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Chapter 16: Problem 29

A certain pipe produces a fundamental frequency of 262 \(\mathrm{Hz}\) in air. (a) If the pipe is filled with helium at the same temperature, what fundamental frequency does it produce? (The molar mass of air is 28.8 \(\mathrm{g} / \mathrm{mol}\) , and the molar mass of helium is 4.00 \(\mathrm{g} / \mathrm{mol}\) . (b) Does your answer to part (a) depend on whether the pipe is open or stopped? Why or why not?

Short Answer

Expert verified
The fundamental frequency in helium is approximately 702 Hz. The answer does not depend on whether the pipe is open or stopped.

Step by step solution

01

Understand the Problem

We are given a pipe that produces a fundamental frequency of 262 Hz in air. We need to find the frequency it will produce when filled with helium, using information about the molar masses of air and helium. Additionally, we will determine whether the answer depends on the type of pipe.
02

Use the Speed of Sound Formula

The speed of sound in a gas is given by the equation \( v = \sqrt{\frac{\gamma R T}{M}} \), where \( \gamma \) is the adiabatic index (ratio of specific heats, approximately 1.4 for diatomic gases like air), \( R \) is the universal gas constant, \( T \) is the temperature in Kelvin, and \( M \) is the molar mass in kg/mol.
03

Determine the Speed of Sound in Air

For air, convert the molar mass to kg/mol, which is \( 28.8\, \text{g/mol} = 0.0288\, \text{kg/mol} \). The speed of sound in air is \( v = \sqrt{\frac{1.4 \cdot R \cdot T}{0.0288}} \). This speed will be used to find the frequency in helium using a proportion.
04

Determine the Speed of Sound in Helium

Convert the molar mass of helium to kg/mol: \( 4.00\, \text{g/mol} = 0.004\, \text{kg/mol} \). Use the formula: \( v_{\text{He}} = \sqrt{\frac{1.4 \cdot R \cdot T}{0.004}} \). The temperature and \( R \) cancel out, so the speed ratio is: \( \frac{v_{\text{He}}}{v_{\text{air}}} = \sqrt{\frac{0.0288}{0.004}} \).
05

Calculate the Frequency in Helium

Since frequency is directly proportional to the speed of sound, we use the ratio: \( f_{\text{He}} = f_{\text{air}} \cdot \sqrt{\frac{0.0288}{0.004}} \). Substituting \( f_{\text{air}} = 262 \) Hz gives \( f_{\text{He}} = 262 \cdot \sqrt{7.2} \approx 262 \cdot 2.68 \approx 702 \) Hz.
06

Analyze Effect of Pipe Type

The type of pipe (open or closed) does not affect the ratio of the molar masses used, because this calculation depends solely on the properties of the gases. Therefore, the type of distribution of nodes and antinodes remains constant in both gases, resulting in the frequency change being independent of the pipe's openness or being stopped.

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Key Concepts

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

Fundamental Frequency
The concept of fundamental frequency is essential in understanding how sound waves behave in various mediums. The fundamental frequency, often referred to as the first harmonic, is the lowest frequency at which a system oscillates. It is significant because it determines the pitch of the sound that is perceptible to us.
In wind instruments like pipes, the fundamental frequency is produced because of the standing wave set up within the pipe. For a pipe with a given length, this determines the longest wavelength that can fit. The relationship between frequency and wavelength is governed by the speed of sound; thus, the speed of sound in any medium directly affects the fundamental frequency. The faster the speed, the higher the frequency, as frequency is inversely proportional to wavelength.
  • Fundamental frequency = lowest frequency of vibration.
  • Determines pitch.
  • Set by speed of sound and physical length of instrument.
Understanding this concept helps in analyzing how changing the medium, like replacing air with helium in a pipe, changes the fundamental frequency.
Helium vs Air
Understanding the difference between helium and air is crucial for predicting how sound behaves in each medium. Helium is much lighter than air due to its lower molar mass. The key properties to focus on include density and the speed of sound.
The speed of sound is faster in helium compared to air. This is because the speed of sound is inversely related to the square root of the molar mass of the gas. Helium, with a molar mass of 4 g/mol, is far less dense than air, at 28.8 g/mol. This results in a significantly higher speed of sound in helium.
  • Helium = lighter, faster speed of sound.
  • Results in higher frequencies.
  • Molar mass significant for speed calculations.
This difference explains why when a pipe filled with helium produces sound, the fundamental frequency is higher compared to when it is filled with air.
Adiabatic Index
The adiabatic index, denoted as \( \gamma \), is a critical parameter when examining the speed of sound in gases. It is defined as the ratio of specific heats at constant pressure and volume, \( \gamma = \frac{C_p}{C_v} \).
This index affects how sound waves propagate through different gases. For diatomic gases like air, the adiabatic index is typically around 1.4. This parameter is essential in the formula to calculate the speed of sound: \( v = \sqrt{\frac{\gamma RT}{M}} \).
  • Adiabatic index affects sound speed.
  • \( \gamma \approx 1.4 \) for most diatomic gases.
  • Essential in calculations involving gas sound speed.
Understanding the adiabatic index helps to see why different gases lead to changes in acoustic properties, such as the fundamental frequency, when they fill a pipe.

