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The longest pipe found in most medium-sized pipe organs is \(4.88 \mathrm{~m}\) (16 ft). What is the wavelength of the note corresponding to the fundamental mode if the pipe is (a) open at both ends and (b) open at one end and closed at the other?

Short Answer

Expert verified
For an open pipe: \( \lambda = 9.76 \text{ m} \), and for a closed-open pipe: \( \lambda = 19.52 \text{ m} \).

Step by step solution

01

Introduction to the Problem

In this problem, we are asked to find the wavelength of the note produced by a pipe organ. The problem specifies two scenarios: one where the pipe is open at both ends (a) and another where it is open at one end and closed at the other (b). We will use the concepts of standing waves and their relation to the pipe length to solve each part.
02

Open Pipe at Both Ends (a)

For a pipe that is open at both ends, the fundamental frequency (first harmonic) has nodes at the ends and an antinode in the middle. The wavelength of the fundamental mode is twice the length of the pipe. This can be expressed as: \[ \lambda = 2L \] where \( L \) is the length of the pipe.Given \( L = 4.88 \) meters, we substitute into the equation:\[ \lambda = 2 \times 4.88 = 9.76 \text{ meters} \]
03

Open Pipe at One End and Closed at the Other (b)

For a pipe that is open at one end and closed at the other, the fundamental frequency (first harmonic) has a node at the closed end and an antinode at the open end. The wavelength of the fundamental mode is four times the length of the pipe. This can be expressed as: \[ \lambda = 4L \] where \( L \) is the length of the pipe.Given \( L = 4.88 \) meters, we substitute into the equation:\[ \lambda = 4 \times 4.88 = 19.52 \text{ meters} \]

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Key Concepts

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

Fundamental Frequency
The fundamental frequency is the lowest frequency at which a system can vibrate to produce a standing wave. In musical terms, it is often associated with the pitch of the note. For a pipe organ, the fundamental frequency generates the base sound of the note when air oscillates within it.
In a pipe, whether open or closed, the fundamental frequency determines the shape and pattern of the standing waves that form. These waves are integral multiples of the fundamental frequency, leading to harmonics or overtones.
To calculate the fundamental frequency, you need to know either the speed of sound in air and the wavelength or use equations specific to the pipe's configuration, such as \( f_n = \frac{nv}{2L} \) for open pipes, where \( n \) is the harmonic number, \( v \) is the speed of sound, and \( L \) is the length of the pipe.
Open Pipe
An open pipe is a resonating tube open at both ends. When sound waves travel through it, nodes form at these open ends. The waves create regions of destructive interference at the ends and constructive interference in between, producing clear waves and tones.
For an open pipe, the fundamental frequency appears as a standing wave with an antinode in the center and nodes at each end, reflecting the air's free movement.
The length of the pipe equals half the wavelength of the fundamental sound wave, described mathematically as \( \lambda = 2L \). Thus, any changes in pipe length directly affect the sound it produces, allowing different musical notes.
Closed Pipe
A closed pipe has one end closed off, limiting air movement and modifying the sound produced. In these pipes, the closed end holds a node, while the open end is an antinode.
Because the length of the pipe equals one-fourth of the wavelength of the fundamental frequency, the relationship is expressed as \( \lambda = 4L \). This configuration restricts the range of notes, causing closed pipes to sound an octave lower than open pipes of the same length.
Consequently, only odd harmonics are present in closed pipes, resulting in a unique sound compared to open pipes.
Wavelength Calculation
Wavelength calculation in standing waves depends on whether the pipe is open or closed, impacting how sound travels.
For an open pipe, use the formula \( \lambda = 2L \). This tells us the wavelength equals twice the pipe's length, facilitating the calculation for open-ended tube scenarios.
For closed pipes, however, the formula changes to \( \lambda = 4L \), as the pipe’s closed end changes how sound travels through, leading to longer wavelengths for the fundamental frequency.
  • In the exercise example, for an open pipe of 4.88 meters: \( \lambda = 2 \times 4.88 = 9.76 \text{ meters} \).
  • For a closed pipe of 4.88 meters: \( \lambda = 4 \times 4.88 = 19.52 \text{ meters} \).
Experimenting with these formulas helps to understand the physical relationship between pipe structure and the sound it produces.

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Most popular questions from this chapter

A violinist is tuning her instrument to concert A \((440 \mathrm{~Hz})\). She plays the note while listening to an electronically generated tone of exactly that frequency and hears a beat of frequency \(3 \mathrm{~Hz},\) which increases to \(4 \mathrm{~Hz}\) when she tightens her violin string slightly. (a) What was the frequency of her violin when she heard the \(3 \mathrm{~Hz}\) beat? (b) To get her violin perfectly tuned to concert A, should she tighten or loosen her string from what it was when she heard the \(3 \mathrm{~Hz}\) beat?

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