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Chapter 17: Problem 52

Many artists sing very high notes in ad lib ornaments and cadenzas. The highest note written for a singer in a published score was F-sharp above high \(\mathrm{C}, 1.480 \mathrm{kHz}\), for Zerbinetta in the original version of Richard Strauss's opera Ariadne auf Naxos. (a) Find the wavelength of this sound in air. (b) Suppose people in the fourth row of seats hear this note with level \(81.0 \mathrm{dB} .\) Find the displacement amplitude of the sound. (c) What If? Because of complaints, Strauss later transposed the note down to \(\mathrm{F}\) above high \(\mathrm{C}, 1.397 \mathrm{kHz}\). By what increment did the wavelength change?

Short Answer

Expert verified
The wavelength of the sound in air is 0.232 m; the displacement amplitude of the sound is 5.63 * 10^{-6} m; the increment in wavelength due to the transposition is 1.4 cm.

Step by step solution

01

Calculate the wavelength

The frequency of the highest note is given as 1480 Hz. We can use the relationship between the speed of sound (\(v\)), frequency (\(f\)), and wavelength (\(位\)) which is \(v = f * 位\). Here, we can assume the speed of sound in air to be approximately 343 m/s. Solving for \(位\), we get: \(位 = v / f = 343 / 1480 = 0.232 m\).
02

Calculate the displacement amplitude

The sound level is given as 81.0 dB. The displacement amplitude of a sound (\(A\)) can be related back to the sound level using the formula: \(尾 = 20 log(A / A_0)\), where \(尾\) is the sound level in dB, \(A_0 = 2*10^{-9} m\) is the reference displacement amplitude. We need to solve for \(A\), which gives us: \(A = A_0 * 10^(尾/20) = 2*10^{-9} * 10^((81.0)/20) = 5.63 * 10^{-6} m\).
03

Calculate the difference in wavelength

The frequency of the transposed note is given as 1397 Hz. Using the formula from step 1, we can calculate the new wavelength: \(位' = v / f' = 343 / 1397 = 0.246 m\). The increment in wavelength due to the transposition is the difference between the final and initial wavelengths: \(螖位 = 位' - 位 = 0.246 - 0.232 = 0.014 m = 1.4 cm\).

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Key Concepts

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

Frequency
Understanding frequency is essential when studying sound waves. Frequency refers to how many cycles of a wave pass a point in one second. We measure this in Hertz (Hz).
In the exercise, the highest note reached 1480 Hz. That means 1480 cycles of the sound wave occur each second.
This high frequency results in a high-pitched sound, like that of a classical soprano's high note.
When sound is transposed to a lower frequency, like in Strauss's modification to 1397 Hz, each cycle takes slightly longer. This results in a lower pitch.
The human ear perceives these frequency changes as variations in the pitch of sound. High frequencies produce high pitches, and low frequencies produce low pitches. Remember that frequency and pitch are directly related.
To summarize:
  • Frequency is measured in Hertz (Hz)
  • Frequency determines pitch: higher frequency = higher pitch
  • A change in frequency changes the note's pitch
Wavelength
Wavelength is another crucial aspect of sound waves. It is the distance between two corresponding points on consecutive cycles of a wave, like from peak to peak.
Wavelength (\(位\)) is related to frequency (\(f\)) and the speed of sound (\(v\)), given by the formula: \[位 = \frac{v}{f}\].Stated simply, as the frequency increases, the wavelength decreases and vice versa.
In the given example, the original frequency of 1480 Hz had a corresponding wavelength of 0.232 meters. Later, when the frequency was modified to 1397 Hz, the wavelength increased to 0.246 meters.
This is an example of when lowering the frequency results in a longer wavelength, as sound waves spread farther apart.
  • Wavelength is the distance between wave peaks
  • Inverse relationship: Higher frequency = shorter wavelength
  • Speed of sound in air is often around 343 m/s at room temperature
Decibels
Decibels (dB) provide a way to measure the intensity of sound. Unlike frequency and wavelength, which deal with the tone, decibels tell us about volume and loudness.
In the exercise, the sound level was 81 dB. This indicates the loudness perceived by the audience. The human ear can detect sounds across an enormous range of amplitudes. Thus, a logarithmic scale was created using decibels to express this range in a more manageable form.
To calculate the displacement amplitude (\(A\)), we use the formula: \[尾 = 20 \log_{10}\left(\frac{A}{A_0}\right)\] , where \(A_0\) is a reference displacement amplitude.In the exercise, this calculation yields a displacement amplitude of 5.63 x 10^-6 meters. This tells us about the wave's actual movement and energy, even when expressed as a higher decibel level.
  • Decibels express sound intensity using a logarithmic scale
  • Higher decibels indicate louder sound
  • Logarithmic scale allows for better handling of vast range in sound intensities

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Most popular questions from this chapter

Find the speed of sound in mercury, which has a bulk modulus of approximately \(2.80 \times 10^{10} \mathrm{N} / \mathrm{m}^{2}\) and a density of \(13600 \mathrm{kg} / \mathrm{m}^{3}\)

A rescue plane flies horizontally at a constant speed searching for a disabled boat. When the plane is directly above the boat, the boat's crew blows a loud horn. By the time the plane's sound detector perceives the horn's sound, the plane has traveled a distance equal to half its altitude above the ocean. If it takes the sound 2.00 s to reach the plane, determine (a) the speed of the plane and (b) its altitude. Take the speed of sound to be \(343 \mathrm{m} / \mathrm{s}\)

A bat, moving at \(5.00 \mathrm{m} / \mathrm{s},\) is chasing a flying insect (Fig. \(P 17.7) .\) If the bat emits a \(40.0 \mathrm{kHz}\) chirp and receives back an echo at \(40.4 \mathrm{kHz},\) at what speed is the insect moving toward or away from the bat? (Take the speed of sound in air to be \(v=340 \mathrm{m} / \mathrm{s} .\))

Ultrasound is used in medicine both for diagnostic imaging and for therapy. For diagnosis, short pulses of ultrasound are passed through the patient's body. An echo reflected from a structure of interest is recorded, and from the time delay for the return of the echo the distance to the structure can be determined. A single transducer emits and detects the ultrasound. An image of the structure is obtained by reducing the data with a computer. With sound of low intensity, this technique is noninvasive and harmless. It is used to examine fetuses, tumors, aneurysms, gallstones, and many other structures. A Doppler ultrasound unit is used to study blood flow and functioning of the heart. To reveal detail, the wavelength of the reflected ultrasound must be small compared to the size of the object reflecting the wave. For this reason, frequencies in the range 1.00 to \(20.0 \mathrm{MHz}\) are used. What is the range of wavelengths corresponding to this range of frequencies? The speed of ultrasound in human tissue is about \(1500 \mathrm{m} / \mathrm{s}\) (nearly the same as the speed of sound in water).

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.
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