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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 \(\mathrm{Hz}\) , a displacement amplitude of\( 1.2 \times 10^{-8} \mathrm{m}\) produces a pressure amplitude of \(3.0 \times 10^{-2} \mathrm{Pa}\) . Water at \(20^{\circ} \mathrm{C}\) has a bulk modulus of \(2.2 \times 10^{9} \mathrm{Pa}\) , and the speed of sound in water at this temperature is 1480 \(\mathrm{m} / \mathrm{s}\) . For \(1000-\mathrm{Hz}\) sound waves in \(20^{\circ} \mathrm{C}\) water, what displacement amplitude is produced if the pressure amplitude is \(3.0 \times 10^{-2}\) Pa? Explain why your answer is much less than \(1.2 \times 10^{-8} \mathrm{Pa}\) .

The shock-wave cone created by the space shuttle at one instant during its reentry into the atmosphere makes an angle of \(58.0^{\circ}\) with its direction of motion. The speed of sound at this altitude is 331 \(\mathrm{m} / \mathrm{s}\) (a) What is the Mach number of the shuttle at this instant, and (b) how fast (in \(\mathrm{m} / \mathrm{s}\) and \(\mathrm{mi} / \mathrm{h} )\) is it traveling relative to the atmosphere? (c) What would be its Mach number and the angle of its shock-wave cone if it flew at the same speed but at low altitude where the speed of sound is 344 \(\mathrm{m} / \mathrm{s} ?\)

(a) What is the sound intensity level in a car when the sound intensity is 0.500\(\mu \mathrm{W} / \mathrm{m}^{2} 7\) (b) What is the sound intensity level in the air near a jackhammer when the pressure amplitude of the sound is 0.150 \(\mathrm{Pa}\) and the temperature is \(20.0^{\circ} \mathrm{C}\) ?

Two loudspeakers, \(A\) and \(B,\) are driven by the same amplifier and emit sinusoidal waves in phase. Speaker \(B\) is 12.0 \(\mathrm{m}\) to the right of speaker \(A\) . The frequency of the waves emitted by each speaker is 688 \(\mathrm{Hz}\) You are standing between the speakers, along the line connecting them, and are at a point of constructive interference. How far must you walk toward speaker \(B\) to move to a point of destructive interference? effects like those in parts (a) and (b) are almost never a factor in listening to home stereo equipment. Why not?

Two identical taut strings under the same tension \(F\) produce a note of the same fundamental frequency \(f_{0}\) . The tension in one of them is now increased by a very small amount \(\Delta F\) . (a) If they are played together in their fundamental, show that the frequency of the beat produced is \(f_{\text { keet }}=f_{0}(\Delta F / 2 F)\) . (b) Two identical violin strings, when in tune and stretched with the same tension, have a fundamental frequency of 440.0 \(\mathrm{Hz}\) . One of the strings is retuned by increasing its tension. When this is done, 1.5 beats per second are heard when both strings are plucked simultaneously at their centers. By what percentage was the string tension changed?
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