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Chapter 17: Problem 42

A tube with a cap on one end, but open at the other end, has a fundamental frequency of \(130.8 \mathrm{Hz}\). The speed of sound is \(343 \mathrm{m} / \mathrm{s}\) (a) If the cap is removed, what is the new fundamental frequency of the tube? (b) How long is the tube?

Short Answer

Expert verified
(a) 261.6 Hz; (b) 0.656 m.

Step by step solution

01

Understand the Problem

The tube is initially closed at one end, making it a closed pipe, but then is opened at both ends, becoming an open pipe. The fundamental frequency of a tube depends on whether it is closed or open.
02

Formula for Closed Pipe Fundamental Frequency

For a closed pipe, the fundamental frequency \( f_c \) is given by:\[ f_c = \frac{v}{4L} \]where \( v \) is the speed of sound and \( L \) is the length of the tube. We are given \( f_c = 130.8 \, \mathrm{Hz} \) and \( v = 343 \, \mathrm{m/s} \).
03

Solve for the Tube Length (L)

Using the formula \( f_c = \frac{v}{4L} \), we solve for \( L \):\[ L = \frac{v}{4f_c} = \frac{343}{4 imes 130.8} \approx 0.656 \, \mathrm{m} \]
04

Formula for Open Pipe Fundamental Frequency

For an open pipe, the fundamental frequency \( f_o \) is given by:\[ f_o = \frac{v}{2L} \]Now that we have the length \( L \), we can find the new frequency.
05

Calculate the New Fundamental Frequency

Substituting \( L = 0.656 \, \mathrm{m} \) into the open pipe formula:\[ f_o = \frac{343}{2 imes 0.656} \approx 261.6 \, \mathrm{Hz} \]

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Key Concepts

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

Fundamental Frequency
Fundamental frequency is the lowest frequency at which a system vibrates naturally. For a musical instrument like a tube, this concept is crucial since it determines the pitch of the sound produced. The fundamental frequency is often referred to as the first harmonic. It is the most dominant frequency emitted when the instrument resonates.

In physics, the fundamental frequency is related to the length of the resonating medium, in this case, a tube, and to the speed of sound in air.
  • For a closed pipe, this frequency can be calculated using the formula: \[ f_{c} = \frac{v}{4L} \]where \( v \) is the speed of sound and \( L \) is the length of the tube.
  • For an open pipe, the formula changes to: \[ f_{o} = \frac{v}{2L} \]
These differences arise because the waveform patterns differ: closed pipes support odd harmonics while open pipes support both odd and even harmonics.
Open and Closed Pipes
In musical acoustics, understanding how open and closed pipes affect sound waves is fundamental. Open and closed pipes refer to tubes that either have both ends open, or one end closed. **The behavior of sound waves inside these pipes differs significantly:**
  • Open Pipe: A tube open at both ends supports both odd and even harmonics. This setup creates antinodes at both ends, necessitating each full wave to involve half of the wavelength inside the tube. Consequently, the fundamental frequency is higher than in a closed pipe of the same length, influencing the pitch of the sound produced. This understanding of pipe behavior allows musicians and engineers to utilize pipe lengths for desired sound output effectively.
Speed of Sound
The speed of sound is a measure of how fast sound waves travel through a medium. In air, this speed is roughly constant under typical conditions, and is approximately \(343 \, \text{m/s}\) at room temperature. This speed can slightly vary with changes in temperature, pressure, and humidity. Sound moves differently through various materials:
  • It travels fastest through solids, due to closely packed molecules which transmit sound energy efficiently.
  • In liquids, the speed is slower than in solids but faster than in gases due to lesser molecular rigidity.
  • Finally, in gases like air, the speed is slowest because the molecules are spread out and less able to transmit the sound waves quickly.
For our discussion on tubes and sound, knowing the speed of sound assists in calculating wave frequencies.

When dealing with open and closed pipes, the known speed of sound allows us to find the fundamental frequency, as it directly influences how quickly sound waves can oscillate within the pipe, affecting the pitch and other harmonics.

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Most popular questions from this chapter

Sound (speed \(=343 \mathrm{m} / \mathrm{s})\) exits a diffraction horn loudspeaker through a rectangular opening like a small doorway. A person is sitting at an angle \(\alpha\) off to the side of a diffraction horn that has a width \(D\) of \(0.060 \mathrm{m} .\) This individual does not hear a sound wave that has a frequency of \(8100 \mathrm{Hz} .\) When she is sitting at an angle \(\alpha / 2,\) the frequency that she does not hear is different. What is this frequency?

A person hums into the top of a well and finds that standing waves are established at frequencies of \(42,70.0,\) and \(98 \mathrm{Hz}\). The frequency of \(42 \mathrm{Hz}\) is not necessarily the fundamental frequency. The speed of sound is \(343 \mathrm{m} / \mathrm{s}\). How deep is the well?

Two waves are traveling in opposite directions on the same string. The displacements caused by the individual waves are given by \(y_{1}=(24.0 \mathrm{mm}) \sin (9.00 \pi t-1.25 \pi x)\) and \(y_{2}=(35.0 \mathrm{mm}) \sin (2.88 \pi t+\) \(0.400 \pi x\) ). Note that the phase angles \((9.00 \pi t-1.25 \pi x)\) and \((2.88 \pi t+\) \(0.400 \pi x\) ) are in radians, \(t\) is in seconds, and \(x\) is in meters. At \(t=4.00 \mathrm{s}\) what is the net displacement (in \(\mathrm{mm}\) ) of the string at (a) \(x=2.16 \mathrm{m}\) and (b) \(x=2.56 \mathrm{m} ?\) Be sure to include the algebraic sign \((+\) or \(-)\) with your answers.

A string has a linear density of \(8.5 \times 10^{-3} \mathrm{kg} / \mathrm{m}\) and is under a tension of \(280 \mathrm{N}\). The string is \(1.8 \mathrm{m}\) long, is fixed at both ends, and is vibrating in the standing wave pattern shown in the drawing. Determine the (a) speed, (b) wavelength, and (c) frequency of the traveling waves that make up the standing wave.

A sound wave with a frequency of 15 kHz emerges through a circular opening that has a diameter of \(0.20 \mathrm{m} .\) Concepts: (i) The diffraction angle for a wave emerging through a circular opening is given by sin \(\theta=1.22 \lambda / D,\) where \(\lambda\) is the wavelength of the sound and \(D\) is the diameter of the opening. What is meant by the diffraction angle? (ii) How is the wavelength related to the frequency of the sound? (iii) Is the wavelength of the sound in air greater than, smaller than, or equal to the wavelength in water? Why? (Note: The speed of sound in air is \(343 \mathrm{m} / \mathrm{s}\) and the speed of sound in water is \(1482 \mathrm{m} / \mathrm{s} .)\) (iv) Is the diffraction angle of the sound in air greater than, smaller than, or equal to the diffraction angle in water? Explain. Calculations: Find the diffraction angle \(\theta\) when the sound travels (a) in air and (b) in water.
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