Problem 108 - Pearson Physics

Question

An organ pipe \(2.5 \mathrm{~m}\) long is open at one end and closed at the other end. What is the linear distance between a node and the adjacent antinode for the third harmonic in this pipe?

Step-by-Step Solution

Answer

The distance is 0.833 meters.

Step 1: Understanding the Problem

We are given an organ pipe that is open at one end and closed at the other, with a length of \(2.5 \mathrm{~m}\). We need to find the linear distance between a node and the adjacent antinode for the third harmonic. In such a pipe, the pattern of the standing wave will have specific nodes and antinodes.

Step 2: Recalling Harmonic Features in a Closed Pipe

In a closed-end pipe, harmonics are odd multiples: 1st, 3rd, 5th, etc. For each harmonic, the number of quarter wavelengths, \(n\), is given by \(n = ext{harmonic number}\). Thus, for the third harmonic, we will have \(n = 3\) quarter wavelengths within the pipe.

Step 3: Applying Closed Pipe Harmonic Formula

The length \(L\) of the closed pipe is equal to \(n\) quarter wavelengths for the \(n ext{th}\) harmonic. Thus, for the third harmonic, \(L = 3\left(\frac{\lambda}{4}\right)\), where \(\lambda\) is the wavelength. With \(L = 2.5 \mathrm{~m}\), we solve for \(\lambda\).

Step 4: Calculating the Wavelength

Solve for the wavelength: \[2.5 \mathrm{~m} = \frac{3\lambda}{4} \implies \lambda = \frac{4 \times 2.5}{3} = \frac{10}{3} \mathrm{~m} = 3.33 \mathrm{~m}.\]

Step 5: Finding Node-Antinode Distance

The distance between a node and an adjacent antinode is equal to a quarter of the wavelength, \(\frac{\lambda}{4}\). Substitute the calculated \(\lambda\): \[\frac{\lambda}{4} = \frac{3.33}{4} = 0.833 \mathrm{~m}.\]

Step 6: Final Calculation Verification

Recheck the steps to ensure the calculations are correctly done and properly align with harmonic properties and the physical setup of the problem.

Key Concepts

Closed-end Organ Pipe

An organ pipe open at one end and closed at the other functions as a closed-end pipe. This setup is quite special in the study of acoustics and physics harmonics. Closed-end organ pipes are commonly used in wind instruments to create specific sounds or notes. Here’s why they are unique:

- One end of the pipe is closed, preventing air from escaping and reflecting sound waves back down the pipe, creating a pattern known as standing waves.
- The other end is open, allowing air movement and facilitating the creation of the first antinode.

The boundary conditions these pipes exhibit mean that they will only allow certain frequencies to resonate within them, specifically odd harmonics, which leads to the creation of the distinctive sound patterns. The length of the pipe determines the nature of these patterns, as it dictates the possible wavelengths and thus the harmonics that can be formed.

Third Harmonic

When we talk about the third harmonic in the context of a closed-end pipe, we need to think about specific patterns that sound waves create inside the pipe. In closed pipes, only odd harmonics form. This pipe setup results in for example the 1st, 3rd, and 5th harmonics, but not the 2nd or 4th.

For the third harmonic:

• The wave setup inside the pipe will include three quarter-wavelengths because the harmonic number directly correlates to the number of quarter wavelengths present.
• This arrangement implies that there will be additional nodes and antinodes compared to the first harmonic, enhancing the complexity of the sound wave.

Understanding which harmonics can exist in a given medium is crucial for solving problems related to wave patterns in organ pipes, as demonstrated by reaching the third harmonic.

Standing Waves

Standing waves are integral to the understanding of harmonics in any closed-end pipe system. These are the wave patterns formed by the interference of two waves of the same frequency traveling in opposite directions. This results in areas of constructive and destructive interference, creating nodes and antinodes inside the pipe.

In closed-end organ pipes, standing waves:

• Have nodes at the closed end where there is no displacement of air particles, and antinodes at the open end.
• Occur because the pipe efficiently reflects the sound wave back upon itself, allowing it to interfere and form these patterns.

Unlike in open pipes where both ends can be antinodes, the closed end mandates that it always be a node. This difference profoundly affects the wavelengths and frequencies that can resonate within a closed-end pipe, showing why only specific harmonics appear.

Node-Antinode Distance

The node-antinode distance in a standing wave is vital for calculating specific intervals between features in a closed-end pipe. This distance is constant within the same harmonic and equals a quarter of the wave's wavelength.

In the given problem, with the pipe supporting the third harmonic, it’s crucial to:

• Determine the wavelength for this specific harmonic.
• Calculate the node-antinode distance by dividing the wavelength by four.

The calculated wavelength being 3.33 meters leads us to find that the distance between a node and the adjacent antinode is approximately 0.833 meters. This insightful measurement bridges practical understanding and theoretical wave mechanics, aiding in analyzing wave behavior within the pipe effectively.