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A well with vertical sides and water at the bottom resonates at \(9.00 \mathrm{~Hz}\) 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 approximately 9.68 meters down the well.

Step by step solution

01

Determine 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 and \( \rho \) is the density of the air. Substituting the given values, \( B = 1.33 \times 10^5 \) Pa and \( \rho = 1.10 \) kg/m³, we get \( v = \sqrt{\frac{1.33 \times 10^5}{1.10}} \approx 348.7 \) m/s.
02

Use Resonance Condition for a Closed-Open Tube

For a tube open at one end and closed at the other, the fundamental frequency \( f \) is given by \( f = \frac{v}{4L} \). Here, \( L \) is the length of the air column (depth of the water), and \( v \) is the speed of sound. Substitute \( f = 9.00 \) Hz and \( v = 348.7 \) m/s into this equation to solve for \( L \).
03

Solve for the Length of the Air Column

Rearranging the equation for the fundamental frequency, we have \( L = \frac{v}{4f} \). Substituting in the known values \( v = 348.7 \) m/s and \( f = 9.00 \) Hz, we get \( L = \frac{348.7}{4 \times 9.00} \approx 9.68 \) m.

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Key Concepts

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

Resonance
In physics, resonance is a fascinating phenomenon where a system vibrates with greater amplitude at a particular frequency. This frequency is known as the natural or resonant frequency. Imagine playing on a swing; when you match the timing of your pushes with the swing’s natural frequency, you swing higher. Frequency is the number of vibrations per second and is measured in hertz (Hz).

When we talk about resonance in the context of acoustic systems like a musical instrument or a well, understanding resonance is key to knowing how sound waves behave and amplify. In the exercise, the well resonates at 9 Hz. This means it is vibrating in harmony with a sound wave of 9 Hz, creating a louder sound.

  • Resonance amplifies sound waves, which can be both a desirable and undesirable effect, depending on the circumstance.
  • In closed-open tubes, such as the well in this problem, the resonance occurs at the lowest frequency, called the fundamental frequency.
  • The formula used, \( f = \frac{v}{4L} \), is a simple representation of how these harmonics work in a tube that is closed on one end and open on the other.
Speed of Sound
The speed of sound is the rate at which sound waves travel through a medium, like air. It's a crucial part of understanding acoustics because it affects how we hear sounds and how far they can travel. Calculating the speed of sound involves the properties of the medium, mainly its density and the bulk modulus.
The formula \( v = \sqrt{\frac{B}{\rho}} \) is used to determine the speed of sound in air, where:\
  • \( v \) is the speed of sound,
  • \( B \) is the bulk modulus that tells us how compressible the air is,
  • \( \rho \) is the density of the air.

In the exercise, using the given values for the bulk modulus and density, we calculated a speed of sound around 348.7 m/s. Why does this matter? Because this speed alongside the shape and characteristics of the well helps determine the resonant frequency. Any change in speed can drastically alter the tones produced or heard.
Waves in Air
Sound waves in air are longitudinal waves that consist of compressions and rarefactions as they move through the medium. When something like a guitar string vibrates, it pushes on the air particles nearby, creating regions of high and low pressure that move outward forming sound waves.

In closed-open systems, such as a partially filled well, these waves behave in unique patterns. The air column in the well only allows certain frequencies to resonate based on its length and the speed of sound. This very principle is what explains why we observe specific pitches or tones in different types of tubes and enclosed spaces.
  • The air acts as an elastic medium transmitting sound from one end to the other.
  • Since the well's one end is closed, the waves reflect back, creating standing waves.
  • These standing waves are integral in creating resonance at the 9 Hz frequency, where the length of the air column is perfectly set to amplify the sound.

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Most popular questions from this chapter

Organ pipe \(A\), with both ends open, has a fundamental frequency of \(425 \mathrm{~Hz}\). The fifth harmonic of organ pipe \(B\), with one end open, has the same frequency as the second harmonic of pipe \(A\). How long are (a) pipe \(A\) and (b) pipe \(B ?\)

Two identical piano wires have a fundamental frequency of 600 Hz when kept under the same tension. What fractional increase in the tension of one wire will lead to the occurrence of \(8.0\) beats/s when both wires oscillate simultaneously?

An ambulance with a siren emitting a whine at \(1620 \mathrm{~Hz}\) 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 point source emits \(30.0 \mathrm{~W}\) of sound isotropically. A small microphone intercepts the sound in an area of \(0.750 \mathrm{~cm}^{2}, 180 \mathrm{~m}\) from the source. Calculate (a) the sound intensity there and (b) the power intercepted by the microphone.

One of the harmonic frequencies of tube \(A\) with two open ends is \(400 \mathrm{~Hz}\). The next-highest harmonic frequency is \(480 \mathrm{~Hz}\). (a) What harmonic frequency is next highest after the harmonic frequency \(160 \mathrm{~Hz}\) ? (b) What is the number of this next-highest harmonic? One of the harmonic frequencies of tube \(B\) with only one open end is \(1080 \mathrm{~Hz}\). The next-highest harmonic frequency is \(1320 \mathrm{~Hz}\). (c) What harmonic frequency is next highest after the harmonic frequency \(600 \mathrm{~Hz}\) ? (d) What is the number of this nexthighest harmonic?

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