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The fundamental frequency of a pipe that is open at both ends is 594 Hz (a) How long is this pipe? If one end is now closed, find (b) the wavelength and (c) the frequency of the new fundamental.

Short Answer

Expert verified
(a) 0.289 m; (b) 1.156 m; (c) 296.9 Hz

Step by step solution

01

Understanding the Problem: Fundamental Frequency of Open Pipe

For a pipe open at both ends, the fundamental frequency corresponds to the first harmonic where the length of the pipe (L) is half the wavelength of the wave. The formula is given by \( f = \frac{v}{2L} \), where \( f \) is the frequency and \( v \) is the speed of sound (approximately 343 m/s at room temperature).
02

Calculate the Length of the Open Pipe

Rearrange the formula to solve for \( L \): \( L = \frac{v}{2f} \). Given \( f = 594 \) Hz, substitute \( v = 343 \) m/s to find \( L = \frac{343}{2 \times 594} \approx 0.289 \) meters.
03

Wavelength with One End Closed

When one end is closed, the pipe supports quarter-wavelength resonances. The fundamental frequency corresponds to the first harmonic, where \( L = \frac{\lambda}{4} \). So the wavelength (\( \lambda \)) is \( \lambda = 4L \). Using \( L = 0.289 \) meters, we find \( \lambda = 4 \times 0.289 \approx 1.156 \) meters.
04

Frequency of the New Fundamental

With a closed end, the fundamental frequency is given by \( f = \frac{v}{\lambda} \). Substitute \( \lambda = 1.156 \) meters and \( v = 343 \) m/s to find the new frequency: \( f = \frac{343}{1.156} \approx 296.9 \) Hz.

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Key Concepts

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

Open Pipe
An open pipe is a type of resonating air column where both ends are open to the surrounding air. In terms of sound, this setup supports standing waves where the ends of the pipe are always antinodes. Antinodes are points of maximum amplitude, making the sound loud and clear at these points. This characteristic is important as it determines the wavelengths of the sound waves that can resonate within the pipe.
For an open pipe, the fundamental frequency is the lowest frequency at which the pipe resonates. The length of the pipe corresponds to half the wavelength (\( \frac{1}{2} \lambda \)). Therefore, in mathematical terms, the frequency of the wave is given by the formula:
  • \( f = \frac{v}{2L} \)
  • Where \( f \) is the frequency
  • \( v \) is the speed of sound (approximately 343 m/s)
  • \( L \) is the length of the pipe.
This relationship helps you calculate important characteristics of sound in musical instruments or acoustics.
Closed Pipe
A closed pipe is similar to an open pipe, but only one end is open while the other is closed. This structure changes the way sound waves travel, producing a different set of harmonics compared to an open pipe. In a closed pipe, the closed end is a node (point of no motion), while the open end remains an antinode (point of maximum motion).
Due to this setup, a closed pipe supports quarter-wavelength resonances. The fundamental frequency in a closed pipe corresponds to a wave whose length is four times the pipe's length. This means the pipe length (\( L \)) is one-fourth of the wavelength (\( \frac{\lambda}{4} \)). As a result, the formula for the wavelength becomes:
  • \( \lambda = 4L \)
Compared to open pipes, closed pipes have a lower fundamental frequency because the air column vibrates at longer wavelengths.
Wavelength Calculation
The wavelength of the sound wave is an essential factor in determining the behavior of sound in pipes. For open and closed pipes, it's crucial to calculate this property because it influences the frequency and harmonics produced. The wavelength represents the distance between two consecutive points in phase, like peaks or troughs, in a wave.In practical situations, knowing the length of a pipe helps in calculating the wavelength. For an open pipe, where the length is half the wavelength, and a closed pipe, where the length is a quarter of the wavelength, these relationships are key:
  • Open Pipe: \( \lambda = 2L \)
  • Closed Pipe: \( \lambda = 4L \)
By understanding these formulas, you can accurately determine the wavelength of sound in different contexts, which aids in designing musical instruments and understanding acoustic phenomena.
Frequency Calculation
Frequency is a critical aspect of sound, defining how high or low a note is perceived. It's the number of wave cycles that pass a point per second, measured in Hertz (Hz). Understanding frequency calculation is essential, especially in the context of sound traveling through pipes.The fundamental frequency of a pipe is dependent on its length and whether it is open or closed. Using the relationship between speed, frequency, and wavelength (\( f = \frac{v}{\lambda} \)), you can determine the frequency of sound for both configurations:
  • For Open Pipe: \( f = \frac{v}{2L} \)
  • For Closed Pipe: \( f = \frac{v}{\lambda} \) with \( \lambda = 4L \)
By applying these formulas, you can easily compute the fundamental frequency and explore how altering the pipe’s length or configuration affects the sound it produces. This knowledge is particularly useful in the fields of music and acoustics engineering.

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Most popular questions from this chapter

An oscillator vibrating at 1250 Hz produces a sound wave that travels through an ideal gas at 325 \(\mathrm{m} / \mathrm{s}\) when the gas temperature is \(22.0^{\circ} \mathrm{C}\) . For a certain experiment, you need to have the same uscillator produce sound of waveleugth 28.5 \(\mathrm{cm}\) in this gas. What should the gas temperature be to achieve this wavelength?

A New Musical Tnstrument. 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 \(v_{v}\) (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 calculatod 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?

(a) Show that the fractional change in the speed of sound \((d v / v)\) due to a very small temperature change \(d T\) is given by \(d v / v=\frac{1}{2} d T / T .\) (Hint: Start with Eq. \(16.10 .\) (b) The speed of sound in air at \(20^{\circ} \mathrm{C}\) is found to be 344 \(\mathrm{m} / \mathrm{s}\) . Use the result in part (a) to find the change in the speed of sound for a \(1.0^{\circ} \mathrm{C}\) change in air temperature.

Standing sound waves are produced in a pipe that is 1.20 \(\mathrm{m}\) long. For the fundamental and first two overtones, determine the 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 woman stands at rest in front of a large, smooth wall. She holds a vibrating tuning fork of frequency \(f_{0}\) directly in front of her (between her and the wall). (a) The woman now runs toward the wall with speed \(v_{\mathrm{W}}\) . She detects beats due to the interference between the sound waves reaching her directly from the fork and those reaching her after being reflected from the wall. How many beats per second will she detect? (Note: If the beat frequency is too large, the woman may have to use some instrumentation other than her ears to detect and count the beats.) (b) If the woman instead runs away from the wall, holding the tuning fork at her back so it is between her and the wall, how many beats per second will she detect?

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