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Review Conceptual Example 5 before attempting this problem. As the drawing shows, the length of a guitar string is \(0.628\) \(\mathrm{m} .\) The frets are numbered for convenience. A performer can play a musical scale on a single string because the spacing between the frets is designed according to the following rule: When the string is pushed against any fret \(j,\) the fundamental frequency of the shortened string is larger by a factor of the twelfth root of two \((\sqrt[12]{2})\) than it is when the string is pushed against the fret \(j-1 .\) Assuming that the tension in the string is the same for any note, find the spacing (a) between fret 1 and fret 0 and (b) between fret 7 and fret \(6 .\)

Short Answer

Expert verified
Spacing between fret 0 and 1 is approx. 0.036 m; between fret 6 and 7 is approx. 0.0396 m.

Step by step solution

01

Understanding the Problem

We are given the length of a guitar string and a rule concerning how the spacing of frets relates to the frequency of notes. When the string is pressed against a fret, the note is produced by shortening the string, which increases the frequency by a factor of \( \sqrt[12]{2} \) compared to the previous fret.
02

Formula for Frequency Ratio

The frequency ratio between successive frets on the guitar, as provided, is \[\text{frequency ratio between frets } j \text{ and } j-1 = \sqrt[12]{2}.\]
03

Length Ratio Between Frets

Since the frequency is inversely proportional to the string length, the relationship between frequencies and lengths gives \[\frac{L_{j-1}}{L_{j}} = \sqrt[12]{2}.\] Rearranging this, we find the length ratio: \[\frac{L_j}{L_{j-1}} = \frac{1}{\sqrt[12]{2}}.\]
04

Spacing Between Fret 0 and Fret 1

Let the length of the open string without pressing any frets be \(L_0 = 0.628\) m. The length when pressing fret 1 is \[L_1 = L_0 \times \frac{1}{\sqrt[12]{2}}.\] Substitute the given value of \(L_0\) into the formula:\[L_1 = 0.628 \times \frac{1}{\sqrt[12]{2}}.\] Thus, \[L_1 \approx 0.592 \text{ m}.\] The spacing between fret 0 and fret 1 is:\[0.628 - 0.592 = 0.036 \text{ m}.\]
05

Spacing Between Fret 6 and Fret 7

Consider the same ratio principle between frets 7 and 6:\[L_7 = L_6 \times \frac{1}{\sqrt[12]{2}}.\] Using the length ratio for successive frets, the spacing between fret 6 and fret 7 follows the same calculation:\[L_{6} = L_0 \times \left( \frac{1}{\sqrt[12]{2}} \right)^6,\]\[L_{7} = L_0 \times \left( \frac{1}{\sqrt[12]{2}} \right)^7.\]Therefore, the spacing is \[L_6 - L_7 = L_0 \times \left( \frac{1}{\sqrt[12]{2}} \right)^6 - L_0 \times \left( \frac{1}{\sqrt[12]{2}} \right)^7.\]Plugging in the values:\[\approx 0.0396 \text{ m}.\]
06

Conclusion

The spacing between fret 0 and fret 1 is approximately \(0.036\) m. The spacing between fret 6 and fret 7 is approximately \(0.0396\) m.

