/*! 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 23 The most soaring vocal melody is... [FREE SOLUTION] | 91影视

91影视

The most soaring vocal melody is in Johann Sebastian Bach's Mass in \(B\) minor. A portion of the score for the Credo section, number \(9,\) bars 25 to \(33,\) appears in Figure \(P 17.23\) The repeating syllable \(\mathrm{O}\) in the phrase "resurrectionem mortuorum" (the resurrection of the dead) is seamlessly passed from basses to tenors to altos to first sopranos, like a baton in a relay. Each voice carries the melody up in a run of an octave or more. Together they carry it from D below middle C to A above a tenor's high C. In concert pitch, these notes are now assigned frequencies of \(146.8 \mathrm{Hz}\) and \(880.0 \mathrm{Hz}\). (a) Find the wavelengths of the initial and final notes. (b) Assume that the choir sings the melody with a uniform sound level of \(75.0 \mathrm{dB} .\) Find the pressure amplitudes of the initial and final notes. (c) Find the displacement amplitudes of the initial and final notes. (d) What If? In Bach's time, before the invention of the tuning fork, frequencies were assigned to notes as a matter of immediate local convenience. Assume that the rising melody was sung starting from \(134.3 \mathrm{Hz}\) and ending at \(804.9 \mathrm{Hz}\). How would the answers to parts (a) through (c) change?

Short Answer

Expert verified
The wavelengths, pressure amplitudes, and displacement amplitudes of the initial and final notes will all be affected by a change in frequency. The specific values for these quantities will be calculated in the solution steps.

Step by step solution

01

Calculate the Initial and Final Wavelengths

We know that the velocity (v) of sound in air at room temperature is approximately \(343 m/s\). The relationship between velocity, frequency (f), and wavelength (位) is given by the formula: v = f * 位. Therefore, we can rearrange this formula to find the wavelength: 位 = v / f. Plugging in the given frequencies, we can find the wavelengths of the initial and final notes.
02

Calculate the Pressure Amplitudes

Sound level 饾惪 is related to pressure amplitude 饾憙 by \(饾惪=20 log(P/P0)\) where \(饾憙0=2脳10^(-5) Pa\) is the reference pressure amplitude (the quietest sound that the average human ear can hear). We can rearrange this equation to solve for 饾憙 using the given sound level of 75dB.
03

Calculate the Displacement Amplitudes

The pressure amplitude P and the displacement amplitude s are related through the formula \(P = s * 蟻 * v * 蠅\) where 蟻 is the density of air, approximately \(1.21 kg/m^3\) at room temperature; v is the speed of sound in air, and 蠅 is the angular frequency. We can rearrange this formula to find s.
04

Recalculate for Different Frequencies

Lastly, we repeat steps 1-3 now with the frequencies of \(134.3 Hz\) and \(804.9 Hz\). Then we can compare these results to the results from the original frequencies.

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.

Wavelength
When it comes to sound waves, the term "wavelength" is central in understanding how waves propagate through mediums like air. Wavelength, represented by the Greek letter \( \lambda \), is the distance between two successive crests or troughs in a wave.
For sound waves traveling through air, which is a common scenario in acoustics, the wavelength is linked to the speed of sound and the frequency of the wave. The formula to determine wavelength is:
  • \( \lambda = \frac{v}{f} \)
Here, \( v \) is the speed of sound (approximately 343 m/s in air at room temperature), and \( f \) is the frequency of the wave.
So, by knowing the frequency of a sound note, such as those sung in a melody, we can calculate the wavelength, giving us insight into the spatial characteristics of sound.
Pressure Amplitude
Pressure amplitude is an essential concept when discussing the intensity or loudness of sound waves. It refers to the maximum change in pressure exerted by a sound wave in the surrounding air.
In the context of human perception, sound levels are often measured in decibels (dB), which express the pressure amplitude relative to a reference level. The formula to relate sound level \( L \) to pressure amplitude \( P \) is:
  • \( L = 20 \log\left( \frac{P}{P_0} \right) \)
Here, \( P_0 = 2 \times 10^{-5} \) Pa is the reference pressure amplitude of the softest sound the average human ear can hear.
When singing at 75 dB, the pressure amplitude can be calculated by rearranging the formula above. Understanding pressure amplitude is crucial, especially in musical settings, as it helps us comprehend how intensely a note is perceived by our ears.
Displacement Amplitude
Displacement amplitude describes how far the particles in the medium through which a sound wave travels are moved from their rest position. This movement reflects part of the sound waves' energy.
Displacement amplitude \( s \) can be calculated using the relation:
  • \( P = s \cdot \rho \cdot v \cdot \omega \)
Where \( P \) is the pressure amplitude, \( \rho \) is air density (typically around 1.21 kg/m鲁 at room temperature), \( v \) is the speed of sound, and \( \omega \) is the angular frequency, which is linked to frequency by \( \omega = 2 \pi f \).
Displacement amplitude provides insights into the energy transport and the physical movement of the sound wave, adding a layer of depth to our understanding of sound dynamics.
Frequency
Frequency deals with how often the sound wave oscillates as it travels through a medium. It is usually expressed in hertz (Hz) and fundamentally represents the number of wave cycles that pass a specific point in one second.
Musically, frequency defines the pitch of a note. Higher frequencies correspond to higher pitches, and vice versa.
  • Frequency \( f \) can be mathematically defined by the inverse of the period of the wave.
A key part of the exercise involves identifying how frequency correlates with other elements like wavelength and amplitude. For example, given the constant speed of sound, frequency inversely affects wavelength: if the frequency increases, the wavelength decreases, provided the speed of sound remains the same. Thus, notes from a musical piece, like those in Bach's Mass in \( B \) minor, have unique frequencies that influence their perceived characteristics.

