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\(\cdot\) Find the fundamental frequency and the frequency of the first three overtones of a pipe 45.0 \(\mathrm{cm}\) long (a) if the pipe is open at both ends; (b) if the pipe is closed at one end. (c) For each of the preceding cases, what is the number of the highest harmonic that may be heard by a person who can hear frequencies from 20 \(\mathrm{Hz}\) to \(20,000 \mathrm{Hz}\) ?

Short Answer

Expert verified
Open pipe: Fundamental frequency = 381.1 Hz, highest harmonic = 52. Closed pipe: Fundamental frequency = 190.6 Hz, highest harmonic = 103.

Step by step solution

01

Determine the Speed of Sound

First, note the speed of sound in air is approximately \( v = 343 \, \text{m/s} \). This value is needed to calculate the frequencies.
02

Calculate the Wavelength in an Open Pipe

For an open pipe resonator, the fundamental frequency (first harmonic) wavelength, \( \lambda_1 \), is given by the formula \( \lambda_1 = 2L \), where \( L \) is the length of the pipe. For a pipe of length \( 45.0 \, \text{cm} = 0.45 \, \text{m} \), the fundamental wavelength is \[ \lambda_1 = 2 \times 0.45 = 0.9 \, \text{m}. \]
03

Calculate the Fundamental Frequency of the Open Pipe

The fundamental frequency \( f_1 \) can be found using the relationship \[ f_1 = \frac{v}{\lambda_1} = \frac{343}{0.9} = 381.1 \, \text{Hz}. \]
04

Determine the Overtones of the Open Pipe

The first three overtones for an open pipe will have frequencies \( f_2 = 2f_1, \) \( f_3 = 3f_1, \) and \( f_4 = 4f_1. \) Using the fundamental frequency from Step 3:- \( f_2 = 2 \times 381.1 = 762.2 \, \text{Hz} \)- \( f_3 = 3 \times 381.1 = 1143.3 \, \text{Hz} \)- \( f_4 = 4 \times 381.1 = 1524.4 \, \text{Hz} \)
05

Calculate the Wavelength in a Closed Pipe

For a pipe closed at one end, the fundamental wavelength \( \lambda_1 \) is given by \( \lambda_1 = 4L \), so \[ \lambda_1 = 4 \times 0.45 = 1.8 \, \text{m}. \]
06

Calculate the Fundamental Frequency of the Closed Pipe

Using the relationship between frequency, speed, and wavelength:\[ f_1 = \frac{v}{\lambda_1} = \frac{343}{1.8} = 190.6 \, \text{Hz}. \]
07

Determine the Overtones in the Closed Pipe

For a closed pipe, overtones occur at odd integer multiples of the fundamental frequency. Calculate the first three overtones:- \( f_3 = 3 \times 190.6 = 571.8 \, \text{Hz} \)- \( f_5 = 5 \times 190.6 = 953.0 \, \text{Hz} \)- \( f_7 = 7 \times 190.6 = 1334.2 \, \text{Hz} \)
08

Find the Highest Audible Harmonic for the Open Pipe

A person can hear from 20 Hz to 20,000 Hz. For an open pipe, the harmonic frequencies are multiples of the fundamental frequency:- \( f_n = n \times 381.1 \, \text{Hz} \), find \( n \) such that \( n\times 381.1 < 20,000 \). Solve \( n < \frac{20,000}{381.1} \approx 52.5 \). The highest harmonic is \( n = 52 \).
09

Find the Highest Audible Harmonic for the Closed Pipe

For the closed pipe, the harmonic frequencies are \( f_n = (2k-1) \times 190.6 \), for odd \( n \) such that \( (2k-1) \times 190.6 < 20,000 \). Solve \( (2k-1) < \frac{20,000}{190.6} \approx 104.9 \). The highest harmonic is \( n = 103 \).

