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Chapter 17: Problem 46

Pipe \(A\), which is \(1.20 \mathrm{~m}\) long and open at both ends, oscillates at its third lowest harmonic frequency. It is filled with air for which the speed of sound is \(343 \mathrm{~m} / \mathrm{s}\). Pipe \(B\), which is closed at one end, oscillates at its second lowest harmonic frequency. This frequency of \(B\) happens to match the frequency of \(A .\) An \(x\) axis extends along the interior of \(B\), with \(x=0\) at the closed end. (a) How many nodes are along that axis? What are the (b) smallest and (c) second smallest value of \(x\) locating those nodes? (d) What is the fundamental frequency of \(B\) ?

Short Answer

Expert verified
There are 2 nodes inside Pipe B. The smallest x is 0.150 m and the second smallest is 0.450 m. The fundamental frequency of B is 143.042 Hz.

Step by step solution

01

Determine the frequency of Pipe A

Pipe A, which is open at both ends, has its third lowest harmonic frequency. Since it's open at both ends, the harmonics follow the formula for a pipe open at both ends: \[ f_n = \frac{n \cdot v}{2L} \]where \(n\) is the harmonic number, \(v\) is the speed of sound, and \(L\) is the length of the pipe. For the third harmonic (\(n = 3\)), the frequency is:\[ f_3 = \frac{3 \cdot 343}{2 \cdot 1.20} \approx 429.167 \text{ Hz} \]
02

Determine the harmonic number for Pipe B

Pipe B is closed at one end, which means it supports odd harmonics only: \[ f_n = \frac{n \cdot v}{4L} \]Since Pipe B matches the frequency of Pipe A at its second lowest harmonic frequency, and Pipe A's frequency is at its third harmonic. Pipe B's second harmonic is its first odd value (\(n = 3\)), so:\[ \frac{3 \cdot 343}{4L} = 429.167 \text{ Hz} \rightarrow L = \frac{3 \cdot 343}{4 \cdot 429.167} \approx 0.600 \text{ m} \]
03

Calculate the number of nodes in Pipe B

For Pipe B at its second harmonic (\(n=3\)), the pattern would show:1. A node at the closed end (\(x=0\)), and antinodes at the open end (\(x=L\)).2. Additional node positions within the pipe can be computed as: - Half-wavelength patterns would form nodes at regular intervals.Hence, along its length, there will be 2 internal nodes at positions of quarter and three-quarters of the pipe's total length.
04

Determine node positions in Pipe B

Since the pipe has a length \(L = 0.600 \text{ m}\), and the natural node positions for the third harmonic are at one-fourth and three-fourths of the wavelength, * First (smallest) node position away from \(x=0\) is: - \( x = \frac{L}{4} = 0.600 \times \frac{1}{4} = 0.150 \text{ m} \) * Second node position is: - \( x = \frac{3L}{4} = 0.600 \times \frac{3}{4} = 0.450 \text{ m} \)
05

Calculate the fundamental frequency of Pipe B

The fundamental frequency (first harmonic) puts \(n=1\), so:\[ f_1 = \frac{v}{4L} \]Plugging in our known values:\[ f_1 = \frac{343}{4 \times 0.600} = 143.042 \text{ Hz} \]

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Key Concepts

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

Open and Closed Pipes
When dealing with musical pipes, understanding whether a pipe is open or closed at its ends is crucial. This characteristic determines the type of harmonics that can form in the pipe.
- **Open Pipes**: A pipe that is open at both ends supports harmonics where nodes and antinodes alternate. In open pipes, the number of nodes or antinodes goes up as you move to higher harmonics. All harmonics (odd and even) are supported in open pipes. - **Closed Pipes**: A pipe that is closed at one end and open at the other supports only odd harmonics. At the closed end, a node is always present, while an antinode forms at the open end. The length of the closed pipe dictates how many nodes form inside.
These characteristics influence the pipe's fundamental frequency and the structure of subsequent harmonics. In the exercise, Pipe A, an open pipe, has its third harmonic playing, while Pipe B, closed at one end, matches this frequency with its second harmonic.
Harmonic Frequency Calculation
Harmonic frequencies in pipes rely on the pipe's length, the speed of sound, and whether a pipe is open or closed.
For open pipes:
  • Harmonic frequency: \( f_n = \frac{n \cdot v}{2L} \)
  • Where \( n \) is the harmonic number, \( v \) the speed of sound, and \( L \) the length.
For closed pipes:
  • Harmonic frequency: \( f_n = \frac{n \cdot v}{4L} \)
  • Only odd harmonics (1st, 3rd, 5th, etc.) exist here.
In the problem, Pipe A oscillates at its third harmonic while Pipe B matches this frequency with its third harmonic (which is the second odd harmonic). Calculations show the frequency of Pipe A is approximately \(429.167 \text{ Hz}\), which helps determine the length and node positions in Pipe B.
Speed of Sound
The speed of sound is a key factor when working with harmonic frequencies and pipes. In this context, sound speed refers to how fast sound waves travel through the air in a given environment.
- **Typical Value in Air**: At room temperature, the speed of sound in air is about \(343 \text{ m/s}\). This value can change with temperature and the medium through which sound travels.- **Impact on Frequencies**: Harmonic frequencies in pipes are directly related to the speed of sound. If the speed of sound increases, frequencies of the harmonics also increase, assuming the length of the pipe stays the same.
In the exercise, knowing the speed of sound as \(343 \text{ m/s}\) allows us to calculate the specific frequencies for the harmonics in both open and closed pipes. The speed of sound determines how many waves fit within the pipe's length, ultimately affecting how we perceive the sound.

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Most popular questions from this chapter

A certain sound source is increased in sound level by \(30.0 \mathrm{~dB}\). By what multiple is (a) its intensity increased and (b) its pressure amplitude increased?

Suppose a spherical loudspeaker emits sound isotropically at \(10 \mathrm{~W}\) into a room with completely absorbent walls, floor, and ceiling (an anechoic chamber). (a) What is the intensity of the sound at distance \(d=3.0 \mathrm{~m}\) from the center of the source? (b) What is the ratio of the wave amplitude at \(d=4.0 \mathrm{~m}\) to that at \(d=3.0 \mathrm{~m}\) ?

A person on a railroad car blows a trumpet note at \(440 \mathrm{~Hz}\). The car is moving toward a wall at \(20.0 \mathrm{~m} / \mathrm{s}\). Find the sound frequency (a) at the wall and (b) reflected back to the trumpeter.

A point source that is stationary on an \(x\) axis emits a sinusoidal sound wave at a frequency of \(686 \mathrm{~Hz}\) and speed \(343 \mathrm{~m} / \mathrm{s}\). The wave travels radially outward from the source, causing air molecules to oscillate radially inward and outward. Let us define a wavefront as a line that connects points where the air molecules have the maximum, radially outward displacement. At any given instant, the wavefronts are concentric circles that are centered on the source. (a) Along \(x\), what is the adjacent wavefront separation? Next, the source moves along \(x\) at a speed of \(110 \mathrm{~m} / \mathrm{s}\). Along \(x\), what are the wavefront separations (b) in front of and (c) behind the source?

A man strikes one end of a thin rod with a hammer. The speed of sound in the rod is 15 times the speed of sound in air. A woman, at the other end with her ear close to the rod, hears the sound of the blow twice with a \(0.12\) s interval between; one sound comes through the rod and the other comes through the air alongside the rod. If the speed of sound in air is \(343 \mathrm{~m} / \mathrm{s}\), what is the length of the rod?
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