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Key Concepts

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

Harmonic Frequencies
Harmonic frequencies are essential to understanding how musical instruments like the guitar produce sound. When a guitar string vibrates, it doesn't just vibrate at a single frequency. Instead, it produces a series of frequencies that are integer multiples of a fundamental frequency. These are called harmonics. The fundamental frequency is the lowest vibrational mode, and each harmonic frequency is a higher mode, contributing to the overall sound quality.
- The first harmonic is the fundamental frequency, producing the pitch we primarily hear. - Higher harmonics (second, third, etc.) add richness and complexity to the sound.
Understanding these frequencies helps musicians tune their instruments and select strings for the desired sound. On a guitar, pressing a string against different frets changes the length of the vibrating section of the string, altering these harmonic frequencies.
Musical Acoustics
Musical acoustics deals with how sound waves are produced, transmitted, and perceived as music. This field blends physics and music, explaining how instruments like the guitar create sound. The physics of sound involves vibrations, which generate sound waves through mediums like air. These sound waves then reach our ears and are interpreted as music.
- Pitch: The perceived frequency of a sound, influenced by the length and tension of a string, such as a guitar string. - Timbre: The quality or color of a sound, determined by the combination of harmonic frequencies. - Volume: The loudness of sound, related to the amplitude of the vibrations.
In the context of a guitar, when a string is plucked, it vibrates at specific frequencies, primarily determined by its length, tension, and mass. Pressing on frets shortens the effective string length, modifying the pitch through a predictable pattern, like the twelfth root of two ratio.
Guitar String Physics
The physics of guitar strings involves understanding how they produce sound through vibrations. A guitar string is fixed at both ends, allowing it to vibrate and create sound waves when plucked. This brings us to several key points:
- **String Length and Pitch**: By altering the length of the vibrating part of the string (i.e., pressing down on a fret), the pitch changes. The shorter the string, the higher the pitch. - **Tension and Mass**: Tension refers to how tight the string is, and mass refers to how heavy it is. Both factors influence how fast the string vibrates and, therefore, the pitch: more tension and less mass lead to higher pitches.
Using these principles, the design of guitar frets is carefully spaced to ensure each successive fret raises the pitch by a precise musical interval, such as the twelfth root of two, creating a harmonious scale across the string.

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Most popular questions from this chapter

A tube is open only at one end. A certain harmonic produced by the tube has a frequency of 450 Hz. The next higher harmonic has a frequency of \(750\) \(\mathrm{Hz}\). The speed of sound in air is \(343\) \(\mathrm{m} / \mathrm{s}\). (a) What is the integer \(n\) that describes the harmonic whose frequency is \(450\) \(\mathrm{Hz} ?\) (b) What is the length of the tube?

A vertical tube is closed at one end and open to air at the other end. The air pressure is \(1.01 \times 10^{5}\) Pa. The tube has a length of \(0.75\) \(\mathrm{m} .\) Mercury (mass density \(=13600\) \(\mathrm{kg} / \mathrm{m}^{3}\) ) is poured into it to shorten the effective length for standing waves. What is the absolute pressure at the bottom of the mercury column, when the fundamental frequency of the shortened, air-filled tube is equal to the third harmonic of the original tube?

A string has a linear density of \(8.5 \times 10^{-3}\) \(\mathrm{kg} / \mathrm{m}\) and is under a tension of \(280\) \(\mathrm{N}\). The string is \(1.8\) \(\mathrm{m}\) long, is fixed at both ends, and is vibrating in the standing wave pattern shown in the drawing. Determine the (a) speed, (b) wavelength, and (c) frequency of the traveling waves that make up the standing wave.

To review the concepts that play roles in this problem, consult Multiple- Concept Example 4. Sometimes, when the wind blows across a long wire, a low- frequency "moaning" sound is produced. This sound arises because a standing wave is set up on the wire, like a standing wave on a guitar string. Assume that a wire (linear density \(=0.0140\) \(\mathrm{kg} / \mathrm{m}\) ) sustains a tension of 323 N because the wire is stretched between two poles that are \(7.60\) \(\mathrm{m}\) apart. The lowest frequency that an average, healthy human ear can detect is 20.0 Hz. What is the lowest harmonic number \(n\) that could be responsible for the "moaning" sound?

Standing waves are set up on two strings fixed at each end, as shown in the drawing. The two strings have the same tension and mass per unit length, but they differ in length by \(0.57\) \(\mathrm{cm} .\) The waves on the shorter string propagate with a speed of \(41.8\) \(\mathrm{m} / \mathrm{s},\) and the fundamental frequency of the shorter string is \(225\) \(\mathrm{Hz}\). Determine the beat frequency produced by the two standing waves.

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