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Chapter 16: Problem 31

You blow across the open mouth of an empty test tube and produce the fundamental standing wave of the air column inside the test tube. The speed of sound in air is 344 \(\mathrm{m} / \mathrm{s}\) and the test tube acts as a stopped pipe. (a) If the length of the air column in the test tube is 14.0 \(\mathrm{cm}\) , what is the frequency of this standing wave? (b) What is the frequency of the nfundamental standing wave in the air column if the test tube is half filled with water?

Short Answer

Expert verified
(a) 614.29 Hz, (b) 1228.57 Hz

Step by step solution

01

Identify Key Information

We have a test tube acting as a stopped pipe, which means one end is closed and one end is open. The speed of sound in air is given as 344 m/s, and the length of the air column is 14.0 cm (or 0.14 m).
02

Understand the Stopped Pipe Concept

For a stopped pipe, the fundamental frequency corresponds to a quarter of a wavelength fitting in the pipe. Thus, the length of the air column, L, equals \( \frac{\lambda}{4} \), where \( \lambda \) is the wavelength.
03

Calculate the Wavelength for Fundamental Frequency

Using the relationship \( L = \frac{\lambda}{4} \), solve for \( \lambda \) (wavelength): \( \lambda = 4 \times 0.14 \ m = 0.56 \ m \).
04

Find the Frequency Using the Wavelength

The frequency \( f \) can be found using \( f = \frac{v}{\lambda} \), where \( v \) is the speed of sound. Substituting the values gives \( f = \frac{344 \ m/s}{0.56 \ m} = 614.29 \ Hz \).
05

Adjust for the Half-Filled Test Tube

If the test tube is half-filled with water, the effective length of the air column is halved. The new length is \( 0.07 \ m \).
06

Recalculate Wavelength and Frequency for Half-Filled Tube

Recalculate \( \lambda \) with the new length using \( L = \frac{\lambda}{4} \). Thus, \( \lambda = 4 \times 0.07 \ m = 0.28 \ m \). The new frequency is \( f = \frac{344 \ m/s}{0.28 \ m} = 1228.57 \ Hz \).

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Key Concepts

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

Understanding Standing Waves
Standing waves are fascinating patterns that occur when waves bounce back and forth within a bounded or fixed space. Imagine a wave traveling down a string and reflecting back; the resulting interference can create stable patterns that appear to "stand" still, hence the name. These patterns are formed by the combination of two waves traveling in opposite directions with the same frequency and amplitude.
In acoustics, standing waves are crucial as they correspond to the resonant frequencies of musical instruments and other systems. One can observe nodes, points of zero displacement, and antinodes, points of maximum displacement, in standing waves.
  • Nodes: Points of no movement due to destructive interference.
  • Antinodes: Points of maximum vibration due to constructive interference.
In musical instruments like pipes or strings, only specific wavelengths will fit in the instrument's dimensions, creating distinct musical notes or tones.
Characteristics of a Stopped Pipe
In acoustics, a stopped pipe refers to a pipe closed at one end and open at the other. This configuration allows the formation of a specific type of standing wave known as the fundamental mode. The closed end of the pipe acts as a displacement node while the open end acts as a displacement antinode.
For a stopped pipe, the fundamental frequency is characterized by the pattern where one-quarter of a wavelength (位/4) fits into the length of the pipe (L). This means the length of the column is directly related to the wavelength by L = 位/4. This relation affects how we calculate the frequencies and harmonics of such a pipe.
  • First harmonic (fundamental frequency): L = 位/4
  • Second harmonic: Not supported in stopped pipes, as it would require another node.
  • Third harmonic: L = 3位/4, etc.
Understanding the stopped pipe is essential in calculating the frequency of acoustic systems that mix both closed and open ends.
Calculating Frequency of a Standing Wave
Calculating the frequency (f) of a standing wave in a stopped pipe involves understanding the relationship between wavelength, speed of sound, and the physical length of the pipe. The formula we use is:\[ f = \frac{v}{\lambda} \]where \( v \) is the speed of sound, and \( \lambda \) is the wavelength.
For a stopped pipe, using the fundamental frequency, the length of the pipe allows \( \frac{\lambda}{4} \) to fit within it. So the wavelength \( \lambda \) can be calculated as:\[ \lambda = 4 \times L \]With the wavelength known, simply substitute it into the frequency formula to find:\[ f = \frac{v}{4 \times L} \]By inserting the known values, it's possible to calculate specific frequencies, allowing us to understand the acoustic properties of the pipe as it relates to the sound it will produce.
Wavelength and Speed of Sound
The wavelength of a sound wave is an essential concept in acoustics, as it helps determine the frequency and pitch of the sound that we hear. Wavelength is the physical length of one complete cycle of a wave, and it's directly related to the speed of sound in the medium through which it travels.
In air, at room temperature, the speed of sound is approximately 344 meters per second. This value changes slightly with variations in air temperature and pressure but provides a good basis for calculations. The relationship between speed, frequency, and wavelength is given by:\[ v = f \times \lambda \]where \( v \) is the speed of sound, \( f \) is the frequency, and \( \lambda \) is the wavelength.
  • Higher frequency results in shorter wavelength.
  • Lower frequency results in longer wavelength.
Understanding how wavelength and the speed of sound interact helps in visualizing how musical notes are generated in various environments, aiding in both mundane applications and more complex acoustic engineering challenges.

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Most popular questions from this chapter

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 standing wave with a frequency of 1100 \(\mathrm{Hz}\) in a column of methane \(\left(\mathrm{CH}_{4}\right)\) at \(20.0^{\circ} \mathrm{C}\) produces nodes that are 0.200 \(\mathrm{m}\) apart. What is the value of \(\gamma\) for methane? (The molar mass of methane is 16.0 \(\mathrm{g} / \mathrm{mol}\) )

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?

Singing in the Showen. 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?

An oscillator vibrating at 1250 Hz produces a sound wave that travels through an ideal gas at 325 \(\mathrm{m} / \mathrm{s}\) when the gas temperature is \(22.0^{\circ} \mathrm{C}\) . For a certain experiment, you need to have the same uscillator produce sound of waveleugth 28.5 \(\mathrm{cm}\) in this gas. What should the gas temperature be to achieve this wavelength?
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