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Chapter 17: Problem 27

Suppose the strings on a violin are stretched with the same tension and each has the same length between its two fixed ends. The musical notes and corresponding fundamental frequencies of two of these strings are \(\mathrm{G}(196.0 \mathrm{~Hz})\) and \(\mathrm{E}(659.3 \mathrm{~Hz}) .\) The linear density of the E string is \(3.47 \times 10^{-4} \mathrm{~kg} / \mathrm{m}\). What is the linear density of the \(\mathrm{G}\) string?

Short Answer

Expert verified
The linear density of the G string is approximately \( 3.94 \times 10^{-3} \text{ kg/m} \).

Step by step solution

01

Understand the Relationship Between Frequency and Linear Density

When strings are under the same tension and have the same length, the frequency of a string is inversely proportional to the square root of the linear density. This can be expressed as \( f \propto \frac{1}{\sqrt{\mu}} \), where \( f \) is the frequency and \( \mu \) is the linear density of the string.
02

Determine the Frequency and Linear Density Ratio

Given that the frequency of the \( E \) string is \( 659.3 \text{ Hz} \) and the frequency of the \( G \) string is \( 196.0 \text{ Hz} \), you can use the formula \( \frac{f_G}{f_E} = \sqrt{\frac{\mu_E}{\mu_G}} \). The linear density of the \( E \) string, \( \mu_E \), is given as \( 3.47 \times 10^{-4} \text{ kg/m} \).
03

Calculate the Ratio of Frequencies

Calculate the ratio by dividing the frequencies: \( \frac{196.0 \text{ Hz}}{659.3 \text{ Hz}} \approx 0.297 \). This ratio represents \( \sqrt{\frac{\mu_E}{\mu_G}} \).
04

Find the Ratio of Linear Densities

Square the frequency ratio to find the ratio of linear densities: \( \left(0.297\right)^2 \approx 0.088 \). This value is the ratio \( \frac{\mu_E}{\mu_G} \).
05

Solve for the Linear Density of the G String

Rearrange the previous ratio to solve for \( \mu_G \): \[ \mu_G = \frac{\mu_E}{0.088} \]. Substitute \( \mu_E = 3.47 \times 10^{-4} \text{ kg/m} \) to obtain \( \mu_G \approx 3.94 \times 10^{-3} \text{ kg/m} \).

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Key Concepts

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

String Tension
String tension is an important factor in determining the sound produced by a string instrument. Tension refers to the force applied to stretch the string between its two fixed endpoints. When all strings are under the same tension, they experience the same stretching force.
  • Higher tension increases the frequency (pitch) of the note produced.
  • Lower tension results in a deeper, lower note.

In this problem, the violin strings are under the same tension, simplifying the comparison of their frequencies and densities. Understanding tension helps explain why strings of different thicknesses and materials can produce the same note if adjusted accordingly.
Linear Density
Linear density, denoted as \( \mu \), is the mass per unit length of a string. It significantly affects how a string vibrates and thus the sound it produces.
  • Expressed in kilograms per meter (\( \text{kg/m} \)).
  • A string with higher linear density has more mass per meter.

The thicker or denser a string, the lower the natural frequency it can produce, given the same tension. In our exercise, the linear density of the \( \mathrm{E} \) string is known, and we're tasked with finding it for the \( \mathrm{G} \) string by using the relationship between linear density and frequency.
Frequency Ratio
Frequency ratio is the comparison of two frequencies, providing insight into how their associated string properties differ. It is calculated by dividing one frequency by another.
  • In this exercise, the frequencies of the \( \mathrm{E} \) and \( \mathrm{G} \) strings are given.
  • The ratio of \( \frac{196.0 \text{ Hz}}{659.3 \text{ Hz}} \approx 0.297 \) is derived.

This ratio is crucial because it relates directly to the square root of the ratio of their linear densities, as given by the formula \( \frac{f_G}{f_E} = \sqrt{\frac{\mu_E}{\mu_G}} \). Understanding this ratio allows us to connect the theoretical concepts to practical calculations.
Fundamental Frequency
The fundamental frequency of a string is the lowest frequency at which the string vibrates when plucked. It is key to the characteristic sound of the string.
  • Depends on tension, length, and linear density of the string.
  • Higher frequencies mean higher sounds.

For the violin strings in the exercise, the given fundamental frequencies help us apply the physical concepts to solve the problem. By knowing that \( f \propto \frac{1}{\sqrt{\mu}} \), we can relate frequencies to string densities, allowing us to find the unknown linear density of the \( \mathrm{G} \) string when its fundamental frequency is given alongside another known string.

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Most popular questions from this chapter

A sound wave is traveling in seawater, where the adiabatic bulk modulus and density are \(2.31 \mathrm{X} 10^{9} \mathrm{~Pa}\) and \(1025 \mathrm{~kg} / \mathrm{m}^{3}\), respectively. The wavelength of the sound is \(3.35 \mathrm{~m}\). A tuning fork is struck underwater and vibrates at \(440.0 \mathrm{~Hz} .\) What would be the beat frequency heard by an underwater swimmer?

The E string on an electric bass guitar has a length of \(0.628 \mathrm{~m}\) and, when producing the note E, vibrates at a fundamental frequency of \(41.2 \mathrm{~Hz}\). Players sometimes add to their instruments a device called a "D-tuner." This device allows the E string to be used to produce the note \(\mathrm{D}\), which has a fundamental frequency of \(36.7 \mathrm{~Hz}\). The D-tuner works by extending the length of the string, keeping all other factors the same. By how much does a D-tuner extend the length of the E string?

The fundamental frequency of a vibrating system is \(400 \mathrm{~Hz}\). For each of the following systems, give the three lowest frequencies (excluding the fundamental) at which standing waves can occur: (a) a string fixed at both ends, (b) a cylindrical pipe with both ends open, and (c) a cylindrical pipe with only one end open.

The approach to solving this problem is similar to that taken in Multiple- Concept Example 4 . On a cello, the string with the largest linear density \(\left(1.56 \times 10^{-2} \mathrm{~kg} / \mathrm{m}\right)\) is the C string. This string produces a fundamental frequency of \(65.4 \mathrm{~Hz}\) and has a length of \(0.800 \mathrm{~m}\) between the two fixed ends. Find the tension in the string.

The \(\mathrm{E}\) string on an electric bass guitar has a length of \(0.628 \mathrm{~m}\) and, when producing the note E, vibrates at a fundamental frequency of \(41.2 \mathrm{~Hz}\). Players sometimes add to their instruments a device called a "D-tuner." This device allows the E string to be used to produce the note D, which has a fundamental frequency of \(36.7 \mathrm{~Hz}\). The D-tuner works by extending the length of the string, keeping all other factors the same. By how much does a D-tuner extend the length of the \(\mathrm{E}\) string?
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