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Chapter 22: Problem 29

Steel and silver wires of the same diameter and same length are stretched with equal tension. Their densities are \(7.80 \mathrm{~g} / \mathrm{cm}^{3}\) and \(10.6 \mathrm{~g} / \mathrm{cm}^{3}\), respectively. What is the fundamental frequency of the silver wire if that of the steel is \(200 \mathrm{~Hz}\) ?

Short Answer

Expert verified
The fundamental frequency of the silver wire is approximately 172 Hz.

Step by step solution

01

Understand the Relationship Between Frequency and Density

The fundamental frequency \( f \) of a wire under tension is given by the formula \( f = \frac{1}{2L} \sqrt{\frac{T}{\mu}} \), where \( T \) is the tension, \( L \) is the length, and \( \mu \) is the linear mass density (mass per unit length). The linear mass density can be calculated from the material's density and the wire's cross-sectional area.
02

Determine Linear Mass Density

The linear mass density \( \mu \) is calculated as \( \mu = \rho A \), where \( \rho \) is the density and \( A \) is the cross-sectional area of the wire. For two wires of the same diameter, \( A \) is the same for both. Thus, \( \mu_{\text{steel}} = \rho_{\text{steel}} A \) and \( \mu_{\text{silver}} = \rho_{\text{silver}} A \).
03

Relate Frequencies with Mass Densities

Given that the frequencies are related by \( \frac{f_{\text{steel}}}{f_{\text{silver}}} = \sqrt{\frac{\mu_{\text{silver}}}{\mu_{\text{steel}}}} \), we use the fact that \( \mu_{\text{silver}} = \rho_{\text{silver}} A \) and \( \mu_{\text{steel}} = \rho_{\text{steel}} A \). Therefore, \( \frac{f_{\text{steel}}}{f_{\text{silver}}} = \sqrt{\frac{\rho_{\text{silver}}}{\rho_{\text{steel}}}} \).
04

Solve for Frequency of Silver Wire

Using \( f_{\text{steel}} = 200 \) Hz, \( \rho_{\text{steel}} = 7.80 \) g/cm鲁, and \( \rho_{\text{silver}} = 10.6 \) g/cm鲁, substitute into the relationship: \[ f_{\text{silver}} = f_{\text{steel}} \times \sqrt{\frac{\rho_{\text{steel}}}{\rho_{\text{silver}}}} = 200 \times \sqrt{\frac{7.80}{10.6}} \]. Calculate to find \( f_{\text{silver}} \).
05

Calculation Result

Calculate \( \sqrt{\frac{7.80}{10.6}} \) to find the ratio, which gives approximately \( 0.860 \). Therefore, \( f_{\text{silver}} = 200 \times 0.860 \approx 172 \) Hz.

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Key Concepts

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

Linear Mass Density
Linear mass density, denoted as \( \mu \), is an important concept when dealing with waves on strings or wires, especially in relation to vibrational frequency. It is defined as the mass per unit length of a wire or string:\[\mu = \frac{m}{L}\]where \( m \) is the mass and \( L \) is the length of the wire. This can also be expressed in terms of the material's density (\( \rho \)) and the cross-sectional area (\( A \)):\[\mu = \rho A\]
  • The formula depends on both the density of the material the wire is made of and its thickness or diameter.
  • In problems where wires have the same dimensions, their cross-sectional area \( A \) remains constant, simplifying calculations.
Understanding how to find linear mass density helps in examining how materials influence the vibration characteristics of wires.
Tension in Wires
The tension in a wire, often represented as \( T \), is crucial in determining the frequency at which the wire vibrates. Tension is the force exerted along the wire to stretch and keep it taut. It is a vital part of the vibrational frequency formula:\[ f = \frac{1}{2L} \sqrt{\frac{T}{\mu}} \]Some important points about tension:
  • Higher tension results in higher frequency vibrations.
  • In many controlled environments, wires are set to equal tension to compare other variables like material density.
Tension ensures that the wire remains straight and taut, allowing consistent wave propagation, thereby affecting the sound or vibrational characteristics.
Density of Materials
Density of materials, represented as \( \rho \), is the mass per unit volume of a material:\[\rho = \frac{m}{V}\]For wires, knowing the material density is key to calculating linear mass density. Different metals have different densities which significantly affect their vibrational properties:
  • Steel has a density of \( 7.80 \text{ g/cm}^3 \).
  • Silver is denser with a density of \( 10.6 \text{ g/cm}^3 \).
The price, weight, and how these materials behave under tension in the form of a wire, are influenced by these densities. A denser material generally means more mass, affecting the natural frequency of the wire.
Cross-Sectional Area
The cross-sectional area \( A \) of a wire plays a significant role in calculating the linear mass density. For cylindrical wires, \( A \) can be calculated if you know the diameter \( d \) of the wire:\[A = \frac{\pi d^2}{4}\]
  • All wires of the same diameter will have the same cross-sectional area.
  • It is constant for wires with the same geometry regardless of the material.
Understanding \( A \) helps in simplifying exercises where wires have equal diameters but are made of various materials, allowing comparisons rooted in linear mass density calculations.
Relation Between Frequency and Density
The relationship between frequency and density is crucial for understanding how different materials affect sound and wave propagation. The fundamental frequency \( f \) of a wire can be affected by its mass density \( \mu \):\[ f = \frac{1}{2L} \sqrt{\frac{T}{\mu}} \]In cases where wires are made of different materials but have equal geometric dimensions and tensions, the relationship simplifies to:\[ \frac{f_{\text{wire1}}}{f_{\text{wire2}}} = \sqrt{\frac{\rho_{\text{wire2}}}{\rho_{\text{wire1}}}} \]
  • Density affects the wire鈥檚 frequency inversely. A denser wire vibrates at a lower frequency.
  • By comparing density ratios, students can predict frequencies of similar wires made from different materials.
This relation helps in experimenting and understanding how material properties translate into observable vibrational characteristics.

