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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
Pipe B has 1 node at 0.30 m; its fundamental frequency is 143.75 Hz.

Step by step solution

01

Determine the frequency of Pipe A

Pipe A, open at both ends, resonates at its third lowest harmonic frequency. In an open pipe, the resonant frequencies for a pipe are given by the formula: \[ f_n = \frac{n \, v}{2L} \] where \( n \) is the harmonic number, \( v \) is the speed of sound, and \( L \) is the length of the pipe. For the third harmonic (\( n = 3 \)), we have:\[ f_3 = \frac{3 \, \times \, 343 \, \text{m/s}}{2 \, \times \, 1.20 \, \text{m}} \approx 428.75 \, \text{Hz} \]
02

Relate to Pipe B's frequency

Pipe B is closed at one end and resonates at its second lowest harmonic frequency, which matches the frequency of Pipe A. For a closed pipe, the resonant frequencies are given by:\[ f_n = \frac{(2n-1) \, v}{4L} \] For the second harmonic (first overtone), \( n = 2 \), and it matches the frequency of Pipe A at 428.75 Hz.Thus,\[ \frac{3 \, v}{4L} = 428.75 \, \text{Hz} \] We can use this equation to find the properties of the pipe.
03

Calculate Pipe B's length

Using the previous step:\[ \frac{3 \, \times \, 343 \, \text{m/s}}{4L} = 428.75 \, \text{Hz} \] Solving for \( L \), we get:\[ L = \frac{3 \, \times \, 343}{4 \, \times \, 428.75} \approx 0.60 \, \text{m} \]
04

Find nodes in Pipe B

In a closed pipe of length \( 0.60 \, \text{m} \), the second harmonic (first overtone) supports nodes at points along the pipe. For the second harmonic:- 1 node appears halfwayStarting from 0 to the length \( L \), the number of nodes is 1.
05

Calculate the smallest value of x

For the second harmonic in a closed pipe, the first node appears at:\[ x_1 = \frac{L}{2} = \frac{0.60}{2} = 0.30 \, \text{m} \] This is the smallest location of an x-value node.
06

Determine the fundamental frequency of Pipe B

The fundamental frequency has only one node at the open end of a closed pipe. Using the formula for the first harmonic:\[ f_1 = \frac{v}{4 \times L} = \frac{343}{4 \times 0.60} \approx 143.75 \, \text{Hz} \]

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Key Concepts

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

Resonant Frequency
The resonant frequency of a pipe determines how it naturally vibrates when sound waves pass through it. For both open and closed pipes, this frequency is when the natural vibrational frequency matches one of the pipe’s harmonic frequencies. The formula for finding the resonant frequency differs based on whether the pipe is open or closed.
In an open pipe, the resonant frequencies are calculated using the formula \( f_n = \frac{n \, v}{2L} \), where \( n \) is the harmonic number, \( v \) is the speed of sound, and \( L \) is the pipe's length. For a closed pipe, the resonant frequencies are given by \( f_n = \frac{(2n-1) \, v}{4L} \). This different formula accounts for the unique pattern of nodes and antinodes in closed pipes.
Nodes are points along the pipe where there is no movement. Antinodes, meanwhile, are points of maximum movement. Understanding resonant frequencies is crucial for determining how sound behaves within different types of pipes.
Open and Closed Pipes
Open and closed pipes have unique characteristics for generating sound waves and forming resonant frequencies. An open pipe is open at both ends. This allows air to vibrate freely, which means the ends act as antinodes. A pipe open at both ends supports harmonics at all integer values of \( n \).
A closed pipe, contrarily, has one end closed, creating a node at the closed end. This arrangement supports only odd harmonics, expressed by increasing odd multiples of one-fourth the wavelength. The resonant frequencies are given by different formulas, as noted above, depending on being open or closed.
Understanding the differences helps to explain why an open pipe can produce a wider range of harmonic frequencies compared to a closed pipe. Additionally, the physical dimensions of the pipe impact the wavelengths and frequencies it can support.
Sound Waves
Sound travels through pipes as waves, which are longitudinal and consist of compressions and rarefactions. These waves travel at a speed determined by the medium—for air, it's about 343 m/s under normal conditions. In musical instruments, these sound waves are manipulated by the dimensions and type of the pipe to create musical tones.
The speed of sound is critical for calculating the frequencies at which pipes resonate. Whether a pipe is open or closed alters how sound waves move. The distinction between nodes (points of no displacement) and antinodes (points of maximum displacement) is central to how sound waves establish particular resonant frequencies in pipes.
By controlling these factors, musicians can produce various pitches and tones, illustrating the foundational principles of acoustics.
Fundamental Frequency
The fundamental frequency is the lowest frequency at which a pipe resonates. It's also known as the first harmonic. For an open pipe, the formula \( f_1 = \frac{v}{2L} \) is used to find this frequency, where a quarter of the wavelength fits into the length of the pipe.
In contrast, the fundamental frequency in a closed pipe is \( f_1 = \frac{v}{4L} \), because only odd harmonics are possible in closed pipes. This means the pipe supports a quarter of a wavelength at the fundamental frequency. The lower end of the harmonic series in a closed pipe starts with a frequency that's effectively half the next odd harmonic in the open counterpart.
Understanding fundamental frequencies is essential for recognizing how instruments like flutes and clarinets produce their base notes, with each instrument’s unique design emphasizing different harmonic properties.

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Most popular questions from this chapter

The speed of sound in a certain metal is \(v_{m} .\) One end of a long pipe of that metal of length \(L\) is struck a hard blow. A listener at the other end hears two sounds, one from the wave that travels along the pipe's metal wall and the other from the wave that travels through the air inside the pipe. (a) If \(v\) is the speed of sound in air, what is the time interval \(\Delta t\) between the arrivals of the two sounds at the listener's ear? (b) If \(\Delta t=1.00 \mathrm{~s}\) and the metal is steel, what is the length \(L ?\)

One of the harmonic frequencies of tube \(A\) with two open ends is \(325 \mathrm{~Hz}\). The next-highest harmonic frequency is \(390 \mathrm{~Hz}\). (a) What harmonic frequency is next highest after the harmonic frequency \(195 \mathrm{~Hz}\) ? (b) What is the number of this next-highest harmonic? One of the harmonic frequencies of tube \(B\) with only one open end is \(1080 \mathrm{~Hz}\). The next-highest harmonic frequency is 1320 \(\mathrm{Hz}\). (c) What harmonic frequency is next highest after the harmonic frequency \(600 \mathrm{~Hz} ?\) (d) What is the number of this next-highest harmonic?

A sound wave in a fluid medium is reflected at a barrier so that a standing wave is formed. The distance between nodes is \(3.8\) \(\mathrm{cm}\), and the speed of propagation is \(1500 \mathrm{~m} / \mathrm{s}\). Find the frequency of the sound wave.

A sound wave travels out uniformly in all directions from a point source. (a) Justify the following expression for the displacement \(s\) of the transmitting medium at any distance \(r\) from the source: $$ s=\frac{b}{r} \sin k(r-v t) $$ where \(b\) is a constant. Consider the speed, direction of propagation, periodicity, and intensity of the wave. (b) What is the dimension of the constant \(b ?\)

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?

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