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Chapter 14: Problem 17

Two steel guitar strings have the same length. String A has a diameter of \(0.50 \mathrm{mm}\) and is under \(410.0 \mathrm{N}\) of tension. String \(\mathrm{B}\) has a diameter of \(1.0 \mathrm{mm}\) and is under a tension of \(820.0 \mathrm{N}\). Find the ratio of the wave speeds, \(v_{\mathrm{A}} / v_{\mathrm{H}}\) in these two strings.

Short Answer

Expert verified
The ratio of wave speeds is approximately 0.354.

Step by step solution

01

Formula for Wave Speed in a String

The wave speed \( v \) in a stretched string is given by \( v = \sqrt{\frac{T}{\mu}} \), where \( T \) is the tension in the string and \( \mu \) is the linear mass density of the string.
02

Calculating Linear Mass Density

The linear mass density \( \mu \) for a string with mass \( m \) and length \( L \) is \( \mu = \frac{m}{L} \). Since both strings have the same length and are made of the same material, let's use the formula \( \mu = \rho A \), where \( \rho \) is the density of the material and \( A \) is the cross-sectional area of the string. Given the diameter of the string, \( A = \pi (\frac{d}{2})^2 \).
03

Compute Cross-Sectional Area for Strings A and B

For String A: \( d_A = 0.50 \; \text{mm} = 0.0005 \; \text{m} \), \( A_A = \pi \left( \frac{0.0005 \; \text{m}}{2} \right)^2 = 1.9635 \times 10^{-7} \; \text{m}^2 \).For String B: \( d_B = 1.0 \; \text{mm} = 0.001 \; \text{m} \), \( A_B = \pi \left( \frac{0.001 \; \text{m}}{2} \right)^2 = 7.854 \times 10^{-7} \; \text{m}^2 \).
04

Calculate Linear Density for Strings A and B

For String A, \( \mu_A = \rho \times A_A = \rho \times 1.9635 \times 10^{-7} \). For String B, \( \mu_B = \rho \times A_B = \rho \times 7.854 \times 10^{-7} \). Since \( \rho \) is the same for both strings, the density ratio simplifies as \( \mu_A / \mu_B = 1.9635 / 7.854 \).
05

Calculate Wave Speed for Strings A and B

Using the wave speed formula, for String A: \( v_A = \sqrt{\frac{410 \; \text{N}}{\mu_A}} \), and for String B: \( v_B = \sqrt{\frac{820 \; \text{N}}{\mu_B}} \).
06

Finding the Ratio of the Wave Speeds

We need the ratio \( \frac{v_A}{v_B} = \sqrt{\frac{T_A}{\mu_A} \times \frac{\mu_B}{T_B}} \). Substituting values, we get:\[ \frac{v_A}{v_B} = \sqrt{\frac{410}{820} \times \frac{1.9635}{7.854}} = \sqrt{\frac{1}{2} \times \frac{1}{4}} = \sqrt{\frac{1}{8}} = \frac{1}{\sqrt{8}} \approx 0.3536. \]

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Key Concepts

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

Tension in a String
Tension is a fundamental concept when it comes to understanding wave speed in a string. The tension in a string is simply the force applied along its length. It influences how fast waves, such as sound waves in a guitar string, travel through the string.
  • When you apply more tension to a string, it becomes tighter.
  • This increased tension allows the wave to travel faster because the string is less flexible.
As seen in the problem, String A is under 410 N of tension while String B has 820 N. This difference in tension directly affects the speed with which waves can travel through each string. Thus, higher tension leads to higher wave speeds.
Linear Mass Density
Linear mass density, denoted as \( \mu \), is a measure of the mass of the string per unit length. It is a significant factor in determining the wave speed in a string because it relates to how heavy the string is.
  • The linear mass density \( \mu \) can be calculated using the formula \( \mu = \rho A \), where \( \rho \) is the material density, and \( A \) is the cross-sectional area of the string.
  • For two strings made of the same material, their relative mass densities depend only on their cross-sectional areas.
In our example, due to the different diameters of the strings, their cross-sectional areas and thus their linear mass densities also differ. String B has a larger diameter and therefore a greater linear mass density resulting in a reduced wave speed despite its higher tension.
Cross-Sectional Area
The cross-sectional area is a critical measurement for understanding how it affects both tension and linear mass density in strings. It requires calculating the area of a string's cross-section, which is usually circular for string instruments.
  • The formula for cross-sectional area \( A \) is \( \pi (\frac{d}{2})^2 \), where \( d \) is the diameter.
  • A larger cross-sectional area means a greater linear mass density, assuming the material's density is constant.
For String A, the diameter is 0.50 mm, yielding a smaller cross-sectional area than String B, which has a diameter of 1.0 mm. This larger area in String B increases its linear mass density and affects how tension translates to wave speed. The overall mechanics suggest that both tension and cross-sectional area must be considered together to fully understand their impact on wave speed.

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Most popular questions from this chapter

Waves on a particular string travel with a speed of \(16 \mathrm{m} / \mathrm{s}\). By what factor should the tension in this string be changed to produce waves with a speed of \(32 \mathrm{m} / \mathrm{s} ?\)

Hearing a Pin Drop The ability to hear a "pin drop" is the sign of sensitive hearing. Suppose a \(0.55-g\) pin is dropped from a height of \(28 \mathrm{cm},\) and that the pin emits sound for \(1.5 \mathrm{s}\) when it lands. Assuming all of the mechanical energy of the pin is converted to sound energy, and that the sound radiates uniformly in all directions, find the maximum distance from which a person can hear the pin drop. (This is the ideal maximum distance, but atmospheric absorption and other factors will make the actual maximum distance considerably smaller.)

A particular jet engine produces a tone of \(495 \mathrm{Hz}\). Suppose that one jet is at rest on the tarmac while a second identical jet flies overhead at \(82.5 \%\) of the speed of sound. The pilot of each jet listens to the sound produced by the engine of the other jet. (a) Which pilot hears a greater Doppler shift? Explain. (b) Calculate the frequency heard by the pilot in the moving jet. (c) Calculate the frequency heard by the pilot in the stationary jet.

An organ pipe that is open at both ends is \(3.5 \mathrm{m}\) long. What is its fundamental frequency?

Toning a Piano To tune middle C on a piano, a tuner hits the key and at the same time sounds a \(261-\mathrm{Hz}\) tuning fork. If the tuner hears 3 beats per second, what are the possible frequencies of the piano key?
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