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Chapter 17: Problem 29

Consider a sound wave modeled with the equation \(s(x, t)=4.00 \mathrm{nm} \cos \left(3.66 \mathrm{m}^{-1} x-1256 \mathrm{s}^{-1} t\right) .\) What is the maximum displacement, the wavelength, the frequency, and the speed of the sound wave?

Short Answer

Expert verified
The maximum displacement (amplitude) is \(4.00\,\text{nm}\), the wavelength is approximately \(1.72\,\text{m}\), the frequency is approximately \(200\,\text{Hz}\), and the speed of the sound wave is approximately \(344\,\text{m/s}\).

Step by step solution

01

Identify the Amplitude

In the wave equation \(s(x,t) = 4.00\,\text{nm} \cos(3.66\,\text{m}^{-1}x - 1256\,\text{s}^{-1}t)\), the maximum displacement (amplitude) is the coefficient of the cosine function. Thus, the amplitude is: \[A = 4.00\,\text{nm}\]
02

Identify the Wave Number and Angular Frequency

From the wave equation, we can identify the wave number \(k\) and the angular frequency \(\omega\) as the coefficients of \(x\) and \(t\), respectively. Therefore, we have: \[k = 3.66\,\text{m}^{-1}\] \[\omega = 1256\,\text{s}^{-1}\]
03

Calculate the Wavelength

The wavelength \(\lambda\) is related to the wave number \(k\) by the formula: \[\lambda = \frac{2\pi}{k}\] Substitute the value of \(k\) in the formula to get the wavelength: \[\lambda = \frac{2\pi}{3.66\,\text{m}^{-1}} \approx 1.72\,\text{m}\]
04

Calculate the Frequency

Frequency \(f\) is related to the angular frequency \(\omega\) by the formula: \[f = \frac{\omega}{2\pi}\] Plugging in the angular frequency, we can calculate the frequency: \[f = \frac{1256\,\text{s}^{-1}}{2\pi} \approx 200\,\text{Hz}\]
05

Calculate the Speed of the Sound Wave

The speed \(v\) of a wave is related to its wavelength and frequency through the equation: \[v = \lambda f\] Now, substituting the calculated wavelength and frequency values, we can find the speed: \[v = (1.72\,\text{m})(200\,\text{Hz}) \approx 344\,\text{m/s}\] So, the maximum displacement (amplitude) is \(4.00\,\text{nm}\), the wavelength is approximately \(1.72\,\text{m}\), the frequency is approximately \(200\,\text{Hz}\), and the speed of the sound wave is approximately \(344\,\text{m/s}\).

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Key Concepts

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

Wave Amplitude
The amplitude of a wave is a measure of the maximum displacement of particles from their equilibrium position when a wave passes. In the context of sound waves, amplitude is related to the loudness of the sound. The higher the amplitude, the louder the sound will be, because louder sounds move the particles in the medium (such as air) more from their rest positions.

For a wave modeled by the equation \( s(x, t) = A \cos(kx - \omega t) \), like the one we're considering for sound, the amplitude (\( A \)) is represented by the coefficient before the cosine function.
Wavelength Calculation
Wavelength is the distance between corresponding points of two consecutive waves. It is usually denoted by the Greek letter lambda (\( \lambda \)). The relationship between wavelength and wave number (\( k \)), as given by the equation \( \lambda = \frac{2\pi}{k} \), allows us to calculate the wavelength of a wave.

In simple terms, the wave number is like a spatial frequency, telling us how many wave cycles fit into a unit of distance. Thus, finding the wavelength involves taking the reciprocal of this spatial frequency (after adjusting for \( 2\pi \)), since frequency and wavelength are inversely related.
Frequency of a Wave
The frequency of a wave (\( f \)) is the number of occurrences of a repeating event per unit time, which in the case of a wave is the number of cycles that pass a point per second. It is measured in hertz (Hz). Calculating frequency from angular frequency (\( \omega \)), which is given in radians per second, uses the formula \( f = \frac{\omega}{2\pi} \).

Understanding frequency is important because it is directly related to the pitch of a sound鈥攖he higher the frequency, the higher the pitch. This concept is crucial in music and acoustic engineering, and it has practical implications for sound wave physics.
Speed of Sound
The speed of sound (\( v \)) is the distance that sound travels through an elastic medium per unit time. It depends on the medium and the state (temperature, pressure) of the medium. For a wave, its speed can be found by the product of its wavelength (\( \lambda \)) and frequency (\( f \)), resulting in the equation \( v = \lambda f \).

In different materials, the speed of sound varies; for instance, it is faster in solids than in liquids, and faster in liquids than in gases. This property of sound is explored in different fields, from designing musical instruments to understanding Earth's interior through seismic waves.

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Most popular questions from this chapter

Suppose a bat uses sound echoes to locate its insect prey, 3.00 m away. (See Figure 17.6.) (a) Calculate the echo times for temperatures of \(5.00^{\circ} \mathrm{C}\) and \(35.0^{\circ} \mathrm{C}\). (b) What percent uncertainty does this cause for the bat in locating the insect? (c) Discuss the significance of this uncertainty and whether it could cause difficulties for the bat. (In practice, the bat continues to use sound as it closes in, eliminating most of any difficulties imposed by this and other effects, such as motion of the prey.)

Sound is more effectively transmitted into a 2. stethoscope by direct contact rather than through the air, and it is further intensified by being concentrated on the smaller area of the eardrum. It is reasonable to assume that sound is transmitted into a stethoscope 100 times as effectively compared with transmission though the air. What, then, is the gain in decibels produced by a stethoscope that has a sound gathering area of \(15.0 \mathrm{cm}^{2}\) and concentrates the sound onto two eardrums with a total area of \(0.900 \mathrm{cm}^{2}\) with an efficiency of \(40.0 \%\) ?

Twin jet engines on an airplane are producing an average sound frequency of \(4100 \mathrm{Hz}\) with a beat frequency of \(0.500 \mathrm{Hz}\). What are their individual frequencies?

What is the decibel level of a sound that is twice as intense as a 90.0 -dB sound? (b) What is the decibel level of a sound that is one-fifth as intense as a 90.0-dB sound?

How long must a flute be in order to have a fundamental frequency of 262 Hz (this frequency corresponds to middle C on the evenly tempered chromatic scale) on a day when air temperature is \(20.0^{\circ} \mathrm{C}\) ? It is open at both ends.
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