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You have a stopped pipe of adjustable length close to a taut 62.0-cm, 7.25-g wire under a tension of 4110 N. You want to adjust the length of the pipe so that, when it produces sound at its fundamental frequency, this sound causes the wire to vibrate in its second \(overtone\) with very large amplitude. How long should the pipe be?

Short Answer

Expert verified
The pipe should be 0.0602 m long.

Step by step solution

01

Calculate the Mass Per Unit Length of the Wire

First, calculate the mass per unit length (linear density) of the wire using the formula: \( \mu = \frac{m}{L} \). The mass \( m \) is 7.25 g, which we need to convert to kilograms, giving us 0.00725 kg. The wire length \( L \) is 62.0 cm, which converts to 0.62 m. So, \( \mu = \frac{0.00725}{0.62} = 0.0117 \ \text{kg/m} \).
02

Determine the Fundamental Frequency of the Wire

The frequency of a vibrating string is determined by the equation: \( f = \frac{1}{2L} \sqrt{\frac{T}{\mu}} \). Here, \( T = 4110 \ \text{N}\), \( L = 0.62 \ \text{m} \), and \( \mu = 0.0117 \ \text{kg/m} \). Substituting these into the equation gives the fundamental frequency \( f = \frac{1}{1.24} \times \sqrt{\frac{4110}{0.0117}} \approx 474\text{ Hz} \).
03

Calculate the Frequency of the Second Overtone

The second overtone of the wire corresponds to the third harmonic, which is three times the fundamental frequency. Thus, \( f_{3} = 3 \times 474 \approx 1422\text{ Hz} \).
04

Match Pipe's Fundamental Frequency with Wire's Second Overtone

To create resonance, the pipe's fundamental frequency should match the wire's second overtone frequency. For a stopped pipe, the fundamental frequency is \( f = \frac{v}{4L} \), where \( v \) is the speed of sound, approximately \( 343 \ \text{m/s} \). Setting \( f = 1422 \), we have \( 1422 = \frac{343}{4L} \).
05

Solve for the Pipe Length

Rearrange the equation from Step 4: \( L = \frac{343}{4 \times 1422} \). Calculating this gives \( L = 0.0602 \ \text{m} \).

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Key Concepts

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

Stopped Pipe
A stopped pipe, also known as a closed pipe, is characterized by having one end closed while the other is open. This configuration gives rise to specific resonant frequencies due to the formation of standing waves.
In a stopped pipe, the fundamental frequency occurs when there is a quarter-wave within the pipe. This sets the foundation for harmonic frequencies as subsequent overtones are formed by fitting additional quarter-waves inside the same pipe.
These resonate at odd multiples of the fundamental frequency. This unique property makes stopped pipes integral to various musical instruments like clarinets and organs.
Harmonic Frequencies
Harmonic frequencies in physics are specific frequencies at which standing waves form on a medium such as a string, or within a column of air like a pipe.
For stopped pipes, the harmonics occur at odd multiples of the fundamental frequency. Hence, the first harmonic (or fundamental frequency) equals the speed of sound divided by 4 times the length of the pipe.
The second harmonic frequency, in this case, is three times the fundamental frequency, showcasing how specific structures limit the types of harmonics that can manifest.
Fundamental Frequency
The fundamental frequency, also known as the first harmonic, is the lowest frequency at which an object or system vibrates. For a stopped pipe, this frequency is calculated using the formula: \[ f = \frac{v}{4L} \]Where:
  • \( f \) is the fundamental frequency,
  • \( v \) is the speed of sound,
  • \( L \) is the length of the pipe.
The fundamental frequency sets the base for other vibrational frequencies or overtones within the system, influencing the sound produced by musical instruments.
Second Overtone
The second overtone refers to the third harmonic in a system of standing waves. In the context of both vibrating strings and stopped pipes, it is essential for creating complex sound patterns.
To find the frequency for the second overtone, we multiply the fundamental frequency by three (for a vibrating string, this factor depends on the type of pipe or string and how its modes are set).
The second overtone's frequency is pivotal in achieving resonance conditions and is crucial for the analysis of sound systems and musical acoustics.
Vibrating String
A vibrating string, like a wire, suspended in tension produces sound through its harmonics. Its fundamental frequency is influenced by several factors, including tension, length, and linear mass density (mass per unit length).
The formula for the fundamental frequency of a vibrating string is given by:\[ f = \frac{1}{2L} \sqrt{\frac{T}{\mu}} \]Where:
  • \( f \) is the frequency,
  • \( L \) is the length of the string,
  • \( T \) is the tension,
  • \( \mu \) is the mass per unit length.
This demonstrates how the string vibrates in segments, creating complex patterns and harmonics.
Speed of Sound
The speed of sound is critical for calculating the resonant frequencies in both stopped pipes and vibrating strings.
Typically, the speed of sound in air is around 343 m/s under standard temperature and pressure conditions.
This speed can vary based on atmospheric conditions such as temperature and pressure. In many exercises involving acoustic resonators, this constant helps calculate the length of pipes or strings necessary to produce a particular frequency.

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Most popular questions from this chapter

Example 16.1 (Section 16.1) showed that for sound waves in air with frequency 1000 Hz, a displacement amplitude of 1.2 \(\times\) 10\(^{-8}\) m produces a pressure amplitude of 3.0 \(\times\) 10\(^{-2}\) Pa. (a) What is the wavelength of these waves? (b) For 1000-Hz waves in air, what displacement amplitude would be needed for the pressure amplitude to be at the pain threshold, which is 30 Pa? (c) For what wavelength and frequency will waves with a displacement amplitude of 1.2 \(\times\) 10\(^{-8}\) m produce a pressure amplitude of 1.5 \(\times\) 10\(^{-3}\) Pa?

Many opera singers (and some pop singers) have a range of about 2\(\frac{1}{2}\) octaves or even greater. Suppose a soprano's range extends from A below middle C (frequency 220 Hz) up to E-flat above high C (frequency 1244 Hz). Although the vocal tract is quite complicated, we can model it as a resonating air column, like an organ pipe, that is open at the top and closed at the bottom. The column extends from the mouth down to the diaphragm in the chest cavity, and we can also assume that the lowest note is the fundamental. How long is this column of air if \(v =\) 354 m/s? Does your result seem reasonable, on the basis of observations of your own body?

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