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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
(a) Helium produces a higher frequency; specifically \( \approx 2.74 \times 262 \ \mathrm{Hz} \). (b) The answer does not depend on the pipe being open or stopped.

Step by step solution

01

Understand the Relation between Speed of Sound and Frequency

The frequency of sound produced by a pipe depends on the speed of sound in the medium inside the pipe. The fundamental frequency \( f \) is given by \( f = \frac{v}{2L} \) for an open pipe and \( f = \frac{v}{4L} \) for a closed pipe, where \( v \) is the speed of sound and \( L \) is the length of the pipe.
02

Calculate the Speed of Sound in Air

The speed of sound in a gas is calculated using the formula \( v = \sqrt{\frac{\gamma R T}{M}} \), where \( \gamma \) is the adiabatic index, \( R \) is the ideal gas constant, \( T \) is the temperature in Kelvin, and \( M \) is the molar mass of the gas. For air, \( \gamma \approx 1.4 \) and \( M = 28.8 \ g/mol \).
03

Calculate the Speed of Sound in Helium

For helium, \( \gamma \approx 1.66 \) and \( M = 4.00 \ g/mol \). Using the same formula as in Step 2, calculate the speed of sound in helium.
04

Compare the Fundamental Frequencies

The frequency changes proportionally with the speed of sound. If \( v_{\text{helium}} \) is the speed of sound in helium and \( v_{\text{air}} \) is in air, then the frequency in helium \( f_{\text{helium}} = f_{\text{air}} \times \frac{v_{\text{helium}}}{v_{\text{air}}} \). Use the values from Steps 2 and 3.
05

Determine Dependence on Pipe Type

The calculation formula for frequency depends on the factor \( \gamma \) in the speed of sound equation, which does not change with the pipe's open or closed nature. Thus, the fundamental frequency change is independent of whether the pipe is open or stopped.

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Key Concepts

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

Speed of Sound
The speed of sound in a medium is crucial in determining the frequency of sound it produces. When sound travels through a substance, it propagates as a wave.
  • The speed at which this wave travels is known as the speed of sound, denoted by \( v \).
  • This speed varies depending on the medium's properties, such as its density and elasticity.
In gases, the speed of sound can be calculated using the formula \( v = \sqrt{\frac{\gamma R T}{M}} \), where:
  • \( \gamma \) is the adiabatic index.
  • \( R \) is the ideal gas constant (8.314 J/mol路K).
  • \( T \) is the temperature in Kelvin.
  • \( M \) is the molar mass of the gas.
This formula shows that sound travels faster in lighter gases and at higher temperatures.
In this exercise, understanding the speed of sound helps explain changes in fundamental frequency when we switch from air to helium.
Open and Closed Pipes
Pipes can be either open at both ends or closed at one end, and this significantly affects the sound they produce.
  • In an **open pipe**, both ends allow air to move freely, creating a node at each end.
  • The fundamental frequency formula for an open pipe is \( f = \frac{v}{2L} \), where \( L \) is the length of the pipe.
On the other hand, a **closed pipe** has only one end open and the other end closed, and air movement is restricted:
  • Thus, it forms a node at the closed end and an antinode at the open end.
  • The fundamental frequency here is \( f = \frac{v}{4L} \), making it lower compared to an open pipe of the same length.
The type of pipe, whether open or closed, doesn't affect the relation between speed of sound and frequency changes. Hence, when filling the same pipe with a different gas, the fundamental frequency changes uniformly for both types.
Molar Mass
Molar mass (\( M \)) is the mass of one mole of a substance, expressed in grams/mol. It influences the speed of sound significantly in gases.
  • A lighter gas, such as helium (4.00 g/mol), allows sound to travel faster than heavier gases, such as air (28.8 g/mol).
The speed of sound in a gas is inversely proportional to the square root of its molar mass, as seen in the formula \( v = \sqrt{\frac{\gamma R T}{M}} \).
  • This means that when using helium instead of air, the lower molar mass causes a notable increase in the speed of sound.
In this exercise, recognizing the role of molar mass helps us calculate the effects of swapping air for helium in a pipe and observe the resulting change in fundamental frequency.
Helium vs Air
Switching from air to helium within a pipe affects both the speed of sound and the resulting sound frequency.
  • Helium, being a lighter gas with a lower molar mass, enables sound to propagate faster.
  • This, in turn, raises the fundamental frequency compared to that of air.
The relationship between the two can be illustrated as:
  • If the speed of sound increases, so does the frequency, as per the formula \( f = \frac{v}{\text{length factor}} \).
In this exercise, applying helium doubles the speed compared to air, significantly altering the pitch of the sound produced. Understanding the differences between these gases allows us to predict changes in the pipe's sound characteristics accurately.
Adiabatic Index
The adiabatic index (\( \gamma \)) is an essential factor in the speed of sound equation. It represents the ratio of specific heats (Cp/Cv) of a gas, indicating how heat capacities change with pressure.
  • For example, air has an adiabatic index of approximately 1.4, while helium's is around 1.66.
A higher adiabatic index results in a faster speed of sound, since \( \gamma \) directly influences the velocity in \( v = \sqrt{\frac{\gamma R T}{M}} \).
  • This illustrates why helium, with a higher index, increases the sound speed more than air.
In this problem, how \( \gamma \) values affect fundamental frequency helps explain why sound changes with different gases, despite keeping the same physical pipe. Recognizing this relationship is key to understanding the acoustic properties of different gases.

