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Standing sound waves are produced in a pipe that is 1.20 m long. For the fundamental and first two overtones, determine the locations along the pipe (measured from the left end) of the displacement 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.

Short Answer

Expert verified
For open pipe nodes: fundamental at 0.6 m; overtone nodes at 0.4 m, 0.8 m. For closed pipe: nodes at 0 m, 0.4 m.

Step by step solution

01

Understanding Open Pipe Harmonics

In an open-ended pipe, both ends are antinodes for displacement. The fundamental frequency has one node in the middle (half-wave) and produces the first harmonic. Overtones are higher harmonics. For the second harmonic, there are two nodes (third harmonic for the next).
02

Calculate Displacement Nodes in Open Pipe

For the fundamental frequency in an open pipe, the displacement node is at L/2: \(1.20/2 = 0.60\,m\). For the first overtone (second harmonic), displacement nodes are at \( L/3 = 0.40 \text{ m} \) and \( 2L/3 = 0.80 \text{ m} \).
03

Identify Pressure Nodes in Open Pipe

The displacement nodes are pressure antinodes, and vice versa. Thus, for the open pipe, pressure nodes occur at the ends (0 and 1.20 m) and midpoints between nodes. For the fundamental, no intermediate pressure node; for the first overtone, at \( L/2 = 0.60 \text{ m} \).
04

Understanding Closed Pipe Harmonics

A closed pipe is closed at one end (node of displacement and antinode of pressure) and open at another (antinode of displacement and node of pressure). The fundamental has one-quarter wavelength, next one adds three-quarters for first overtone.
05

Calculate Displacement Nodes in Closed Pipe

For a closed pipe's fundamental, the displacement node is at the beginning \((0\,m)\) and the first overtone appears at \( L/3 = 0.40 \text{ m} \).
06

Identify Pressure Nodes in Closed Pipe

In a closed pipe, the closed end corresponds to a pressure antinode. The first displacement node is the first pressure node. For the fundamental, no node before the end; for the first overtone, \( 2L/3 = 0.80 \text{ m} \).

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Key Concepts

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

Open Pipe Harmonics
An open pipe is one where both ends are open to the air, allowing sound waves to pass through easily. In such pipes, the ends act as points of maximum air displacement, or antinodes. This is because the air can freely move at these points. For standing wave patterns, this characteristic affects how harmonics, or the natural frequencies of the pipe, are structured.

The fundamental frequency, or first harmonic, creates a wave pattern with just one node, which is a point of zero displacement, located exactly in the middle of the pipe. The distance between each node and antinode in this case is half of the pipe length. For a pipe length of 1.20 m, the node is at the midpoint, specifically at 0.60 m.

If we move to the first overtone, which is the second harmonic, this pattern changes. In this case, there are two additional nodes, dividing the pipe into thirds. This results in displacement nodes located at 0.40 m and 0.80 m along the pipe. As you progress to higher harmonics, more nodes form along the pipe, continuing this pattern.
Closed Pipe Harmonics
In contrast to open pipes, closed pipes have one end sealed and the other open. The closed end of the pipe is a displacement node because the air cannot move freely, while the open end remains a displacement antinode, where maximum movement occurs.

The standing wave pattern for the fundamental frequency in a closed pipe is a quarter wavelength, meaning the pattern from the closed end to the first antinode occupies only a quarter of the wave cycle. For a pipe 1.20 m long, there is a node at the closed end (0 m), and the length accommodates a quarter wave for the first harmonic.

The first overtone in a closed pipe involves adding three quarters of the wavelength to the existing pattern, forming a new displacement node at approximately 0.40 m from the closed end. This pattern means that closed pipes do not have straightforward integer multiples for harmonics, unlike open pipes, and leads to differences in the possible frequencies produced. This unique configuration is why closed pipes are often used in musical instruments like clarinets or certain organ pipes.
Displacement Nodes
Displacement nodes are essential in understanding standing waves, as they are points along the medium where there is no movement at all. These nodes occur due to the destructive interference of waves traveling in opposite directions, causing them to cancel out at specific points.

In open pipes, displacement nodes appear in regular patterns along the pipe. For the fundamental frequency in an open pipe, there's a single displacement node in the center. For the first overtone, nodes appear at one-third and two-thirds along the pipe.

In closed pipes, the first displacement node is always at the closed end because air cannot move here. For the first overtone, an additional node will appear further along the pipe, marked by a decrease in displacement as the wave reflects back towards the open end.
Pressure Nodes
Pressure nodes are different from displacement nodes; these are the points of constant air pressure, where changes in pressure constructively or destructively interfere to create no variation.

In open pipes, pressure nodes occur at the very ends, where there is the greatest air movement, corresponding to displacement antinodes. Between the nodes of displacement, pressure nodes can be located along the pipe equally spaced.

For closed pipes, the closed end is a pressure antinode due to no air fluctuations and a maximum pressure level, while the open end is a node of pressure where pressure varies minimally. In terms of harmonics, a pressure node is halfway between two displacement nodes in an open pipe, whereas, in a closed pipe, it coincides with the first displacement node. These relationships help define how sound waves behave in pipes and dictate the musical notes possible from various pipe instruments.

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Most popular questions from this chapter

A soprano and a bass are singing a duet. While the soprano sings an A-sharp at 932 Hz, the bass sings an A-sharp but three octaves lower. In this concert hall, the density of air is 1.20 kg/m\(^3\) and its bulk modulus is 1.42 \(\times\) 10\(^5\) Pa. In order for their notes to have the same sound intensity level, what must be (a) the ratio of the pressure amplitude of the bass to that of the soprano and (b) the ratio of the displacement amplitude of the bass to that of the soprano? (c) What displacement amplitude (in m and in nm) does the soprano produce to sing her A-sharp at 72.0 dB?

The shock-wave cone created by a 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 m/s. (a) What is the Mach number of the shuttle at this instant, and (b) how fast (in m/s and in mi/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 m/s ?

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 m/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 cm, what is the frequency of this standing wave? (b) What is the frequency of the fundamental standing wave in the air column if the test tube is half filled with water?

Example 16.1 (Section 16.1) showed that for sound waves in air with frequency 1000 Hz, a displacement amplitude of 1.2 \(\times\) 10\(^{-8}\) m produces a pressure amplitude of 3.0 \(\times\) 10\(^{-2}\) Pa. (a) What is the wavelength of these waves? (b) For 1000-Hz waves in air, what displacement amplitude would be needed for the pressure amplitude to be at the pain threshold, which is 30 Pa? (c) For what wavelength and frequency will waves with a displacement amplitude of 1.2 \(\times\) 10\(^{-8}\) m produce a pressure amplitude of 1.5 \(\times\) 10\(^{-3}\) Pa?

A violinist is tuning her instrument to concert A (440 Hz). She plays the note while listening to an electronically generated tone of exactly that frequency and hears a beat frequency of 3 Hz, which increases to 4 Hz when she tightens her violin string slightly. (a) What was the frequency of the note played by her violin when she heard the 3-Hz beats? (b) To get her violin perfectly tuned to concert A, should she tighten or loosen her string from what it was when she heard the 3-Hz beats?

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