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Divers working in underwater chambers at great depths must deal with the danger of nitrogen narcosis (the "bends"), in which nitrogen dissolves into the blood at toxic levels. One way to avoid this danger is for divers to breathe a mixture containing only helium and oxygen. Helium, however, has the effect of giving the voice a high-pitched quality, like that of Donald Duck's voice. To see why this occurs, assume for simplicity that the voice is generated by the vocal cords vibrating above a gas-filled cylindrical tube that is open only at one end. The quality of the voice depends on the harmonic frequencies generated by the tube; larger frequencies lead to higher-pitched voices. Consider two such tubes at \(20{ }^{\circ} \mathrm{C}\). One is filled with air, in which the speed of sound is \(343 \mathrm{~m} / \mathrm{s} .\) The other is filled with helium, in which the speed of sound is \(1.00 \times 10^{3} \mathrm{~m} / \mathrm{s}\) To see the effect of helium on voice quality, calculate the ratio of the \(n^{\text {th }}\) natural frequency of the helium- filled tube to the \(n^{\text {th }}\) natural frequency of the air-filled tube.

Short Answer

Expert verified
The frequency ratio of helium to air is approximately 2.92.

Step by step solution

01

Understand the Problem

We are asked to find the ratio of the natural frequencies of two tubes filled with different gases: air and helium. The frequency is determined by the speed of sound in the medium and the length of the tube. Since both tubes are assumed to have the same length and conditions, the ratio simplifies to comparing the speed of sound.
02

Identify the Formula for Natural Frequencies

The natural frequencies of a tube that is open at one end are given by the formula:\[ f_n = \frac{n v}{4L} \]where \(f_n\) is the frequency of the \(n^{\text{th}}\) harmonic, \(v\) is the speed of sound in the medium, \(L\) is the length of the tube, and \(n\) is an odd integer (1, 3, 5,...).
03

Simplify the Frequency Ratio

Since we are interested in the ratio between the natural frequencies in helium and air, the formula becomes:\[ \text{Ratio} = \frac{f_n(\text{helium})}{f_n(\text{air})} = \frac{\frac{n v_{\text{helium}}}{4L}}{\frac{n v_{\text{air}}}{4L}} \]The \(n\) and \(4L\) terms cancel out, leaving us with the ratio of speeds of sound:\[ \text{Ratio} = \frac{v_{\text{helium}}}{v_{\text{air}}} \]
04

Plug in the Known Values

We know that the speed of sound in helium is \(v_{\text{helium}} = 1.00 \times 10^3 \ \text{m/s}\) and in air is \(v_{\text{air}} = 343 \ \text{m/s}\). Substitute these values into the ratio formula:\[ \text{Ratio} = \frac{1.00 \times 10^3}{343} \approx 2.92 \]
05

Conclude with the Result

The ratio of the \(n^{\text{th}}\) natural frequency of the helium-filled tube to the air-filled tube is approximately 2.92. This indicates that the frequencies in the helium-filled tube are higher than those in the air-filled tube, explaining the high-pitched voice effect.

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Key Concepts

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

Helium-Oxygen Mixture
When divers work at great depths, nitrogen present in regular air can dissolve into the bloodstream at potentially dangerous levels, a condition known as nitrogen narcosis or "the bends." To prevent this, divers use a breathing mixture that comprises helium and oxygen instead of nitrogen. Helium is chosen because it is inert and less soluble in human tissues compared to nitrogen, reducing the risk of narcotic effects. However, there's a side effect to using helium: it changes how a diver's voice sounds.
Helium, being lighter than air, affects the timbre of a voice by influencing the speed of sound in the respiratory pathways. This combination not only provides safety benefits but also introduces unique acoustic properties that have significant effects on speech. Understanding these effects involves diving into how sound travels through different gases.
Speed of Sound
The speed of sound varies depending on the medium through which it propagates. This variance is primarily due to the differences in density and elasticity of gases. For divers using a helium-oxygen mixture, this becomes particularly important.
In air, at room temperature, the speed of sound is approximately 343 meters per second (m/s). However, when the medium is helium, the speed of sound increases dramatically to about 1000 m/s. This is because helium is much less dense than air, allowing sound waves to travel faster through it.
These differences in speed of sound are crucial because they influence the frequency at which sounds and voices are perceived. A higher speed implies that sounds can achieve higher frequencies, which is perceived as a higher pitch.
Natural Frequencies
A concept central to understanding why helium affects voice pitch is natural frequencies. In an open tube, like the vocal tract, the natural frequencies are determined by the speed of sound in the medium as well as the length of the tube. The formula to determine these frequencies is given by:\[ f_n = \frac{n v}{4L} \]where \( f_n \) represents the frequency of the \( n^{\text{th}} \) harmonic, \( v \) is the speed of sound in the medium, \( L \) is the effective length of the tube, and \( n \) is an odd integer.When comparing a helium-filled tube to an air-filled one, if the length \( L \) and the harmonic number \( n \) are kept constant, the natural frequency is directly proportional to the speed of sound. This means that the natural frequencies in helium are significantly higher due to its higher speed of sound compared to air.
High-Pitched Voice Effect
Diving with a helium-oxygen mixture results in divers sounding unusually high-pitched. This is referred to as the high-pitched voice effect and is a direct consequence of the altered speed of sound in helium compared to air.
The increased speed of sound in helium causes the natural frequencies of the vocal tract to be nearly three times higher than they would be in air (approximately a ratio of 2.92 based on the speeds given). Higher frequencies correspond to higher pitches, making the diver's voice sound similar to individuals who are speaking with a cartoonish, high-pitched quality.
This phenomenon, while humorously likened to cartoon voices, is an important factor all divers need to understand. Communication can be affected underwater; thus, awareness and sometimes systems designed to modulate voice frequencies might be employed to compensate for the helium effect.

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Most popular questions from this chapter

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 wires are stretched between two fixed supports and have the same length. On wire A there is a second-harmonic standing wave whose frequency is \(660 \mathrm{~Hz}\). However, the same frequency of \(660 \mathrm{~Hz}\) is the third harmonic on wire \(\mathrm{B}\). (a) Is the fundamental frequency of wire A greater than, less than, or equal to the fundamental frequency of wire \(\mathrm{B}\) ? Explain. (b) How is the fundamental frequency related to the length \(L\) of the wire and the speed \(v\) at which individual waves travel back and forth on the wire? (c) Do the individual waves travel on wire A with a greater, smaller, or the same speed as on wire B? Give your reasoning. The common length of the wires is \(1.2 \mathrm{~m}\). Find the speed at which individual waves travel on each wire. Verify that your answer is consistent with your answers to the Concept Questions.

Two loudspeakers face each other, vibrate in phase, and produce identical 440 -Hz tones. A listener walks from one speaker toward the other at a constant speed and hears the loudness change (loud-soft-loud) at a frequency of \(3.0 \mathrm{~Hz} .\) The speed of sound is \(343 \mathrm{~m} /\) s. What is the walking speed?

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Sound exits a diffraction horn loudspeaker through a rectangular opening like a small doorway. Such a loudspeaker is mounted outside on a pole. In winter, when the temperature is \(273 \mathrm{~K}\), the diffraction angle \(\theta\) has a value of \(15.0^{\circ} .\) What is the diffraction angle for the same sound on a summer day when the temperature is \(311 \mathrm{~K}\) ?

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