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Chapter 16: Problem 26

Many professional singers have a range of \(2 \frac{1}{2}\) octaves or even greater. Suppose a soprano's range extends from A below middle C (frequency \(220 \mathrm{~Hz}\) ) up to E-flat above high \(\mathrm{C}\) (frequency \(1244 \mathrm{~Hz}\) ). Although the vocal tract is complicated, we can model it as a resonating air column, like an organ pipe, that is open at the top and closed at the bottom. The column extends from the mouth down to the diaphragm in the chest cavity. Assume that the lowest note is the fundamental. How long is this column of air if \(v=354 \mathrm{~m} / \mathrm{s} ?\) Does your result seem reasonable, on the basis of observations of your body?

Short Answer

Expert verified
The length of the column of air in the singer's vocal tract is approximately \(0.4m\) or \(40cm\), which seems reasonable given the height of a typical human.

Step by step solution

01

Identify known quantities

From the problem, it is known that the frequency of the lowest note \(f = 220 Hz\) and the speed of sound in the singer's vocal tract \(v = 354 m/s\). These will be used directly as inputs into the formula for the length of the air column in the vocal tract.
02

Apply the frequency formula to find the length of the column of air

Using the formula \(l=\frac{v}{4f}\), one can substitute the given values. So, the length of the column of air \(l\) becomes \(l=\frac{354 m/s}{4 * 220 Hz}\).
03

Calculate the length of the column of air

After filling in the known quantities into the equation, perform the division to calculate the length of the column of air in the singer's vocal tract.
04

Evaluate the result

Check if the result seems reasonable by comparing it to the height of a typical human.

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Key Concepts

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

Fundamental Frequency
In acoustics, the fundamental frequency is the lowest frequency of a sound wave that can resonate in a given medium or structure. It's like the base note that can be produced before all other harmonic frequencies come into play. For a soprano singer, we can consider the lowest note in her range as her fundamental frequency.

When dealing with instruments or the human voice, the fundamental frequency determines the pitch that is perceived. For example:
  • A frequency of 220 Hz corresponds to the A note below middle C typically referred to as A3 in musical terms.
  • This fundamental note is crucial for establishing the full range of harmonics that define the character of a singer's voice.
Understanding the fundamental frequency helps singers and musicians tune their voices or instruments to create harmonious sounds.
Resonating Air Column
A resonating air column is a crucial concept in understanding how sounds are produced in wind instruments and the human vocal tract. In this model, the vocal tract can be thought of as a closed air column, similar to an organ pipe that is closed at one end. This is because the sound wave reflects at the closed end (near the diaphragm), creating standing waves that resonate.

For this type of system, the length of the air column determines the fundamental frequency, with the formula:
  • \[ l = \frac{v}{4f} \]
  • Where \( l \) is the length of the air column, \( v \) is the speed of sound, and \( f \) is the fundamental frequency.
This equation helps us calculate how long the air column must be for a particular fundamental frequency. It's essential because the length impacts the pitch produced, significantly affecting the singer's ability to hit different notes.
Vocal Tract Modeling
Modeling the vocal tract as a resonating air column helps simplify the study of how we produce sound. Although the real human vocal tract is complex and dynamic, it behaves similarly to simple closed-end instruments for the purposes of theoretical calculations.

The vocal tract consists of various parts:
  • The mouth, which shapes sound before it leaves the body.
  • The throat and nasal cavities, which affect the timbre and quality of the voice.
  • The diaphragm and chest cavity, which act as the closed end of our resonating model.
By understanding how these components work together, it is possible to comprehend how different sounds and pitches are formed. This modeling aids in training singers to optimize their vocal capabilities.
Sound Wave Speed
The speed of sound is the rate at which sound waves travel through a medium, like air. For this exercise, the speed of sound in the air within the vocal tract is taken as 354 m/s, slightly different from the typical 343 m/s due to variables like temperature and air density.

The speed of sound affects:
  • The calculation of wave frequencies and the resulting pitch.
  • The perception of the sound in different environments.
This parameter is vital when using the formula for the resonating air column, as it directly influences the calculated length needed to achieve a desired fundamental frequency. Recognizing the differences in sound speed in various conditions helps in making accurate predictions and adjustments in musical settings.

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Most popular questions from this chapter

At point \(A, 3.0 \mathrm{~m}\) from a small source of sound that is emitting uniformly in all directions, the sound intensity level is \(53 \mathrm{~dB}\). (a) What is the intensity of the sound at \(A ?\) (b) How far from the source must you go so that the intensity is one-fourth of what it was at \(A ?\) (c) How far must you go so that the sound intensity level is one-fourth of what it was at \(A ?\) (d) Does intensity obey the inverse-square law? What about sound intensity level?

You have a stopped pipe of adjustable length close to a taut \(62.0 \mathrm{~cm}, 7.25 \mathrm{~g}\) wire under a tension of \(4110 \mathrm{~N}\). You want to adjust the length of the pipe so that, when it produces sound at its fundamental frequency, this sound causes the wire to vibrate in its second overtone with very large amplitude. How long should the pipe be?

Horseshoe bats (genus Rhinolophus) emit sounds from their nostrils and then listen to the frequency of the sound reflected from their prey to determine the prey's speed. (The "horseshoe" that gives the bat its name is a depression around the nostrils that acts like a focusing mirror, so that the bat emits sound in a narrow beam like a flashlight.) A Rhinolophus flying at speed \(v_{\text {bat }}\) emits sound of frequency \(f_{\text {bat }} ;\) the sound it hears reflected from an insect flying toward it has a higher frequency \(f_{\text {refl }}\). (a) Show that the speed of the insect is $$ v_{\text {insect }}=v\left[\frac{f_{\text {ref }}\left(v-v_{\text {bat }}\right)-f_{\text {bat }}\left(v+v_{\text {bat }}\right)}{f_{\text {ref }}\left(v-v_{\text {bat }}\right)+f_{\text {bat }}\left(v+v_{\text {bat }}\right)}\right] $$ where \(v\) is the speed of sound. (b) If \(f_{\text {bat }}=80.7 \mathrm{kHz}, f_{\text {refl }}=83.5 \mathrm{kHz}\) and \(v_{\text {bat }}=3.9 \mathrm{~m} / \mathrm{s},\) calculate the speed of the insect.

A stationary police car emits a sound of frequency \(1200 \mathrm{~Hz}\) that bounces off a car on the highway and returns with a frequency of \(1250 \mathrm{~Hz}\). The police car is right next to the highway, so the moving car is traveling directly toward or away from it. (a) How fast was the moving car going? Was it moving toward or away from the police car? (b) What frequency would the police car have received if it had been traveling toward the other car at \(20.0 \mathrm{~m} / \mathrm{s} ?\)

A long spring such as a Slinkyrs is often used to demonstrate longitudinal waves. (a) Show that if a spring that obeys Hooke's law has mass \(m,\) length \(L\), and force constant \(k^{\prime}\), the speed of longitudinal waves on the spring is \(v=L \sqrt{k^{\prime} / m}\) (see Section 16.2 ). (b) Evaluate \(v\) for a spring with \(m=0.250 \mathrm{~kg}, L=2.00 \mathrm{~m},\) and \(k^{\prime}=1.50 \mathrm{~N} / \mathrm{m}\)
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