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Most popular questions from this chapter

A long tube contains air at a pressure of 1.00 atm and a temperature of \(77.0^{\circ} \mathrm{C}\) . The tube is open at one end and closed at the other by a movable piston. A tuning fork near the open end is vibrating with a frequency of 500 \(\mathrm{Hz}\) . Resonance is produced when the piston is at distances \(18.0,55.5,\) and 93.0 \(\mathrm{cm}\) from the open end. (a) From these measurements, what is the speed of sound in air at \(77.0^{\circ} \mathrm{C} ?\) (b) From the result of part (a), what is the value of \(\gamma ?\) (c) These data show that a displacement antinode is slightly outside of the open end of the tube. How far outside is it?

Moving Source vs. Moving Listener. (a) A sound source producing \(1.00-\mathrm{kHz}\) waves moves toward a stationary listener at one-half the speed of sound. What frequency will the listener hear? (b) Suppose instead that the source is stationary and the listener moves toward the source at one- half the speed of sound. What frequency does the listener hear'? How does your answer compare to that in part (a)? Explain on physical grounds why the two answers differ.

Standing sound waves are produced in a pipe that is 1.20 \(\mathrm{m}\) long. For the fundamental and first two overtones, determine the locations along the pipe (measured from the left end) of the dis- placement nodes and the pressure nodes if (a) the pipe is open at both ends and (b) the pipe is closed at the left end and open at the right end.

singing in the Shower. A pipe closed at both ends can have standing waves inside of it, but you normally don't hear them because little of the sound can get out. But you can hear them if you are inside the pipe, such as someone singing in the shower. (a) Show that the wavelengths of standing waves in a pipe of length \(L\) that is closed at both ends are \(\lambda_{n}=2 L / n\) and the frequencies are given by \(f_{n}=n v / 2 L=n f_{1},\) where \(n=1,2,3, \ldots,\) (b) Modeling it as a pipe, find the frequency of the fundamental and the first two overtones for a shower 2.50 \(\mathrm{m}\) tall. Are these frequencies audible?

BIO For a person with normal hearing, the faintest sound that can be heard at a frequency of 400 \(\mathrm{Hz}\) has a pressure amplitude of about \(6.0 \times 10^{-5}\) Pa. Calculate the (a) intensity; (b) sound intensity level; (c) displacement amplitude of this sound wave at \(20^{\circ} \mathrm{C} .\)
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