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Chapter 17: Problem 47

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 (at the bottom) and one open end (at the top). 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 about 12.42 meters down in the well.

Step by step solution

01

Calculate the speed of sound in air

The speed of sound in a medium is given by the formula: \[ v = \sqrt{\frac{B}{\rho}} \]where \( B \) is the bulk modulus and \( \rho \) is the density of the medium. Here, \( B = 1.33 \times 10^5 \mathrm{~Pa} \) and \( \rho = 1.10 \mathrm{~kg/m^3} \).Let's calculate the speed of sound:\[ v = \sqrt{\frac{1.33 \times 10^5}{1.10}} \approx \sqrt{120909.09} \approx 347.67 \mathrm{~m/s} \].
02

Determine the wavelength of the fundamental frequency

In a closed-end pipe, the fundamental frequency \( f_1 \) corresponds to a quarter wavelength \( \lambda/4 \) fitting into the length \( L \) of the pipe. The wavelength \( \lambda \) can be found by the formula:\[ \lambda = \frac{v}{f} \] Given \( v = 347.67 \mathrm{~m/s} \) and \( f = 7.00 \mathrm{~Hz} \), we find:\[ \lambda = \frac{347.67}{7.00} \approx 49.67 \mathrm{~m} \].
03

Calculate the length of the air column

For a closed-end pipe, the fundamental frequency occurs when the length of the air column (\( L \)) is a quarter of the wavelength \( \lambda \). Hence:\[ L = \frac{\lambda}{4} \]Substitute \( \lambda = 49.67 \mathrm{~m} \):\[ L = \frac{49.67}{4} \approx 12.42 \mathrm{~m} \]Thus, the water surface is approximately 12.42 meters down in the well.

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Key Concepts

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

Resonance
Resonance occurs when an object or system vibrates at its natural frequency due to external energy matching this frequency. It's like when you push a child on a swing at the right moment, the swings get higher. In acoustics, resonance happens when sound waves reinforce each other, making the sound louder.
  • This concept is crucial in musical instruments, allowing them to produce powerful and beautiful sounds.
  • In a physical setting like a well, resonance helps determine the audible frequencies produced.
Understanding resonance is key to grasping how certain frequencies work in harmony with their environments, leading to amplified sound.
Speed of Sound
The speed of sound is the distance that sound waves travel through a medium in a given time. Sound moves at different speeds depending on the medium's properties, such as its density and elasticity.

Through the formula: \[ v = \sqrt{\frac{B}{\rho}} \]which calculates speed using the bulk modulus \( B \) (stiffness of the medium) and the density \( \rho \). In our example, sound travels at approximately 347.67 meters per second in the well due to the given conditions. Remember that:
  • Sound travels faster in liquids and solids than in gases, due to particles being closer together.
  • Factors like temperature can also influence the speed of sound in air.
Being familiar with this concept helps in predicting how sound propagates through various environments.
Closed-End Pipe
A closed-end pipe is a structure with one end sealed, generally producing unique sound characteristics. This set-up allows a quarter wavelength of sound to fit in the pipe, creating a distinctive pattern of harmonics.
  • The closed end acts as a node, where the displacement of particles is minimal.
  • The open end serves as an antinode, where the displacement of particles is maximal.
Understanding how sound behaves in closed-end pipes helps clarify why certain pitch patterns occur and is pivotal in fields such as musical acoustics and engineering sound systems.
Fundamental Frequency
The fundamental frequency is the lowest frequency that can produce resonant standing waves in a given system. Think of it as the root sound of a musical note. This frequency determines the pitch of the sound that we hear.

For a closed-end pipe, this corresponds to a quarter-wavelength fitting into the length of the air column.
  • Formula: \[ \lambda = \frac{v}{f} \]where \( \lambda \) is the wavelength, \( v \) the speed of sound, and \( f \) the frequency.
  • The fundamental frequency sets the base tone from which overtones or harmonics are built.
Understanding the fundamental frequency enables us to calculate and predict the sound characteristics of varied acoustic setups efficiently.

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Most popular questions from this chapter

A sound wave in a fluid medium is reflected at a barrier so that a standing wave is formed. The distance between nodes is \(3.8 \mathrm{~cm},\) and the speed of propagation is \(1500 \mathrm{~m} / \mathrm{s} .\) Find the frequency of the sound wave.

You can estimate your distance from a lightning stroke by counting the seconds between the flash you see and the thunder you later hear. By what integer should you divide the number of seconds to get the distance in kilometers?

At a certain point, two waves produce pressure variations given by \(\Delta p_{1}=\Delta p_{m} \sin \omega t\) and \(\Delta p_{2}=\Delta p_{m} \sin (\omega t-\phi) .\) At this point, what is the ratio \(\Delta p_{r} / \Delta p_{m},\) where \(\Delta p_{r}\) is the pressure amplitude of the resultant wave, if \(\phi\) is (a) \(0,\) (b) \(\pi / 2,\) (c) \(\pi / 3\), and (d) \(\pi / 4 ?\)

A sound wave of frequency \(300 \mathrm{~Hz}\) has an intensity of \(1.00 \mu \mathrm{W} / \mathrm{m}^{2} .\) What is the amplitude of the air oscillations caused by this wave?

Figure 17-48 shows an air filled, acoustic interferometer, used to demonstrate the interference of sound waves. Sound source \(S\) is an oscillating diaphragm; \(D\) is a sound detector, such as the ear or a microphone. Path \(S B D\) can be varied in length, but path \(S A D\) is fixed. At \(D,\) the sound wave coming along path \(S B D\) interferes with that coming along path \(S A D .\) In one demonstration, the sound intensity at \(D\) has a minimum value of 100 units at one position of the movable arm and continuously climbs to a maximum value of 900 units when that arm is shifted by \(1.65 \mathrm{~cm} .\) Find (a) the frequency of the sound emitted by the source and (b) the ratio of the amplitude at \(D\) of the \(S A D\) wave to that of the \(S B D\) wave. (c) How can it happen that these waves have different amplitudes, considering that they originate at the same source?
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