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A sound wave is traveling in seawater, where the adiabatic bulk modulus and density are \(2.31 \times 10^{9}\) \(\mathrm{Pa}\) and \(1025\) \(\mathrm{kg} / \mathrm{m}^{3},\) respectively. The wavelength of the sound is \(3.35\) \(\mathrm{m} .\) A tuning fork is struck under water and vibrates at \(440.0\) \(\mathrm{Hz}\). What would be the beat frequency heard by an underwater swimmer?

Short Answer

Expert verified
The beat frequency is 8.6 Hz.

Step by step solution

01

Calculate Speed of Sound in Seawater

The speed of sound in seawater can be calculated using the formula \( v = \sqrt{\frac{B}{\rho}} \), where \( B \) is the adiabatic bulk modulus and \( \rho \) is the density. Substituting the given values, \( B = 2.31 \times 10^9 \, \text{Pa} \) and \( \rho = 1025 \, \text{kg/m}^3 \), we get: \[v = \sqrt{\frac{2.31 \times 10^9}{1025}} \approx 1500.3 \, \text{m/s}.\]
02

Determine the Frequency of the Sound Wave in Seawater

The frequency of a wave is given by the formula \( f = \frac{v}{\lambda} \), where \( \lambda \) is the wavelength. With the speed \( v \approx 1500.3 \, \text{m/s} \) and wavelength \( \lambda = 3.35 \, \text{m} \), the frequency becomes: \[f = \frac{1500.3}{3.35} \approx 448.6 \, \text{Hz}.\]
03

Calculate the Beat Frequency

The beat frequency is the absolute difference between the frequencies of two sound waves. Here, the frequency of the sound wave in seawater is \( 448.6 \, \text{Hz} \) and the tuning fork's frequency is \( 440.0 \, \text{Hz} \). Thus, the beat frequency is: \[\text{Beat Frequency} = |448.6 - 440.0| = 8.6 \, \text{Hz}.\]

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Key Concepts

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

Speed of Sound
The speed of sound is a fascinating concept that varies based on the medium it travels through. In this exercise, we calculated the speed of sound in seawater, using specific properties of the medium: the adiabatic bulk modulus and the density. The formula for speed of sound is given by:\[ v = \sqrt{\frac{B}{\rho}} \]where:
  • \( v \) is the speed of sound,
  • \( B \) is the adiabatic bulk modulus (a measure of how compressible a medium is),
  • \( \rho \) is the density of the medium.
In our calculation, substituting the values \( B = 2.31 \times 10^9 \, \mathrm{Pa} \) and \( \rho = 1025 \, \mathrm{kg/m^3} \), we determined the speed of sound in seawater to be approximately \( 1500.3 \, \mathrm{m/s} \).
It’s essential to remember that the speed of sound is generally faster in liquids and solids compared to gases. This is due to closer molecule spacing, allowing for quicker transmission of sound waves.
Beat Frequency
Beat frequency is an interesting phenomenon that occurs when two sound waves of different frequencies interfere with each other. When this happens, it creates a new sound pattern that oscillates between being louder and softer, known as beats. The beat frequency is calculated as the absolute difference between the frequencies of the two sound waves involved:\[ \text{Beat Frequency} = |f_1 - f_2| \]Where:
  • \( f_1 \) is the frequency of one sound source,
  • \( f_2 \) is the frequency of the other sound source.
In our example, an underwater swimmer hears the beats produced by a tuning fork of frequency \( 440.0 \, \mathrm{Hz} \) and the sound wave traveling through seawater at a frequency of \( 448.6 \, \mathrm{Hz} \).
The resulting beat frequency is \( 8.6 \, \mathrm{Hz} \). Beat frequencies not only provide insight into sound wave interference but are also used in applications like musical tuning and even sonar technology.
Frequency Calculation
The frequency of a sound wave is an essential property that describes how often sound waves pass a point in one second. It is measured in Hertz (Hz), with one Hertz equivalent to one complete wave cycle per second. To determine frequency, we use the formula:\[ f = \frac{v}{\lambda} \]where:
  • \( f \) is the frequency,
  • \( v \) is the speed of sound in the medium,
  • \( \lambda \) is the wavelength of the sound wave.
For the sound wave traveling in seawater, knowing that \( v \approx 1500.3 \, \mathrm{m/s} \) and the wavelength \( \lambda = 3.35 \, \mathrm{m} \), the frequency was calculated as \( 448.6 \, \mathrm{Hz} \).
Understanding frequency is crucial, as it relates directly to the pitch of the sound we hear. Higher frequencies produce higher pitches, while lower frequencies result in lower pitches. Frequency calculations play a vital role in fields like acoustics, music, and telecommunications.

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Most popular questions from this chapter

A string has a linear density of \(8.5 \times 10^{-3}\) \(\mathrm{kg} / \mathrm{m}\) and is under a tension of \(280\) \(\mathrm{N}\). The string is \(1.8\) \(\mathrm{m}\) long, is fixed at both ends, and is vibrating in the standing wave pattern shown in the drawing. Determine the (a) speed, (b) wavelength, and (c) frequency of the traveling waves that make up the standing wave.

Suppose that the strings on a violin are stretched with the same tension and each has the same length between its two fixed ends. The musical notes and corresponding fundamental frequencies of two of these strings are \(\mathrm{G}(196.0 \mathrm{Hz})\) and \(\mathrm{E}(659.3 \mathrm{Hz}) .\) The linear density of the \(\mathrm{E}\) string is \(3.47 \times\) \(10^{-4} \mathrm{kg} / \mathrm{m} .\) What is the linear density of the G string?

Speakers \(\mathrm{A}\) and \(\mathrm{B}\) are vibrating in phase. They are directly facing each other, are \(7.80 \mathrm{m}\) apart, and are each playing a \(73.0-\mathrm{Hz}\) tone. The speed of sound is \(343 \mathrm{m} / \mathrm{s}\). On the line between the speakers there are three points where constructive interference occurs. What are the distances of these three points from speaker A?

Two speakers, one directly behind the other, are each generating a \(245-\mathrm{Hz}\) sound wave. What is the smallest separation distance between the speakers that will produce destructive interference at a listener standing in front of them? The speed of sound is \(343\) \(\mathrm{m} / \mathrm{s}\).

Divers working in underwater chambers at great depths must deal with the danger of nitrogen narcosis (the "bends"), in which nitrogen dissolves into the blood at toxic levels. One way to avoid this danger is for divers to breathe a mixture containing only helium and oxygen. Helium, however, has the effect of giving the voice a high-pitched quality, like that of Donald Duck's voice. To see why this occurs, assume for simplicity that the voice is generated by the vocal cords vibrating above a gas-filled cylindrical tube that is open only at one end. The quality of the voice depends on the harmonic frequencies generated by the tube; larger frequencies lead to higher-pitched voices. Consider two such tubes at \(20^{\circ} \mathrm{C} .\) One is filled with air, in which the speed of sound is \(343 \mathrm{m} / \mathrm{s} .\) The other is filled with helium, in which the speed of sound is \(1.00 \times 10^{3} \mathrm{m} / \mathrm{s}\) To see the effect of helium on voice quality, calculate the ratio of the \(n\) th natural frequency of the helium-filled tube to the \(n\) th natural frequency of the air-filled tube.

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