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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) 260.4 Hz, (b) 0.657 m.

Step by step solution

01

Understanding the fundamental frequency of a closed tube

For a tube closed at one end, the fundamental frequency is determined when the tube has a length equal to one-fourth of the wavelength \[ \text{Length of tube } L = \frac{\lambda}{4}. \] The formula for the fundamental frequency is \[ f_1 = \frac{v}{4L}, \] where \(f_1\) is the fundamental frequency and \(v\) is the speed of sound, which is given as \(343 \, \mathrm{m/s}\). From this, we can find the length of the tube.
02

Calculating the length of the tube with one end closed

Rearrange the formula for the fundamental frequency to solve for the length \(L\): \[ L = \frac{v}{4f_1}. \] Substitute the given values \(v = 343 \, \mathrm{m/s}\) and \(f_1 = 130.8 \, \mathrm{Hz}\) into the equation:\[ L = \frac{343}{4 \times 130.8} \approx 0.657 \, \mathrm{m}. \]This gives the length of the tube.
03

Understanding the fundamental frequency of an open tube

For a tube open at both ends, the fundamental frequency is determined when the tube has a length equal to one-half of the wavelength: \[ L = \frac{\lambda}{2}. \] The formula for the fundamental frequency in this case is \[ f'_1 = \frac{v}{2L}. \] This information will be useful for determining the new frequency when the cap is removed.
04

Calculating the new fundamental frequency with both ends open

Use the previously calculated length, \(L = 0.657 \, \mathrm{m}\), and substitute it into the formula to calculate the new frequency:\[ f'_1 = \frac{343}{2 \times 0.657} \approx 260.4 \, \mathrm{Hz}. \]This is the new fundamental frequency when the tube is open at both ends.

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Key Concepts

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

Fundamental Frequency
The concept of fundamental frequency is critical in understanding musical instruments like a tube. In acoustics, the fundamental frequency is the lowest frequency at which a system oscillates. When we talk about musical instruments, this frequency is often related to the note they produce.
For a tube that is closed at one end, the fundamental frequency occurs when the length of the tube supports a wave that is one-fourth the wavelength of the sound wave. The formula used to calculate this frequency is \[f_1 = \frac{v}{4L},\]where:
  • \(f_1\) is the fundamental frequency,
  • \(v\) is the speed of sound (for this example, \(343 \, \mathrm{m/s}\)),
  • \(L\) is the length of the tube.
Understanding the impact of the tube's length on frequency helps design musical instruments and adjust for different sound properties.
Sound Waves
Sound waves are essential in the study of acoustics since they are the basis for any sound we hear. They are created by vibrations, which bring about pressure changes in the air. These pressure variations move through the air as waves, reaching our ears and allowing us to perceive sound.
The speed of sound varies depending on the medium through which it travels. In the exercise example, sound travels through air at \(343 \, \mathrm{m/s}\).
Sound waves can have different frequencies and wavelengths which are inversely related. - High-frequency sounds have short wavelengths; this is why they have higher pitches. - Low-frequency sounds have long wavelengths, resulting in lower pitches.
Manipulating a tube's structure, whether having one or both ends open, directly affects these waves and, as a result, the sound we hear.
Tube Resonance
Tube resonance is a fascinating phenomenon that explains why tubes produce sound when air is vibrating within them. In acoustics, resonance occurs when the natural frequency of a particular system aligns with the frequency of the wave applied to it. This creates strong vibrations and hence, a louder sound.
A tube's design, whether open or closed at either end, substantially impacts how sound resonates within it. For a tube open at both ends, the length supports a wave that is half its wavelength. The formula to find the new fundamental frequency when both ends are open is \[f'_1 = \frac{v}{2L},\]where \(f'_1\) is the new fundamental frequency.
The exercise shows that removing the cap from a previously closed tube doubles the fundamental frequency, as the tube can now resonate at a longer wavelength due to both ends being open. Understanding tube resonance is crucial for tuning musical instruments and developing acoustical engineering solutions.

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Most popular questions from this chapter

You and your team are exploring an antiquated research facility in the mountains of southern Argentina that had been abandoned in the 1960 s. You come to a giant locked door that has no visible handles or actuators, but you find a hand-held device nearby that has two buttons, labeled "Open" and "Close." It looks like some kind of crude remote control, but when you push the buttons they just click and nothing happens. You open the device's top cover and inspect it. The internal mechanism resembles an old acoustic remote control called the "Space Command \(600 "\) that your parents had for their ancient TV. Pushing a button on the remote actuated a small hammer on the inside that struck the end of an aluminum rod, about 2 or \(3 \mathrm{cm}\) in length. The rod vibrated and emitted an ultrasonic sound wave that actuated an electrical circuit in the TV that was sensitive to that frequency. The TV remote had three buttons, and therefore three rods that vibrated at different frequencies. The first frequency turned the \(\mathrm{TV}\) on and off, the second made the tuning dial click to the next station, and the third made the dial turn in the opposite direction. You pull off the cover of the device and find that it has places for two \(1 / 4\) -inch diameter rods, but both are missing. However, written on the inside of the cover of the device is the following: "Open \(=95.50 \mathrm{kHz}\) " and "Close \(=102.50 \mathrm{kHz}\) " Your team members search and eventually find a long piece of \(1 / 4\) -inch diameter aluminum rod. (a) To what lengths must you cut the rod in order to get the remote to work properly (i.e., so that the fundamental frequencies of the rods match those utilized by the remote)? (b) Suppose you had instead found a titanium rod. What lengths would be required in that case? (Young's moduli are \(Y_{\mathrm{Al}}=6.9 \times 10^{10} \mathrm{N} / \mathrm{m}^{2}\) and \(Y_{\mathrm{Ti}}=1.2 \times 10^{11} \mathrm{N} / \mathrm{m}^{2} ;\) the mass densities are \(\rho_{\mathrm{Al}}=2700 \mathrm{kg} / \mathrm{m}^{3}\) and \(\left.\rho_{\mathrm{Ti}}=4500 \mathrm{kg} / \mathrm{m}^{3} .\right)\)

A string is fixed at both ends and is vibrating at \(130\) \(\mathrm{Hz},\) which is its third harmonic frequency. The linear density of the string is \(5.6 \times 10^{-3} \mathrm{kg} / \mathrm{m}\), and it is under a tension of \(3.3\) \(\mathrm{N}\). Determine the length of the 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 speakers, one directly behind the other, are each generating a \(245-\mathrm{Hz}\) sound wave. What is the smallest separation distance between the speakers that will produce destructive interference at a listener standing in front of them? The speed of sound is \(343\) \(\mathrm{m} / \mathrm{s}\).

Two wires, each of length \(1.2\) \(\mathrm{m},\) are stretched between two fixed supports. On wire A there is a second-harmonic standing wave whose frequency is 660 Hz. However, the same frequency of \(660\) \(\mathrm{Hz}\) is the third harmonic on wire \(\mathrm{B}\). Find the speed at which the individual waves travel on each wire.
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