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Ch 17: Superposition
Knight Calc - Physics for Scientists and Engineers 5th Edition
Knight Calc5th EditionPhysics for Scientists and EngineersISBN: 9780137344796Not the one you use?Change textbook
All textbooksKnight Calc 5th EditionCh 17: SuperpositionProblem 17
Chapter 17, Problem 17

The fundamental frequency of an open-open tube is 1500 Hz when the tube is filled with 0°C helium. What is its frequency when filled with 0°C air?

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Identify the relationship between the speed of sound in a medium and the frequency of sound. The speed of sound in a medium is given by the formula: \( v = f \lambda \), where \( v \) is the speed of sound, \( f \) is the frequency, and \( \lambda \) is the wavelength.
Understand that the wavelength of the fundamental frequency in an open-open tube is twice the length of the tube, i.e., \( \lambda = 2L \). Since the tube's length and the type of its ends (open-open) remain constant, the wavelength for the fundamental frequency does not change.
Use the formula for the speed of sound in a gas, which is \( v = \sqrt{\frac{\gamma R T}{M}} \), where \( \gamma \) is the adiabatic index, \( R \) is the universal gas constant, \( T \) is the temperature in Kelvin, and \( M \) is the molar mass of the gas. Note that for helium and air at the same temperature, the difference in molar mass will affect the speed of sound.
Calculate the speed of sound in both helium and air using their respective molar masses. Helium has a molar mass of about 4 g/mol, and air has a molar mass of about 29 g/mol. Since the temperature and other constants remain the same, the speed of sound will be higher in helium due to its lower molar mass.
Apply the relationship between speed of sound and frequency to find the new frequency when the tube is filled with air. Since the wavelength remains constant and the speed of sound in air is lower, the frequency of the sound when the tube is filled with air will also be lower compared to when it is filled with helium.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Fundamental Frequency

The fundamental frequency is the lowest frequency at which a system oscillates. In the context of a tube, it is determined by the length of the tube and the speed of sound in the medium inside it. For an open-open tube, the fundamental frequency can be calculated using the formula f = v/2L, where v is the speed of sound in the medium and L is the length of the tube.
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Speed of Sound in Different Gases

The speed of sound varies in different gases due to differences in density and temperature. In general, sound travels faster in lighter gases like helium compared to heavier gases like air. At 0°C, the speed of sound in helium is approximately 972 m/s, while in air, it is about 331 m/s. This difference significantly affects the fundamental frequency of sound waves in a tube.
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Effect of Medium on Frequency

The frequency of sound produced in a tube is directly related to the medium filling the tube. When the medium changes, the speed of sound changes, which in turn alters the fundamental frequency. For an open-open tube, if the medium changes from helium to air, the frequency will decrease due to the lower speed of sound in air compared to helium, demonstrating the relationship between medium properties and sound frequency.
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Related Practice
Textbook Question

A bass clarinet can be modeled as a 120-cm-long open-closed tube. A bass clarinet player starts playing in a 20° C room, but soon the air inside the clarinet warms to where the speed of sound is 352m/s . Does the fundamental frequency increase or decrease? By how much?

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Textbook Question

The two highest-pitch strings on a violin are tuned to 440 Hz (the A string) and 659 Hz (the E string). What is the ratio of the mass of the A string to that of the E string? Violin strings are all the same length and under essentially the same tension.

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Textbook Question

A 170-cm-long open-closed tube has a standing sound wave at 250 Hz on a day when the speed of sound is 340m/s . How many pressure antinodes are there, and how far is each from the open end of the tube?

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Textbook Question

Standing waves on a 1.0-m-long string that is fixed at both ends are seen at successive frequencies of 36 Hz and 48 Hz. Draw the standing-wave pattern when the string oscillates at 48 Hz.

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Textbook Question

A carbon dioxide laser is an infrared laser. A CO2 laser with a cavity length of 53.00 cm oscillates in the m=100,000 mode. What are the wavelength and frequency of the laser beam?

Textbook Question

The lowest note on a grand piano has a frequency of 27.5 Hz. The entire string is 2.00 m long and has a mass of 400 g. The vibrating section of the string is 1.90 m long. What tension is needed to tune this string properly?

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