musical experience retains its richness and complexity.

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Most popular questions from this chapter

In Concept Simulation \(17.2\) at you can explore the concepts that are important in this problem. A 440.0-Hz tuning fork is sounded together with an out-of-tune guitar string, and a beat frequency of \(3 \mathrm{~Hz}\) is heard. When the string is tightened, the frequency at which it vibrates increases, and the beat frequency is heard to decrease. What was the original frequency of the guitar string?

Two loudspeakers are mounted on a merry-go-round whose radius is \(9.01 \mathrm{~m} .\) When stationary, the speakers both play a tone whose frequency is \(100.0 \mathrm{~Hz}\). As the drawing illustrates, they are situated at opposite ends of a diameter. The speed of sound is 343.00 \(\mathrm{m} / \mathrm{s},\) and the merry-go-round revolves once every \(20.0 \mathrm{~s}\). What is the beat frequency that is detected by the listener when the merry-go- round is near the position shown?

The \(\mathrm{E}\) string on an electric bass guitar has a length of \(0.628 \mathrm{~m}\) and, when producing the note E, vibrates at a fundamental frequency of \(41.2 \mathrm{~Hz}\). Players sometimes add to their instruments a device called a "D-tuner." This device allows the E string to be used to produce the note D, which has a fundamental frequency of \(36.7 \mathrm{~Hz}\). The D-tuner works by extending the length of the string, keeping all other factors the same. By how much does a D-tuner extend the length of the \(\mathrm{E}\) string?

Sound enters the ear, travels through the auditory canal, and reaches the eardrum. The auditory canal is approximately a tube open at only one end. The other end is closed by the eardrum. A typical length for the auditory canal in an adult is about \(2.9 \mathrm{~cm} .\) The speed of sound is \(343 \mathrm{~m} / \mathrm{s}\). What is the fundamental frequency of the canal? (Interestingly, the fundamental frequency is in the frequency range where human hearing is most sensitive.)

Two cars have identical horns, each emitting a frequency \(f_{s}\). One of the cars is moving toward a bystander waiting at a corner, and the other is parked. The two horns sound simultaneously. (a) From the moving horn, does the bystander hear a frequency that is greater than, less than, or equal to \(f_{\mathrm{s}} ?\) (b) From the stationary horn, does the bystander hear a frequency that is greater than, less than, or equal to \(f_{\mathrm{s}} ?\) (c) Does the bystander hear a beat frequency from the combined sound of the two horns? Account for your answers. The frequency that the horns emit is \(f_{\mathrm{s}}=395 \mathrm{~Hz}\). The speed of the moving car is \(12.0 \mathrm{~m} / \mathrm{s},\) and the speed of sound is \(343 \mathrm{~m} / \mathrm{s}\). What is the beat frequency heard by the bystander?
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