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Most popular questions from this chapter

A string vibrates in five segments at a frequency of \(460 \mathrm{~Hz} .(a)\) What is its fundamental frequency? \((b)\) What frequency will cause it to vibrate in three segments? Detailed Method If the string is \(n\) segments long, then from Fig. 22-2 we have \(n\left(\frac{1}{2} \lambda\right)=L\). But \(\lambda=v / f_{n}\), so \(L=n\left(v / 2 f_{n}\right)\). Solving for \(f_{n}\) provides $$ f_{n}=n\left(\frac{v}{2 L}\right) $$ We are told that \(f_{5}=460 \mathrm{~Hz}\), and so $$ 460 \mathrm{~Hz}=5\left(\frac{v}{2 L}\right) \quad \text { or } \quad \frac{v}{2 L}=92.0 \mathrm{~Hz} $$ Substituting this in the above relation gives $$ f_{n}=(n)(92.0 \mathrm{~Hz}) $$ (a) \(f_{1}=92.0 \mathrm{~Hz}\). (b) \(f_{3}=(3)(92 \mathrm{~Hz})=276 \mathrm{~Hz}\) Alternative Method Recall that for a string held at both ends, \(f_{n}=n f_{1}\). Knowing that \(f_{5}=460 \mathrm{~Hz}\), it follows that \(f_{1}=92.0 \mathrm{~Hz}\) and \(f_{3}=276 \mathrm{~Hz}\).

Radio station WJR broadcasts at \(760 \mathrm{kHz}\). The speed of radio waves is \(3.00 \times 10^{8} \mathrm{~m} / \mathrm{s}\). What is the wavelength of WJR's waves?

Radar waves with \(3.4 \mathrm{~cm}\) wavelength are sent out from a transmitter. Their speed is \(3.00 \times 10^{8} \mathrm{~m} / \mathrm{s}\). What is their frequency?

A uniform flexible cable is \(20 \mathrm{~m}\) long and has a mass of \(5.0 \mathrm{~kg}\). It hangs vertically under its own weight and is vibrated (perpendicularly) from its upper end with a frequency of \(7.0 \mathrm{~Hz}\). ( \(a\) ) Find the speed of a transverse wave on the cable at its midpoint. (b) What are the frequency and wavelength at the midpoint? (a) We shall use \(v=\sqrt{\text { (Tension/(Mass per unit length) }}\). The midpoint of the cable supports half its weight, so the tension there is $$ F_{T}=\frac{1}{2}(5.0 \mathrm{~kg})\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)=24.5 \mathrm{~N} $$ Further \(\quad\) Mass per unit length \(=\frac{5.0 \mathrm{~kg}}{20 \mathrm{~m}}=0.25 \mathrm{~kg} / \mathrm{m}\) so that $$ v=\sqrt{\frac{24.5 \mathrm{~N}}{0.25 \mathrm{~kg} / \mathrm{m}}}=9.9 \mathrm{~m} / \mathrm{s} $$ (b) Because wave crests do not pile up along a string or cable, the number passing one point must be the same as that for any other point. Therefore, the frequency, \(7.0 \mathrm{~Hz}\), is the same at all points. To find the wavelength at the midpoint, we must use the speed we found for that point, \(9.9 \mathrm{~m} / \mathrm{s}\). That gives us $$ \lambda=\frac{v}{f}=\frac{9.9 \mathrm{~m} / \mathrm{s}}{7.0 \mathrm{~Hz}}=1.4 \mathrm{~m} $$

(a) At what point should a stretched string be plucked to make its fundamental tone most prominent? At what point should it be plucked and then at what point touched \((b)\) to make its first overtone most prominent and \((c)\) to make its second overtone most prominent?
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