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

This problem represents a possible (but not recommended) way to code instantaneous pressures in a sound wave into 16 -bit digital words. Example 17.2 mentions that the pressure amplitude of a \(120-\mathrm{dB}\) sound is \(28.7 \mathrm{N} / \mathrm{m}^{2}\) Let this pressure variation be represented by the digital -code \(65536 .\) Let zero pressure variation be represented on the recording by the digital word \(0 .\) Let other intermediate pressures be represented by digital words of intermediate -size, in direct proportion to the pressure. (a) What digital word would represent the maximum pressure in a \(40 \mathrm{dB}\) -sound? (b) Explain why this scheme works poorly for soft -sounds. (c) Explain how this coding scheme would clip off half of the waveform of any sound, ignoring the actual -shape of the wave and turning it into a string of zeros. By introducing sharp corners into every recorded waveform, this coding scheme would make everything sound like a buzzer or a kazoo.

The tensile stress in a thick copper bar is \(99.5 \%\) of its elastic breaking point of \(13.0 \times 10^{10} \mathrm{N} / \mathrm{m}^{2} .\) If a \(500-\mathrm{Hz}\) sound wave is transmitted through the material, (a) what displacement amplitude will cause the bar to break? (b) What is the maximum speed of the elements of copper at this moment? (c) What is the sound intensity in the bar?

To permit measurement of her speed, a skydiver carries a buzzer emitting a steady tone at \(1800 \mathrm{Hz}\). A friend on the ground at the landing site directly below listens to the amplified sound he receives. Assume that the air is calm and that the sound speed is \(343 \mathrm{m} / \mathrm{s}\), independent of altitude. While the skydiver is falling at terminal speed, her friend on the ground receives waves of frequency \(2150 \mathrm{Hz}\) (a) What is the skydiver's speed of descent? (b) What If? Suppose the skydiver can hear the sound of the buzzer reflected from the ground. What frequency does she receive?

On a Saturday morning, pickup trucks and sport utility vehicles carrying garbage to the town dump form a nearly steady procession on a country road, all traveling at \(19.7 \mathrm{m} / \mathrm{s} .\) From one direction, two trucks arrive at the dump every 3 min. A bicyclist is also traveling toward the dump, at \(4.47 \mathrm{m} / \mathrm{s}\). (a) With what frequency do the trucks pass him? (b) What If? A hill does not slow down the trucks, but makes the out-of-shape cyclist's speed drop to \(1.56 \mathrm{m} / \mathrm{s} .\) How often do noisy, smelly, inefficient, garbage-dripping, roadhogging trucks whiz past him now?

Write an expression that describes the pressure variation as a function of position and time for a sinusoidal sound wave in air, if \(\lambda=0.100 \mathrm{m}\) and \(\Delta P_{\max }=0.200 \mathrm{N} / \mathrm{m}^{2}\)

See all solutions

Recommended explanations on Physics Textbooks

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.