Problem 35 - Physics

Question

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?

Step-by-Step Solution

Answer

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

Step 1: 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.

Step 2: 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.

Step 3: 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: \[ \lambda = \frac{4}{3} \times L = \frac{4}{3} \times 1.5 \mathrm{~m} = 2 \mathrm{~m}. \]

Step 4: 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.

Step 5: Calculate the Node to Antinode Distance

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

Key Concepts

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’s 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.