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Chapter 17: Problem 35

A tube of air is open at only one end and has a length of \(1.5 \mathrm{~m}\). This tube sustains a standing wave at its third harmonic. What is the distance between one node and the adjacent antinode?

Short Answer

Expert verified
The distance between a node and an adjacent antinode is 0.5 meters.

Step by step solution

01

Understand the Tube Configuration

Recognize that the tube is open at only one end. This means it supports standing waves where there is a node at the closed end and an antinode at the open end.
02

Identify the Harmonic Mode

Identify that the problem refers to the third harmonic. In a tube open at one end, the third harmonic corresponds to a pattern where there are three-quarters of a wavelength within the length of the tube.
03

Calculate Wavelength for Third Harmonic

For the third harmonic in an open-closed tube, the length of the tube is equal to \(\frac{3}{4}\) of the wavelength (\(\lambda\)). Thus, the wavelength is calculated as: \[ \lambda = \frac{4}{3} \times L = \frac{4}{3} \times 1.5 \mathrm{~m} = 2 \mathrm{~m}. \]
04

Understand Node to Antinode Distance

In a standing wave pattern, the distance from one node to an adjacent antinode is one quarter of the wavelength. This reflects the distance from the stop/start of a compression or rarefaction cycle to the center.
05

Calculate the Node to Antinode Distance

Using the wavelength calculated, determine the node to antinode distance as one-quarter of the wavelength: \[ d = \frac{1}{4} \times \lambda = \frac{1}{4} \times 2 \mathrm{~m} = 0.5 \mathrm{~m}. \]

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Key Concepts

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

Harmonics in Physics
In physics, harmonics are critical when studying waves and vibrations. They describe the various patterns and frequencies at which standing waves resonate.

Each harmonic is an integer multiple of the fundamental frequency. This means:
  • The first harmonic is the simplest standing wave pattern, with only one wave.
  • The second harmonic has two waves, and so on.
When dealing with tubes or strings, harmonics help us understand how different frequencies correspond to distinct vibrational patterns, each forming nodes and antinodes.

Recognizing these patterns aids in determining characteristics like wave speed and wavelength, essential in various fields from music to acoustics.
Open-Closed Tube
An open-closed tube is unique because it is open at one end and closed at the other.

This configuration affects how standing waves form inside the tube. At the closed end, there is always a node (point of no movement), while at the open end, there is an antinode (point of maximum movement).

For example, in the third harmonic of an open-closed tube:
  • The tube contains three-fourths of a wave.
  • This pattern results in specific node and antinode placements along the tube's length.
Understanding this setup is vital for calculating wave properties like wavelength and frequency using the known length and harmonic number.
Wavelength Calculation
Wavelength calculation in standing waves involves finding the distance a wave travels to complete one full cycle.

For an open-closed tube, knowing the harmonic number helps in determining how much of the wavelength fits inside the tube.

In the third harmonic:
  • The tube鈥檚 length equals three-fourths (\(\frac{3}{4}\)) of the wavelength.
  • Thus, to find the wavelength \(\lambda\), use the equation:\[\lambda = \frac{4}{3} \times L\]
  • For a tube length of 1.5 meters, the wavelength becomes 2 meters.
This approach allows for easy calculation of node-to-antinode distances and other related properties, crucial in wave mechanics.

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Most popular questions from this chapter

Sound enters the ear, travels through the auditory canal, and reaches the eardrum. The auditory canal is approximately a tube open at only one end. The other end is closed by the eardrum. A typical length for the auditory canal in an adult is about \(2.9 \mathrm{~cm} .\) The speed of sound is \(343 \mathrm{~m} / \mathrm{s}\). What is the fundamental frequency of the canal? (Interestingly, the fundamental frequency is in the frequency range where human hearing is most sensitive.)

In Concept Simulation \(17.2\) at you can explore the concepts that are important in this problem. A 440.0-Hz tuning fork is sounded together with an out-of-tune guitar string, and a beat frequency of \(3 \mathrm{~Hz}\) is heard. When the string is tightened, the frequency at which it vibrates increases, and the beat frequency is heard to decrease. What was the original frequency of the guitar string?

The range of human hearing is roughly from twenty hertz to twenty kilohertz. Based on these limits and a value of \(343 \mathrm{~m} / \mathrm{s}\) for the speed of sound, what are the lengths of the longest and shortest pipes (open at both ends and producing sound at their fundamental frequencies) that you expect to find in a pipe organ?

Suppose the strings on a violin are stretched with the same tension and each has the same length between its two fixed ends. The musical notes and corresponding fundamental frequencies of two of these strings are \(\mathrm{G}(196.0 \mathrm{~Hz})\) and \(\mathrm{E}(659.3 \mathrm{~Hz}) .\) The linear density of the E string is \(3.47 \times 10^{-4} \mathrm{~kg} / \mathrm{m}\). What is the linear density of the \(\mathrm{G}\) string?

The fundamental frequency of a vibrating system is \(400 \mathrm{~Hz}\). For each of the following systems, give the three lowest frequencies (excluding the fundamental) at which standing waves can occur: (a) a string fixed at both ends, (b) a cylindrical pipe with both ends open, and (c) a cylindrical pipe with only one end open.
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