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A person hums into the top of a well and finds that standing waves are established at frequencies of \(42,70.0,\) and \(98 \mathrm{~Hz} .\) The frequency of \(42 \mathrm{~Hz}\) is not necessarily the fundamental frequency. The speed of sound is \(343 \mathrm{~m} / \mathrm{s} .\) How deep is the well?

Short Answer

Expert verified
The well is approximately 6.125 meters deep.

Step by step solution

01

Understanding Standing Waves

The problem states that standing waves are established in the well, which is open at the top and closed at the bottom. This means that the well forms a closed-end air column where standing waves occur at odd harmonics of the fundamental frequency.
02

Identifying Harmonics

The frequencies given are 42 Hz, 70 Hz, and 98 Hz. Since these frequencies are not in a straightforward sequence of odd harmonics, we need to determine if they are consecutive odd harmonics of some fundamental frequency. Let's express them in terms of the fundamental frequency \(f_1\):1. For the first harmonic, given \(n=1\), \(f_1 = f_1\)2. For the third harmonic, given \(n=3\), \(f_3 = 3f_1 = 42\, \mathrm{Hz}\)3. For the fifth harmonic, given \(n=5\), \(f_5 = 5f_1 = 70\, \mathrm{Hz}\)4. For the seventh harmonic, given \(n=7\), \(f_7 = 7f_1 = 98 \mathrm{Hz}\)
03

Solving for Fundamental Frequency

From the equation for the third harmonic, we have \(3f_1 = 42\, \mathrm{Hz}\). Solving for \(f_1\), we get:\[f_1 = \frac{42}{3} = 14 \mathrm{Hz}\]
04

Calculate the Fundamental Wavelength

Using the speed of sound \(v = 343 \, \mathrm{m/s}\) and the fundamental frequency \(f_1 = 14 \, \mathrm{Hz}\), we can find the wavelength \(\lambda_1\):\[\lambda_1 = \frac{v}{f_1} = \frac{343}{14} = 24.5 \, \mathrm{m}\]
05

Determine the Length of the Well

For a closed-end air column, the fundamental frequency corresponds to a quarter wavelength fitting within the well. Therefore, the depth \(L\) of the well is:\[L = \frac{\lambda_1}{4} = \frac{24.5}{4} = 6.125 \, \mathrm{m}\]

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Key Concepts

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

Harmonics
Harmonics are integral parts of understanding standing waves. In a closed-end air column, like a well or a pipe that is open at one end and closed at the other, standing waves can be produced. These waves have nodes and antinodes that occur at specific points in the column. Harmonics refer to the integer multiples of the fundamental frequency. In such systems, you only get odd harmonics – meaning the fundamental frequency (first harmonic), third harmonic, fifth harmonic, and so on.
  • The first harmonic or fundamental frequency is the simplest form, usually represented by one antinode and one node.
  • The third harmonic includes an additional node and antinode compared to the first, and is triple the frequency of the fundamental frequency.
  • Similarly, the fifth harmonic has two more nodes and antinodes than the first harmonic and is five times the fundamental frequency.
Consequently, recognizing these patterns of harmonics helps us identify the frequencies and thus solve various acoustic problems.
Fundamental Frequency
The concept of the fundamental frequency is crucial for understanding how sounds and notes are formed in physical spaces like air columns. This is the lowest frequency at which a system can vibrate. It is often referred to as the 'first harmonic'. The entire pattern of harmonics builds around this base frequency. The exercise problem hints that 42 Hz might not be the fundamental frequency. By checking given frequencies (42, 70, and 98 Hz), we find that they align with the third (42 Hz), fifth (70 Hz), and seventh (98 Hz) harmonics, respectively.
The calculation of the fundamental frequency, in this case, comes from the relationship:
\[f_3 = 3f_1 = 42 \, \text{Hz} \]
Solving this equation shows that the fundamental frequency is actually 14 Hz, not initially apparent from the given data. Understanding this concept allows the determination of all other harmonics and the corresponding wavelengths within the air column.
Closed-End Air Column
A closed-end air column appears when one end of the column remains open, while the other is closed. The closed end forces the air to form a node (where displacement is minimal), while the open end forms an antinode (where oscillation is at its maximum). This configuration gives rise to a specific pattern of resonant frequencies, solely consisting of odd harmonics.
In this arrangement, each sound wave fits into the column as an odd multiple of a quarter-wavelength, such as 1/4, 3/4, 5/4, etc. This principle leads to only the odd-numbered harmonics being permissible.
Closed-end columns are found not only in wells but also in many musical instruments like clarinets and some organ pipes, contributing to a unique sound quality. This property significantly impacts the design and sound of these instruments, showing the practical application of closed-end air columns in music.
Wavelength Calculation
Calculating the wavelength in a standing wave scenario is essential to understanding the physical dimensions supporting wave behavior. Wavelength is the distance over which the wave's shape repeats. For a given wave speed and frequency, it can be found using the formula:
\[\lambda = \frac{v}{f} \]
where \(v\) is the speed of sound, and \(f\) is the frequency of the wave. For the fundamental frequency of 14 Hz in the problem, with a sound speed of 343 m/s, the wavelength is calculated as follows:
\[\lambda_1 = \frac{343}{14} = 24.5 \, \text{m} \]
In a closed-end air column, this wavelength corresponds to four times the actual length of the column. Hence, each quarter wavelength fits within the column length, helping us calculate the length of the well:
\[L = \frac{\lambda}{4} = \frac{24.5}{4} = 6.125 \, \text{m} \]
Understanding these calculations allows us to predict and analyze acoustical properties in similar physical systems.

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Most popular questions from this chapter

Two loudspeakers face each other, vibrate in phase, and produce identical 440 -Hz tones. A listener walks from one speaker toward the other at a constant speed and hears the loudness change (loud-soft-loud) at a frequency of \(3.0 \mathrm{~Hz} .\) The speed of sound is \(343 \mathrm{~m} /\) s. What is the walking speed?

A tube of air is open at only one end and has a length of \(1.5 \mathrm{~m}\). This tube sustains a standing wave at its third harmonic. What is the distance between one node and the adjacent antinode?

A tube is open only at one end. A certain harmonic produced by the tube has a frequency of \(450 \mathrm{~Hz}\). The next higher harmonic has a frequency of \(750 \mathrm{~Hz}\). The speed of sound in air is \(343 \mathrm{~m} / \mathrm{s}\). (a) What is the integer \(n\) that describes the harmonic whose frequency is \(450 \mathrm{~Hz} ?\) (b) What is the length of the tube?

Two out-of-tune flutes play the same note. One produces a tone that has a frequency of \(262 \mathrm{~Hz},\) while the other produces \(266 \mathrm{~Hz}\). When a tuning fork is sounded together with the \(262-\mathrm{Hz}\) tone, a beat frequency of \(1 \mathrm{~Hz}\) is produced. When the same tuning fork is sounded together with the 266 -Hz tone, a beat frequency of 3 Hz is produced. What is the frequency of the tuning fork?

When a guitar string is sounded along with a 440 -Hz tuning fork, a beat frequency of \(5 \mathrm{~Hz}\) is heard. When the same string is sounded along with a 436 -Hz tuning fork, the beat frequency is \(9 \mathrm{~Hz}\). What is the frequency of the string?

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