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Key Concepts

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

Fundamental Frequency
The fundamental frequency, often known as the first harmonic, refers to the lowest frequency produced by an oscillating system, such as a pipe resonator. It serves as the base frequency upon which overtones or harmonics build.
An important formula to remember when dealing with pipe resonators is the equation for frequency, given by:
  • \[ f_1 = \frac{v}{\lambda} \]
In this equation, \( f_1 \) is the fundamental frequency, \( v \) is the speed of sound, and \( \lambda \) is the wavelength.
In simpler terms, when you pluck a string of a given length, its vibration creates a sound wave that travels at the speed of sound, producing the fundamental frequency.
Understanding the fundamental frequency is crucial as it sets the stage for determining overtones, and hence, the harmonics of the system.
Open Pipe Resonator
An open pipe resonator is a type of tube with both ends open, allowing air to move freely in and out. This design supports both even and odd harmonics, multiplying the sound experience.
For an open pipe, the length \( L \) of the pipe plays a key role in determining its fundamental frequency. Using the equation:
  • \[ \lambda_1 = 2L \]
One can calculate the fundamental wavelength, and thus, the fundamental frequency using our earlier mentioned formula.
The open pipe's harmonics can be seen in the formula:
  • \( f_n = n \times f_1 \)
where \( n \) represents the order of the harmonic.
In such resonators, the first overtone is twice the fundamental frequency, the second overtone is thrice, and so on. These harmonics create the rich sounds utilized in musical instruments like flutes and organ pipes.
Closed Pipe Resonator
A closed pipe resonator is a tube that is open at one end and closed at the other. This unique setup produces sound waves that only reflect odd harmonics, unlike their open-ended counterparts.
For closed pipes, the fundamental wavelength is given by:
  • \[ \lambda_1 = 4L \]
where \( L \) is the length of the pipe.
The fundamental frequency is also determined using our standard formula, but the harmonics here follow a distinct pattern:
  • \( f_n = (2k-1) \times f_1 \)
where \( k \) is an integer starting from 1, representing only odd-numbered harmonics.
This means the pipe supports only the first, third, fifth harmonics, and so on, making instruments like clarinets and bassoons sound unique compared to open pipe instruments.
Speed of Sound
The speed of sound is a critical component in the propagation of sound waves through a medium. On a standard day at sea level, the speed of sound in air is approximately 343 meters per second.
Factors affecting the speed of sound include:
  • Temperature: Sound travels faster in warmer air because molecules move more quickly.
  • Medium: Sound waves travel at different speeds through air, water, and solid materials due to their density and elasticity.
  • Humidity: More humid air increases the speed of sound because water vapor is less dense than dry air.
In calculating frequencies of resonators, the speed of sound allows us to translate the physical properties of the pipe into the musical qualities of pitch and tone.
Understanding the speed of sound's dependency on the environment aids in accurately predicting and designing musical instruments and audio arrays.

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Most popular questions from this chapter

\(\bullet\) The sound source of a ship's sonar system operates at a frequency of 22.0 \(\mathrm{kHz}\) . The speed of sound in water (assumed to be at a uniform \(20^{\circ} \mathrm{C}\) ) is 1482 \(\mathrm{m} / \mathrm{s}\) . (a) What is the wavelength of the waves emitted by the source? (b) What is the difference in frequency between the directly radiated waves andthe waves reflected from a whale traveling straight toward the ship at 4.95 \(\mathrm{m} / \mathrm{s} ?\) The ship is at rest in the water.

Transverse waves on a string have wave speed \(8.00 \mathrm{m} / \mathrm{s},\) amplitude \(0.0700 \mathrm{m},\) and wavelength 0.320 \(\mathrm{m} .\) These waves travel in the \(x\) direction, and at \(t=0\) the \(x=0\) end of the string is at \(y=0\) and moving downward. (a) Find the frequency, period, and wave number of these waves. (b) Write the equation for \(y(x, t)\) describing these waves. (c) Find the transverse displacement of a point on the string at \(x=0.360 \mathrm{m}\) at time \(t=0.150 \mathrm{s}\) .

Ultrasound and infrasound. (a) Whale communication. Blue whales apparently communicate with each other using sound of frequency 17 \(\mathrm{Hz}\) , which can be heard nearly 1000 \(\mathrm{km}\) away in the ocean. What is the wavelength of such a sound in seawater, where the speed of sound is 1531 \(\mathrm{m} / \mathrm{s} ?\) (b) Dolphin clicks. One type of sound that dolphins emit is a sharp click of wavelength 1.5 \(\mathrm{cm}\) in the ocean. What is the frequency of such clicks? (c) Dog whistles. One brand of dog whistles claims a frequency of 25 \(\mathrm{kHz}\) for its product. What is the wavelength of this sound? (d) Bats. While bats emit a wide variety of sounds, one type emits pulses of sound having a frequency between 39 \(\mathrm{kHz}\) and 78 \(\mathrm{kHz}\) . What is the range of wavelengths of this sound? (e) Sonograms. Ultrasound is used to view the interior of the body, much as x rays are utilized. For sharp imagery, the wavelength of the sound should be around one-fourth (or less) the size of the objects to be viewed. Approximately what frequency of sound is needed to produce a clear image of a tumor that is 1.0 \(\mathrm{mm}\) across if the speed of sound in the tissue is 1550 \(\mathrm{m} / \mathrm{s} ?\)

\(\cdot\) Two guitarists attempt to play the same note of wavelength 6.50 \(\mathrm{cm}\) at the same time, but one of the instruments is slightly out of tune and plays a note of wavelength 6.52 \(\mathrm{cm}\) instead. What is the frequency of the beat these musicians hear when they play together?

\(\cdot\) Mapping the ocean floor. The ocean floor is mapped by sending sound waves (sonar) downward and measuring the time it takes for their echo to return. From this information, the ocean depth can be calculated if one knows that sound travels at 1531 \(\mathrm{m} / \mathrm{s}\) in seawater. If a ship sends out sonar pulses and records their echo 3.27 s later, how deep is the ocean floor at that point, assuming that the speed of sound is the same at all depths?

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