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Chapter 16: Problem 18

The heaviest and lightest strings on a certain violin have linear densities of \(3.0\) and \(0.29 \mathrm{~g} / \mathrm{m}\). What is the ratio of the diameter of the heaviest string to that of the lightest string, assuming that the strings are of the same material?

Short Answer

Expert verified
The ratio of the diameters is approximately 3.217.

Step by step solution

01

Understanding Materials' Density

Since the strings are made from the same material, they have the same mass density (\(\rho\)). This density is defined as \(\rho = \frac{m}{V}\) where \(m\) is mass and \(V\) is volume. For strings, linear density (\( \mu \)) is used, which is mass per unit length: \(\mu = \frac{m}{L}\). Since \(\mu\) directly depends on the string's diameter \(d\), we relate it to \(\rho\) and \(d\).
02

Express Linear Density in Terms of Diameter

For a string with circular cross-section, the volume \(V\) for length \(L\) is \(V = A \cdot L \), where \( A = \pi \left( \frac{d}{2} \right)^2 \). The linear density is \(\mu = \frac{m}{L} = \frac{ \rho \cdot V}{L} = \rho \cdot \pi \left( \frac{d}{2} \right)^2 = \rho \cdot \frac{\pi d^2}{4}\).
03

Set Up the Ratio of Diameters

Given two strings with linear densities \(\mu_1 = 3.0 \, \text{g/m}\) and \(\mu_2 = 0.29 \, \text{g/m}\), their diameters \(d_1\) and \(d_2\) can be related. Using the formula \(\mu = \rho \cdot \frac{\pi d^2}{4}\), we can write \(\mu_1 = \rho \cdot \frac{\pi d_1^2}{4}\) and \(\mu_2 = \rho \cdot \frac{\pi d_2^2}{4}\).
04

Calculate Diameter Ratio

To find the ratio of the diameters, divide the two linear density equations: \(\frac{\mu_1}{\mu_2} = \frac{d_1^2}{d_2^2}\). Solving for the diameter ratio, we have \(\frac{d_1}{d_2} = \sqrt{\frac{\mu_1}{\mu_2}} = \sqrt{\frac{3.0}{0.29}}\).
05

Execute Calculation

Perform the calculation: \(\frac{d_1}{d_2} = \sqrt{\frac{3.0}{0.29}} \approx \sqrt{10.3448} \approx 3.217\). This is the ratio of the diameters of the heaviest string to the lightest string.

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Key Concepts

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

Linear Density
Linear density, represented as \( \mu \), is a measure of how much mass a string has per unit length. It is essentially the amount of mass that spans over one meter of the string. The higher the linear density, the heavier the string. For strings like those on a violin, linear density plays a crucial role in the sound they produce. It can be calculated using the formula:\[ \mu = \frac{m}{L} \]where \( m \) is the mass and \( L \) is the length of the string. This directly ties into the string's diameter because a thicker string holds more mass.
  • Linear Density increases with more mass in a given length.
  • It affects the flexibility and tension requirement of the string.
  • Different linear densities lead to diverse sound characteristics.
Material Density
Material density (\( \rho \)) is a fundamental property of the material, defining how much mass exists within a given volume. Since the strings in the exercise are made from the same material, they share the same density. This uniformity simplifies calculations when comparing strings of identical materials.
Material density is expressed as:\[ \rho = \frac{m}{V} \]Where \( V \) is the volume. In practical terms, it tells us how compact the material structure is. Lower density might imply a material is less massive per unit volume, potentially being more flexible.
  • Material Density is constant for materials of the same kind.
  • It helps determine how mass is distributed over the string's volume.
  • Vital in calculating derived properties like linear density.
Diameter Ratio
The diameter ratio of two objects offers insight into their relative thicknesses. In this exercise, understanding the diameter ratio of the violin strings provides information on their mechanical and acoustic properties. The diameter influences both the string's linear density and its vibrational characteristics.
To find this ratio, use the relationship:\[ \frac{d_1}{d_2} = \sqrt{\frac{\mu_1}{\mu_2}} \]Here, \( \mu_1 \) and \( \mu_2 \) represent the linear densities of the strings. The larger the ratio, the thicker the heavier string compared to the lighter one, affecting its ability to produce lower pitches.
  • Larger diameter implies a string can be less flexible and carry more tension.
  • Influences how sound waves propagate along the string.
  • Informs about the design and tuning of musical instruments.
String Mass
The string mass is the total mass of a violin string, an important factor dictated by both linear density and overall length. To find the mass, the formula for linear density comes into play:\[ m = \mu \times L \]Where \( m \) is mass, \( \mu \) is the linear density, and \( L \) is the length. Hence, even a string with small diameter but high linear density can result in a significant mass, influencing its tension and vibrational frequency.
Understanding string mass helps in several areas:
  • Determines the amount of tension needed to produce a particular tone.
  • Indicates how durable and strong the string needs to be under tension.
  • Essential for adjusting brevity and sonority of string notes.

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Most popular questions from this chapter

A transverse sinusoidal wave is generated at one end of a long, horizontal string by a bar that moves up and down through a distance of \(1.00 \mathrm{~cm}\). The motion is continuous and is repeated regularly 120 times per second. The string has linear density 120 \(\mathrm{g} / \mathrm{m}\) and is kept under a tension of \(90.0 \mathrm{~N}\). Find the maximum value of (a) the transverse speed \(u\) and (b) the transverse component of the tension \(\tau\). (c) Show that the two maximum values calculated above occur at the same phase values for the wave. What is the transverse displacement \(y\) of the string at these phases? (d) What is the maximum rate of energy transfer along the string? (e) What is the transverse displacement \(y\) when this maximum transfer occurs? (f) What is the minimum rate of energy transfer along the string? (g) What is the transverse displacement \(y\) when this minimum transfer occurs?

A string that is stretched between fixed supports separated by \(75.0 \mathrm{~cm}\) has resonant frequencies of 420 and \(315 \mathrm{~Hz}\), with no intermediate resonant frequencies. What are (a) the lowest resonant frequency and (b) the wave speed?

If a wave \(y(x, t)=(6.0 \mathrm{~mm}) \sin (k x+(600 \mathrm{rad} / \mathrm{s}) t+\phi)\) travels along a string, how much time does any given point on the string take to move between displacements \(y=+2.0 \mathrm{~mm}\) and \(y=-2.0 \mathrm{~mm}\) ?

What are (a) the lowest frequency, (b) the second lowest frequency, and (c) the third lowest frequency for standing waves on a wire that is \(10.0 \mathrm{~m}\) long, has a mass of \(100 \mathrm{~g}\), and is stretched under a tension of \(250 \mathrm{~N}\) ?

The speed of electromagnetic waves (which include visible light, radio, and \(x\) rays) in vacuum is \(3.0 \times 10^{8} \mathrm{~m} / \mathrm{s}\). (a) Wavelengths of visible light waves range from about \(400 \mathrm{~nm}\) in the violet to about \(700 \mathrm{~nm}\) in the red. What is the range of frequencies of these waves? (b) The range of frequencies for shortwave radio (for example, FM radio and VHF television) is \(1.5\) to \(300 \mathrm{MHz}\). What is the corresponding wavelength range? (c) X-ray wavelengths range from about \(5.0 \mathrm{~nm}\) to about \(1.0 \times 10^{-2} \mathrm{~nm} .\) What is the frequency range for \(\mathrm{x}\) rays?
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