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Chapter 14: Problem 77

An organ pipe open at both ends has a harmonic with a frequency of \(440 \mathrm{Hz}\). The next higher harmonic in the pipe has a frequency of \(495 \mathrm{Hz}\). Find \((\mathrm{a})\) the frequency of the fundamental and (b) the length of the pipe.

Short Answer

Expert verified
The fundamental frequency is 55 Hz. The pipe length is approximately 3.12 m.

Step by step solution

01

Understand the Problem

We have an organ pipe open at both ends. It has a harmonic at 440 Hz and the next higher harmonic at 495 Hz. We need to find the frequency of the fundamental harmonic and the length of the pipe.
02

Write the Frequency Relationship

For an organ pipe open at both ends, the harmonics are multiples of the fundamental frequency. If the fundamental frequency is \(f_1\), then harmonics are \(f_2 = 2f_1, f_3 = 3f_1\), and so on. We have \(f_n = 440\, \mathrm{Hz}\) and \(f_{n+1} = 495\, \mathrm{Hz}\).
03

Determine the Harmonic Numbers

From the frequencies, we know two consecutive harmonics. The difference in frequency between consecutive harmonics for a pipe open at both ends is equal to the fundamental frequency, \(f_1\). So, \(f_1 = f_{n+1} - f_n = 495\, \mathrm{Hz} - 440\, \mathrm{Hz} = 55\, \mathrm{Hz}\).
04

Solve for the Fundamental Frequency

The fundamental frequency \(f_1\) is the difference between the two given frequencies. Thus, \(f_1 = 55\, \mathrm{Hz}\).
05

Use the Frequency to find the Pipe's Length

The speed of sound \(v\) in air is approximately \(343\, \mathrm{m/s}\). For the fundamental frequency of a pipe open at both ends, \( f_1 = \frac{v}{2L} \). Using \( f_1 = 55\, \mathrm{Hz} \), we rearrange to find \[ L = \frac{v}{2f_1} = \frac{343}{2 \times 55} \approx 3.12\, \mathrm{m} \].
06

Provide the Solution

The fundamental frequency of the organ pipe is \(55\, \mathrm{Hz}\). The length of the organ pipe is approximately \(3.12\, \mathrm{m}\).

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Key Concepts

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

Harmonics
Harmonics are the natural frequencies at which an organ pipe, or any musical instrument, vibrates. They produce beautiful, organized sounds. These are specific frequencies that are integer multiples of the lowest frequency, which is known as the fundamental frequency.
For example, if the fundamental frequency is denoted by \( f_1 \), then the harmonics will be \( 2f_1, 3f_1, 4f_1 \), and so on. In the case of the organ pipe that is open at both ends, each frequency is a higher harmonic of the fundamental frequency.
  • The harmonics are significant because they define the tonal quality or timbre of the sound.
  • Consecutive harmonics help in identifying the fundamental frequency since they are evenly spaced by the fundamental frequency amount.
The fundamental frequency determines the increments by which harmonics appear. In the given exercise, if an organ pipe produces a harmonic at \(440 \mathrm{Hz}\) and the next one at \(495 \mathrm{Hz}\), then their difference provides the fundamental frequency of the pipe, which is \(55 \mathrm{Hz}\).
Fundamental Frequency
Fundamental frequency is the lowest frequency at which a system like a musical instrument naturally vibrates. It is a crucial characteristic in acoustics, as it serves as the base, or "first harmonic," upon which all other harmonics build.
For pipes open at both ends, the fundamental frequency can be calculated using the formula: \[ f_1 = \frac{v}{2L} \]where \(v\) is the speed of sound in air (approximately \(343 \, \mathrm{m/s}\)) and \(L\) is the length of the pipe.
In our exercise:
  • Given two consecutive harmonics, the difference between them is the fundamental frequency.
  • The calculated fundamental from the problem is \(55 \, \mathrm{Hz}\).
Knowing the fundamental frequency helps in understanding how the organ pipe behaves and resonates when air passes through it.
Length of Pipe
The length of an organ pipe is a key factor that influences its acoustical performance. The length determines the wavelengths that the pipe can support, thereby affecting the pipe's fundamental frequency and its harmonics.
When a pipe is open at both ends, it effectively forms an air column that resonates at frequencies determined by its own natural lengths.
To calculate the length \(L\) of an organ pipe given its fundamental frequency \(f_1\), you can rearrange the formula for the fundamental frequency:\[ L = \frac{v}{2f_1} \]Substituting \(f_1 = 55 \, \mathrm{Hz}\) and \(v = 343 \, \mathrm{m/s}\), you find that the length is approximately \(3.12 \, \mathrm{m}\).
This calculation shows how the physical dimensions of an organ pipe correlate to the frequencies it can produce.
  • Shorter pipes yield higher pitch sounds since they have a higher fundamental frequency.
  • Longer pipes produce deeper tones because of their longer wavelengths.
Understanding this relationship is essential for designing and tuning musical instruments.

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Most popular questions from this chapter

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