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Chapter 12: Problem 40

A pipe in air at 23.0\(^\circ\)C is to be designed to produce two successive harmonics at 280 Hz and 320 Hz. How long must the pipe be, and is it open or closed?

Short Answer

Expert verified
The pipe must be 4.32 meters long, and it is open.

Step by step solution

01

Understanding Harmonics in Pipes

In pipes, especially in acoustics, harmonics refer to the multiples of the fundamental frequency at which the system resonates. An open pipe supports harmonics at all multiples (odd and even) of the fundamental frequency, while a closed pipe supports only odd harmonics.
02

Identifying the Type of Pipe

Given that the frequencies 280 Hz and 320 Hz are consecutive harmonics, it suggests that the harmonics might be successive odd harmonics. However, calculating the ratio \( \frac{320}{280} = \frac{8}{7} \times 2.5 \), these cannot be two successive odd harmonics (as the ratio is not odd as required for a closed pipe), thus the pipe is likely open.
03

Expressing Harmonics for an Open Pipe

For an open pipe, consecutive harmonics are found at \( n \cdot f_1 \) and \( (n+1)\cdot f_1 \). Here \( f_1 \) is the fundamental frequency, \( n \) is an integer. The difference between the harmonics given is 40 Hz (i.e., \( 320 - 280 = 40 \)). Thus, the fundamental frequency \( f_1 \) is 40 Hz.
04

Calculating the Speed of Sound

The speed of sound \( v \) at a given temperature \( T \) in degrees Celsius is given by \( v = 331 + 0.6T \). At 23.0掳C, \( v = 331 + 0.6 \times 23 = 345.8 \text{ m/s} \).
05

Determining the Length of the Pipe

For an open pipe, the fundamental frequency is given by \( f_1 = \frac{v}{2L} \). Solving for \( L \), we have \( L = \frac{v}{2f_1} \). Using \( v = 345.8 \text{ m/s} \) and \( f_1 = 40 \text{ Hz} \), we find \[ L = \frac{345.8}{2 \times 40} = 4.3225 \text{ meters} \].

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Key Concepts

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

Open Pipe Harmonics
When we talk about open pipe harmonics, we're referring to the natural frequencies at which an open pipe resonates. The harmonics, or overtones, in an open pipe consist of both odd and even harmonics. This means that in an open pipe, harmonics occur at integer multiples of the fundamental frequency. For example, if the fundamental frequency is a certain value, the harmonics would be at 1, 2, 3 times this frequency, and so forth.

In contrast, a closed pipe (where one end is blocked) only supports odd harmonics, such as 1, 3, 5 times the fundamental frequency. This difference in harmonic structure helps us deduce whether a pipe is open or closed based on the frequencies it resonates at.
Fundamental Frequency
The fundamental frequency, often denoted as \( f_1 \), is the lowest frequency at which a system like a pipe resonates. It serves as the building block for all other harmonics in the system. In the context of an open pipe, the first harmonic is equal to the fundamental frequency, and the other harmonics are integer multiples of this frequency.

If we consider two consecutive harmonics in an open pipe, they occur at \( n \cdot f_1 \) and \( (n+1) \cdot f_1 \). The fact that the difference between the given frequencies is constant, specifically 40 Hz in this exercise, directly tells us that the fundamental frequency is 40 Hz.
Speed of Sound
The speed of sound in air depends on the temperature of the air. This is because warmer air molecules move more quickly, allowing sound waves to travel faster. The formula to calculate the speed of sound \( v \) at a given temperature \( T \) is \( v = 331 + 0.6T \).

So, for air at 23.0掳C, the speed of sound can be calculated as \( v = 331 + 0.6 \times 23 = 345.8 \text{ m/s} \). This speed of sound is crucial when determining the fundamental frequency and, subsequently, the length of the pipe for a given frequency.
Consecutive Harmonics
Consecutive harmonics are harmonics that follow one another closely. For an open pipe, these can be expressed as \( n \cdot f_1 \) and \( (n+1) \cdot f_1 \). The difference in frequency values between consecutive harmonics helps identify the fundamental frequency when it is not directly given.

In the given problem, the frequencies of 280 Hz and 320 Hz are consecutive harmonics. The 40 Hz difference between these harmonics means that this value corresponds to the fundamental frequency. By identifying the fundamental frequency and using the speed of sound, we can then calculate the length of the pipe required for these harmonics.

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Most popular questions from this chapter

A \(\textbf{Doppler flow meter}\) uses ultrasound waves to measure blood-flow speeds. Suppose the device emits sound at 3.5 MHz, and the speed of sound in human tissue is about 1540 m\(/\)s. What is the expected beat frequency if blood is flowing in large leg arteries at 3.0 cm\(/\)s directly away from the sound source?

Workers around jet aircraft typically wear protective devices over their ears. Assume that the sound level of a jet airplane engine, at a distance of 30 m, is 130 dB, and that the average human ear has an effective radius of 2.0 cm.What would be the power intercepted by an unprotected ear at a distance of 30 m from a jet airplane engine?

The intensity at the threshold of hearing for the human ear at a frequency of about 1000 Hz is \(I_0 =\) 1.0 \(\times\) 10\(^{-12}\) W\(/\)m\(^2\), for which \(\beta\), the sound level, is 0 dB. The threshold of pain at the same frequency is about 120 dB, or \(I =\) 1.0 W\(m^2\), corresponding to an increase of intensity by a factor of 10\(^{12}\). By what factor does the displacement amplitude, \(A\), vary?

The predominant frequency of a certain fire truck's siren is 1650 Hz when at rest.What frequency do you detect if you move with a speed of 30.0 m\(/\)s (a) toward the fire truck, and (b) away from it?

Two automobiles are equipped with the same singlefrequency horn. When one is at rest and the other is moving toward the first at 18 m\(/\)s, the driver at rest hears a beat frequency of 4.5 Hz. What is the frequency the horns emit? Assume \(T =\) 20\(^\circ\)C.
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