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Chapter 16: Problem 65

An organ pipe has two successive harmonics with frequencies 1372 and 1764 \(\mathrm{Hz}\) (a) Is this an open or a stopped pipe? Explain. (b) What two harmonics are these? (c) What is the length of the pipe?

Short Answer

Expert verified
(a) Open pipe. (b) 3rd and 4th harmonics. (c) Pipe length is approximately 0.438 m.

Step by step solution

01

Identify the Fundamental Frequency

The two successive harmonic frequencies are given as 1372 Hz and 1764 Hz. The difference between these two frequencies is 1764 - 1372 = 392 Hz. This difference represents the fundamental frequency because, in a harmonic series, consecutive harmonics differ by the fundamental frequency.
02

Determine Pipe Type

In an open pipe, harmonics occur at all integer multiples of the fundamental frequency, while in a stopped pipe, harmonics occur at odd integer multiples. The first harmonic for both types is the fundamental frequency. Here, 1372 Hz and 1764 Hz can be written as 1372 = 3 * 392 and 1764 = 4 * 392, respectively. This indicates that the odd and even multiples are consecutive, meaning these harmonics belong to an open pipe.
03

Identify the Harmonics

Since it is an open pipe and the fundamental frequency is 392 Hz, the frequencies of the harmonics can be expressed as n * 392 Hz where n is an integer. Therefore, 1372 Hz = 3 * 392 Hz (3rd harmonic), and 1764 Hz = 4 * 392 Hz (4th harmonic). Thus, the two harmonics are the 3rd and 4th harmonics of an open pipe.
04

Calculate the Length of the Pipe

For an open pipe, the wavelength of the fundamental frequency (first harmonic) is given by \(\lambda = \frac{v}{f}\), where \(v\) is the speed of sound (approximately 343 m/s at room temperature), and \(f\) is the fundamental frequency (392 Hz). Thus, \(\lambda = \frac{343}{392} \approx 0.8755\,m\). The length of the pipe is half the wavelength of the fundamental frequency, or \(L = \frac{\lambda}{2} \approx \frac{0.8755}{2} \approx 0.438\,m\).

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Key Concepts

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

Open vs Stopped Pipe
Understanding open and stopped pipes is crucial for figuring out the nature of harmonics in musical instruments and sound systems.

Open pipes, like a flute, have both ends open. This allows air to vibrate freely within the tube. The harmonics produced in open pipes occur at all integer multiples of the fundamental frequency. So if the fundamental frequency is, say, 100 Hz, the harmonics would be at 200 Hz, 300 Hz, 400 Hz, and so on.
  • Open pipes: Harmonics at all integer multiples.
  • Example: Fundamental frequency of 100 Hz, harmonics at 200 Hz, 300 Hz, etc.
Stopped pipes, on the other hand, have one closed end, like a clarinet. This affects the vibration patterns such that only odd multiples of the fundamental frequency are present. With a fundamental frequency of 100 Hz, the harmonics would be at 300 Hz, 500 Hz, etc.
  • Stopped pipes: Harmonics at odd integer multiples.
  • Example: Fundamental frequency of 100 Hz, harmonics at 300 Hz, 500 Hz, etc.
In our exercise, the presence of consecutive harmonics at 1372 Hz and 1764 Hz suggests an open pipe. This is because they are integer multiples of the fundamental frequency and do not follow the pattern of stopped pipes.
Fundamental Frequency
The fundamental frequency is the lowest frequency produced by any vibrating object, and it's the first harmonic. In our exercise, finding the fundamental frequency was essential to solving the problem.

The two given harmonics were 1372 Hz and 1764 Hz.
  • Consecutive harmonics have a difference equal to the fundamental frequency.
By subtracting these two, 1764 Hz - 1372 Hz, we found the fundamental frequency to be 392 Hz.

This frequency sets the base value for all harmonics in the series. Knowing that the fundamental is the first harmonic in open pipes, the harmonics can be represented as multiples of this frequency.
  • Fundamental frequency: 392 Hz.
  • Controls all harmonic frequencies of the pipe.
This illustrates why identifying the fundamental frequency helps in predicting the specific harmonics formed within the pipe.
Pipe Length Calculation
Calculating the length of the pipe ties together our understanding of frequency, waves, and harmonics. This section explains the process using the fundamental frequency, which we calculated as 392 Hz.
For an open pipe, the fundamental frequency corresponds to the first harmonic.
  • The formula involves sound speed: ewline \( \lambda = \frac{v}{f} \) where \( \lambda \) is the wavelength, \( v = 343 \, \text{m/s} \, \text{at room temperature}, \) and \( f \) is the frequency.
Substituting values, we get:
  • \( \lambda = \frac{343}{392} \approx 0.8755 \, \text{m} \)
In open pipes, the length of the pipe is half the wavelength of the fundamental frequency.
This gives us:
  • Pipe Length \( L = \frac{\lambda}{2} = \frac{0.8755}{2} \approx 0.438 \, \text{m} \)
Understanding this calculation helps illustrate the relationship between the physical dimensions of a pipe and the sound waves it produces.

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Most popular questions from this chapter

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

A soprano und a bass are singing a duet. While the soprano sings an \(A\) -sharp at 932 \(\mathrm{Hz}\) , the bass sings an \(A\) -sharp but three octaves lower. In this concert hall, the density of air is 1.20 \(\mathrm{kg} / \mathrm{m}^{3}\) and its bulk modulus is \(1.42 \times 10^{5} \mathrm{Pa.}\) 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 \(\mathrm{dB} ?\)

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Ep A New Musical Instrument. You have designed a new musical instrument of very simple construction. Your design consists of a metal tube with length \(L\) and diameter \(L / 10 .\) You have stretched a string of mass per unit length \(\mu\) across the open end of the tube. The other end of the tube is closed. To produce the musical effect you're looking for, you want the frequency of the third-harmonic standing wave on the string to be the same as the fundamental frequency for sound waves in the air column in the tube. The speed of sound waves in this air column is \(v_{\mathrm{s}}\) . (a) What must be the tension of the string to produce the desired effect? (b) What happens to the sound produced by the instrument if the tension is changed to twice the value calculated in part (a)? (c) For the tension calculated in part (a), what other harmonics of the string, if any, are in resonance with standing waves in the air column?

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