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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 12.44 meters down in the well.

Step by step solution

01

Understand the Problem

We need to find the distance from the top of the well to the water surface at the bottom by considering the air column inside the well as an air column in a tube with one end closed. It resonates at a frequency of 7.00 Hz.
02

Use the Formula for Density and Bulk Modulus

The speed of sound is given by \( v = \sqrt{\frac{B}{\rho}} \), where \( B = 1.33 \times 10^5 \) Pa is the bulk modulus, and \( \rho = 1.10 \) kg/m³ is the density. Substituting these values gives: \[ v = \sqrt{\frac{1.33 \times 10^5}{1.10}} \].
03

Calculate Speed of Sound

Calculate \( v = \sqrt{\frac{1.33 \times 10^5}{1.10}} \approx 348.43 \) m/s. This is the speed of sound in the air column inside the well.
04

Apply the Resonance in a Closed Tube

For a tube closed at one end, the fundamental frequency (first harmonic) is \( f = \frac{v}{4L} \), where \( L \) is the length of the air column. Since the fundamental frequency given is 7.00 Hz, we have \( 7.00 = \frac{348.43}{4L} \).
05

Solve for the Length of the Air Column

Rearranging \( 7.00 = \frac{348.43}{4L} \), we get \( L = \frac{348.43}{4 \times 7.00} \).
06

Calculate the Length L

Calculate \( L \approx \frac{348.43}{28} \approx 12.44 \) meters. This is the distance from the top of the well to the water surface.

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Key Concepts

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

Resonance
Resonance is an essential concept in acoustics and refers to the tendency of a system to vibrate with increased amplitude at certain frequencies. When a system such as an air column in a well is subjected to oscillations, resonance can occur when the frequency of these oscillations matches the system's natural frequency. This matching amplifies the sound waves, resulting in louder and more distinct sounds.
In the context of our exercise, the air column inside the well resonates exactly at 7 Hz, indicating that 7 Hz is a natural frequency of this system. The resonance of such a system is influenced by various factors including the shape of the air column and the physical properties of the air itself.
Speed of Sound
The speed of sound in a medium is a crucial factor determining how quickly sound waves travel through that medium. It plays a significant role in calculating the resonance frequency of air columns. In our exercise, this speed plays a key role in determining the resonance pattern inside the well.
To find the speed of sound in air, we use the formula derived from the properties of the air itself: \[ v = \sqrt{\frac{B}{\rho}} \]This formula relates the speed of sound \(v\) to the bulk modulus \(B\) (a measure of the medium's resistance to uniform compression) and the air's density \(\rho\). In practical terms, a higher speed of sound results in smaller wavelengths for the same frequency, affecting how resonances form within the column.
Harmonics
Harmonics are overtones that accompany the fundamental frequency in resonance phenomena. In closed tubes like the air column in the well, the fundamental frequency (the lowest frequency of resonance) is the first harmonic. The first harmonic can be understood by the relationship \( f = \frac{v}{4L} \) for a tube closed at one end, where \(L\) is the length of the tube.
Harmonics in closed tubes are odd multiples of the fundamental frequency, such as the third harmonic, fifth harmonic, and so on. Each harmonic represents a different mode of vibration, with nodes and antinodes appearing at particular points along the length of the air column. In the context of a well, these harmonics can affect the depth to which the water level is perceived based on the vibration pattern.
Bulk Modulus
The bulk modulus is a fundamental property of materials that describes their response to uniform pressure applied from all sides. It quantifies how compressible or incompressible a material is. In the context of acoustics, the bulk modulus of air (\(1.33 \times 10^{5} \text{ Pa}\) in the given exercise) is used alongside density to determine how swiftly sound waves will move through the air.
This property is intrinsic to the molecular structure of the medium and directly affects sound propagation. A higher bulk modulus typically means that sound travels faster through the medium because molecules re-establish equilibrium positions more quickly after being disturbed by sound waves.
Density of Air
Density is another critical property that influences sound propagation in air. It is defined as the mass per unit volume and, in this exercise, the air's density is given as \(1.10 \text{ kg/m}^3\).
Density plays a dual role: while a higher density might suggest a slower speed of sound due to increased mass, the actual speed of sound in a gas is more intricately linked to temperature, pressure, and molecular makeup. In our case, alongside the bulk modulus, density is vital for calculating the velocity of sound in air, as previously discussed, through the relationship\[ v = \sqrt{\frac{B}{\rho}} \].
This understanding is essential to comprehending sound behavior in different environmental conditions.

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Most popular questions from this chapter

A sperm whale (Fig. \(17-43 a\) ) vocalizes by producing a series of clicks. Actually, the whale makes only a single sound near the front of its head to start the series. Part of that sound then emerges from the head into the water to become the first click of the series. The rest of the sound travels backward through the spermaceti sac (a body of fat), reflects from the frontal sac (an air layer), and then travels forward through the spermaceti sac. When it reaches the distal sac (another air layer) at the front of the head, some of the sound escapes into the water to form the second click, and the rest is sent back through the spermaceti sac (and ends up forming later clicks). Figure \(17-43 b\) shows a strip-chart recording of a series of clicks. A unit time interval of \(1.0 \mathrm{~ms}\) is indicated on the chart. Assuming that the speed of sound in the spermaceti sac is \(1372 \mathrm{~m} / \mathrm{s}\), find the length of the spermaceti sac. From such a calculation, marine scientists estimate the length of a whale from its click series.

What is the bulk modulus of oxygen if \(32.0 \mathrm{~g}\) of oxygen occupies \(22.4 \mathrm{~L}\) and the speed of sound in the oxygen is \(317 \mathrm{~m} / \mathrm{s}\) ?

A pipe \(0.60 \mathrm{~m}\) long and closed at one end is filled with an unknown gas. The third lowest harmonic frequency for the pipe is 750 Hz. (a) What is the speed of sound in the unknown gas? (b) What is the fundamental frequency for this pipe when it is filled with the unknown gas?

A sinusoidal sound wave moves at \(343 \mathrm{~m} / \mathrm{s}\) through air in the positive direction of an \(x\) axis. At one instant, air molecule \(A\) is at its maximum displacement in the negative direction of the axis while air molecule \(B\) is at its equilibrium position. The separation between those molecules is \(15.0 \mathrm{~cm}\), and the molecules between \(A\) and \(B\) have intermediate displacements in the negative direction of the axis. (a) What is the frequency of the sound wave? In a similar arrangement, for a different sinusoidal sound wave, air molecule \(C\) is at its maximum displacement in the positive direction while molecule \(D\) is at its maximum displacement in the negative direction. The separation between the molecules is again \(15.0 \mathrm{~cm}\), and the molecules between \(C\) and \(D\) have intermediate displacements. (b) What is the frequency of the sound wave?

Two sounds differ in sound level by \(1.00 \mathrm{~dB}\). What is the ratio of the greater intensity to the smaller intensity?

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