/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 34 When you voice the vowel sound i... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 16: Problem 34

When you voice the vowel sound in "hat," you narrow the opening where your throat opens into the cavity of your mouth so that your vocal tract appears as two connected tubes. The first is in your throat, closed at the vocal cords and open at the back of the mouth. The second is the mouth itself, open at the lips and closed at the back of the mouth-a different condition than for the throat because of the relatively larger size of the cavity. The corresponding formant frequencies are \(800 \mathrm{Hz}\) (for the throat) and \(1500 \mathrm{Hz}\) (for the mouth). What are the lengths of these two cavities? Assume a sound speed of \(350 \mathrm{m} / \mathrm{s}\).

Short Answer

Expert verified
The length of the throat cavity is approximately 0.219m, while the length of the mouth cavity is approximately 0.117m.

Step by step solution

01

Rearrange the formula

Rearrange the equation \(f = v/(2L)\) to solve for \(L\), giving \(L = v/(2f)\). This is because you need to compute the length of the cavities and the speed of sound and frequency are provided.
02

Calculate the length of the throat cavity

Substitute the given values for the frequency of the throat cavity \(f = 800 \mathrm{Hz}\) and the speed of sound \(v = 350 \mathrm{m}/\mathrm{s}\) into the rearranged formula. Hence \(L_{throat} = 350 \mathrm{m} / \mathrm{s} / (2 \cdot 800 \mathrm{Hz})\).
03

Calculate the length of the mouth cavity

Repeat the same process as in Step 2, but using the given frequency for the mouth cavity \(f = 1500 \mathrm{Hz}\). So, \(L_{mouth} = 350 \mathrm{m} / \mathrm{s} / (2 \cdot 1500 \mathrm{Hz})\).
04

Simplify to find the final results

Simplify the expressions in Step 2 and 3 to obtain the lengths for the throat and mouth cavities respectively.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Vocal Tract
The vocal tract plays a crucial role in shaping the sounds we produce, including vowels like the one in "hat." It consists of several key components: the throat, mouth, and nasal passages. When we articulate sounds, these parts form a continuous tube-like structure.
It starts at the vocal cords and ends at the lips, significantly influencing the resonant properties of the sound.
  • The throat acts like a tube closed at the vocal cords and open at the back of the mouth.
  • The mouth can also be seen as a tube, but it is closed at the back of the mouth and open at the lips.
These anatomical features determine the resonant frequencies or formants, which give character to different vowel sounds.
This is why different shapes of tubes affect different frequency levels and lengths, as shown in this exercise.
Formant Frequencies
Formant frequencies are critical in determining how we perceive different vowel sounds. They result from the resonances that occur in the vocal tract when we produce speech sounds.
In our exercise, the formant frequencies of the vowel in "hat" are given as:
  • 800 Hz for the throat cavity
  • 1500 Hz for the mouth cavity
These frequencies are a direct product of the vocal tract's shape and size.
A formant is essentially a peak in the sound spectrum, and it results when a particular part of the vocal tract amplifies certain frequencies. Therefore, the formant frequencies provide important diagnostic information regarding the cavities, allowing us to determine their corresponding lengths through a sequence of calculations.
Sound Speed
The speed of sound is a vital component in our calculations involving the vocal tract. Under normal conditions, sound travels at approximately 343 m/s in air, but can vary based on temperature and medium.
In the problem at hand, sound speed is approximated to 350 m/s. This slight increase is typical for warmer air.
  • The speed of sound depends on factors like ambient temperature and air composition.
  • It influences how quickly sound waves travel through different mediums.
When evaluating the length of resonating cavities, sound speed helps in tying together frequency and physical dimensions via the formula: \[ f = \frac{v}{2L} \] where \( v \) is sound speed, \( f \) is frequency, and \( L \) is the length of the cavity.
Physics of Speech
The physics of speech encompasses the understanding of how sound is produced, transmitted, and perceived by our ears.
This involves various physical principles, including acoustics, which study sound waves and how they behave in different environments.
  • Vibration of the vocal cords initiates sound, acting like a source of sound waves.
  • The vocal tract then modifies these waves, enhancing certain frequencies and dampening others.
These modifications allow us to produce distinct sounds that form syllables and words. During speech, different muscular configurations adjust the shape of the vocal tract.
This action determines the resonant frequencies to produce various vowels.
In sum, speech is a complex but intriguing interaction between biological structures and the immutable principles of physics, elegantly highlighted in the exercise.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

In addition to producing images, ultrasound can be used to heat tissues of the body for therapeutic purposes. An emitter is placed against the surface of the skin; the amplitude of the ultrasound wave at this point is quite large. When a sound wave hits the boundary between soft tissue and bone, most of the energy is reflected. The boundary acts like the closed end of a tube, which can lead to standing waves. Suppose \(0.70 \mathrm{MHz}\) ultrasound is directed through a layer of tissue with a bone \(0.55 \mathrm{cm}\) below the surface. Will standing waves be created? Explain.

A drainage pipe running under a freeway is \(30.0 \mathrm{m}\) long. Both ends of the pipe are open, and wind blowing across one end causes the air inside to vibrate. a. If the speed of sound on a particular day is \(340 \mathrm{m} / \mathrm{s},\) what will be the fundamental frequency of air vibration in this pipe? b. What is the frequency of the lowest harmonic that would be audible to the human ear? c. What will happen to the frequency in the later afternoon as the air begins to cool?

The fundamental frequency of a standing wave on a \(1.0-\mathrm{m}-\) long string is \(440 \mathrm{Hz} .\) What would be the wave speed of a pulse moving along this string?

Piano tuners tune pianos by listening to the beats between the harmonics of two different strings. When properly tuned, the note A should have the frequency \(440 \mathrm{Hz}\) and the note E should be at \(659 \mathrm{Hz}\). The tuner can determine this by listening to the beats between the third harmonic of the A and the second harmonic of the E. A tuner first tunes the A string very precisely by matching it to a \(440 \mathrm{Hz}\) tuning fork. She then strikes the A and \(\mathrm{E}\) strings simultaneously and listens for beats between the harmonics. What beat frequency indicates that the E string is properly tuned?

Ocean waves of wavelength \(26 \mathrm{m}\) are moving directly toward a concrete barrier wall at \(4.4 \mathrm{m} / \mathrm{s}\). The waves reflect from the wall, and the incoming and reflected waves overlap to make a lovely standing wave with an antinode at the wall. (Such waves are a common occurrence in certain places.) A kayaker is bobbing up and down with the water at the first antinode out from the wall. How far from the wall is she? What is the period of her up-and-down motion?
See all solutions

Recommended explanations on Physics Textbooks

Waves Physics

Read Explanation

Measurements

Read Explanation

Physical Quantities And Units

Read Explanation

Rotational Dynamics

Read Explanation

Linear Momentum

Read Explanation

Nuclear Physics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.