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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 long tube contains air at a pressure of 1.00 atm and a temperature of 77.0\(^\circ\)C. The tube is open at one end and closed at the other by a movable piston. A tuning fork that vibrates with a frequency of 500 Hz is placed near the open end. Resonance is produced when the piston is at distances 18.0 cm, 55.5 cm, and 93.0 cm from the open end. (a) From these values, what is the speed of sound in air at 77.0\(^\circ\)C? (b) From the result of part (a), what is the value of \(\gamma\)? (c) These results show that a displacement antinode is slightly outside the open end of the tube. How far outside is it?

A turntable 1.50 m in diameter rotates at 75 rpm. Two speakers, each giving off sound of wavelength 31.3 cm, are attached to the rim of the table at opposite ends of a diameter. A listener stands in front of the turntable. (a) What is the greatest beat frequency the listener will receive from this system? (b) Will the listener be able to distinguish individual beats?

The intensity due to a number of independent sound sources is the sum of the individual intensities. (a) When four quadruplets cry simultaneously, how many decibels greater is the sound intensity level than when a single one cries? (b) To increase the sound intensity level again by the same number of decibels as in part (a), how many more crying babies are required?

Consider a sound wave in air that has displacement amplitude 0.0200 mm. Calculate the pressure amplitude for frequencies of (a) 150 Hz; (b) 1500 Hz; (c) 15,000 Hz. In each case compare the result to the pain threshold, which is 30 Pa.

Singing in the Shower. 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 =\) \(2L /n\) and the frequencies are given by \({f_n}\) \(= n$$v/2L = nf$$_1\), where \(n =\) 1, 2, 3, . . . . (b) Modeling it as a pipe, find the frequency of the fundamental and the first two overtones for a shower 2.50 m tall. Are these frequencies audible?

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