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Chapter 18: Problem 33

A well with vertical sides and water at the bottom resonates at \(7.00 \mathrm{~Hz}\) and at no lower frequency. (The air-filled portion of the well acts as a tube with one closed end and one open end.) The air in the well has a density of \(1.10 \mathrm{~kg} / \mathrm{m}^{3}\) and a bulk modulus of \(1.33 \times 10^{5} \mathrm{~Pa}\). How far down in the well is the water surface?

Short Answer

Expert verified
The water surface is approximately 12.45 meters down in the well.

Step by step solution

01

Identify the type of resonator

The well acts as a tube with one closed end and one open end. This type of tube supports standing waves with nodes at the closed end and antinodes at the open end.
02

Determine the formula for the fundamental frequency

For a tube with one closed end, the fundamental frequency (the lowest frequency at which resonance occurs) is given by: \[ f_1 = \frac{v}{4L} \] where \(v\) is the speed of sound in air, and \(L\) is the length of the tube (depth of the well).
03

Calculate the speed of sound in air

The speed of sound in air can be calculated using the formula: \[ v = \sqrt{\frac{B}{\rho}} \] where \(B\) is the bulk modulus of the air, and \(\rho\) is the density of air. Substituting the given values: \[ v = \sqrt{\frac{1.33 \times 10^5 \; \text{Pa}}{1.10 \; \text{kg/m}^3}} = \sqrt{1.21 \times 10^5} \approx 348.71 \; \text{m/s} \]
04

Calculate the length of the well

Using the calculated speed of sound and the fundamental frequency, substitute into the formula for the fundamental frequency: \[ f_1 = \frac{348.71 \; \text{m/s}}{4L} \]. Given the fundamental frequency (\(f_1\)) is 7.00 Hz: \[ 7.00 = \frac{348.71}{4L} \]. Solving for \(L\): \[ L = \frac{348.71}{4 \cdot 7.00} = \frac{348.71}{28} \approx 12.45 \; \text{m} \]

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Key Concepts

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

Fundamental Frequency
The fundamental frequency is the lowest frequency at which a system naturally vibrates. In the case of a tube with one closed end, this frequency is determined by the length of the tube and the speed of sound within it. For such a tube, the fundamental frequency can be calculated using the formula: \( f_1 = \frac{v}{4L} \) where \( f_1 \) is the fundamental frequency, \( v \) is the speed of sound, and \( L \) is the length of the tube. This relationship shows that as the length of the tube increases, the fundamental frequency decreases. This is because it takes sound waves longer to travel through a longer tube.
Speed of Sound
The speed of sound is how fast sound waves travel through a medium. Air is a common medium, and the speed of sound in air depends on factors like temperature, pressure, and humidity. In this exercise, the air in the well has a density of \(1.10 \; \text{kg/m}^3\) and a bulk modulus of \(1.33 \times 10^5 \; \text{Pa}\). We calculate the speed of sound in air using the formula: \( v = \sqrt{\frac{B}{\rho}} \) where \( B \) is the bulk modulus and \( \rho \) is the density. Substituting in the given values: \( v = \sqrt{\frac{1.33 \times 10^5}{1.10}} = \sqrt{1.21 \times 10^5} \approx 348.71 \; \text{m/s} \). Understanding the speed of sound is crucial since it directly influences the fundamental frequency of the air column.
Tube with One Closed End
A tube with one closed end, also known as a quarter-wave resonator, is a special kind of resonant system. In this setup, one end of the tube is closed, creating a node (where there is no movement of air particles), while the open end creates an antinode (where the air particles move the most). This unique setup allows the tube to have a fundamental frequency that is a quarter wavelength of the sound wave. The relationship between the length of the tube \( L \), the speed of sound \( v \), and the fundamental frequency \( f_1 \) is: \( f_1 = \frac{v}{4L} \). For practical applications, understanding how the length of the tube and the properties of air affect the resonance frequency is important in designing musical instruments and various acoustic devices.

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Most popular questions from this chapter

An ambulance with a siren emitting a whine at \(1600 \mathrm{~Hz}\) overtakes and passes a cyclist pedaling a bike at \(2.44 \mathrm{~m} / \mathrm{s}\). After being passed, the cyclist hears a frequency of \(1590 \mathrm{~Hz}\). How fast is the ambulance moving?

A tuning fork of unknown frequency makes three beats per second with a standard fork of frequency \(384 \mathrm{~Hz}\). The beat frequency decreases when a small piece of wax is put on a prong of the first fork. What is the freauency of this fork?

A French submarine and a U.S. submarine move toward each other during maneuvers in motionless water in the North Atlantic (Fig. 18-33). The French sub moves at \(50.0 \mathrm{~km} / \mathrm{h}\), and the U.S. sub at \(70.0 \mathrm{~km} / \mathrm{h}\). The French sub sends out a sonar signal (sound wave in water) at \(1000 \mathrm{~Hz}\). Sonar waves travel at \(5470 \mathrm{~km} / \mathrm{h}\). (a) What is the signal's frequency as detected by the U.S. sub? (b) What frequency is detected by the French sub in the signal reflected back to it by the U.S. sub?

A violin string \(30.0 \mathrm{~cm}\) long with linear density \(0.650 \mathrm{~g} / \mathrm{m}\) is placed near a loudspeaker that is fed by an audio oscillator of variable frequency. It is found that the string is set into oscillation only at the frequencies 880 and \(1320 \mathrm{~Hz}\) as the frequency of the oscillator is varied over the range \(500-1500 \mathrm{~Hz}\). What is the tension in the string?

A violin string \(15.0 \mathrm{~cm}\) long and fixed at both ends oscillates in its \(n=1\) mode. The speed of waves on the string is \(250 \mathrm{~m} / \mathrm{s}\), and the speed of sound in air is \(348 \mathrm{~m} / \mathrm{s}\). What are (a) the frequency and (b) the wavelength of the emitted sound wave?
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