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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 172 Hz.

Step by step solution

01

Understand the Relationship between Frequency and Linear Density

The frequency of a wire under tension is given by the formula \( f = \frac{1}{2L} \sqrt{\frac{T}{\mu}} \), where \( f \) is the frequency, \( L \) is the length of the wire, \( T \) is the tension, and \( \mu \) is the linear mass density of the wire. The linear density \( \mu \) is related to the volume density \( \rho \) via the wire's cross-sectional area \( A \): \( \mu = \rho \cdot A \). In this problem, as both wires have the same diameter, the cross-sectional area cancels out in the comparison.
02

Calculate the Linear Mass Density Ratio

Since both wires have the same diameter and length, we calculate the linear mass density \( \mu \) for each wire: \[ \mu_{steel} \propto \rho_{steel}, \quad \mu_{silver} \propto \rho_{silver} \]The ratio of the linear densities is then \( \frac{\mu_{silver}}{\mu_{steel}} = \frac{\rho_{silver}}{\rho_{steel}} = \frac{10.6}{7.8} \approx 1.36 \).
03

Analyze the Frequency Proportions

The frequency formula \( f = \frac{1}{2L} \sqrt{\frac{T}{\mu}} \) implies that the frequency is inversely proportional to the square root of the linear density \( \mu \):\( f \propto \frac{1}{\sqrt{\mu}} \).Thus the proportion of the fundamental frequencies is:\[ \frac{f_{silver}}{f_{steel}} = \sqrt{\frac{\mu_{steel}}{\mu_{silver}}} \approx \sqrt{\frac{1}{1.36}} \approx 0.86 \].
04

Calculate the Fundamental Frequency of the Silver Wire

Given that the fundamental frequency of the steel wire is \( f_{steel} = 200 \text{ Hz} \), we find the frequency of the silver wire using the ratio derived:\[ f_{silver} = f_{steel} \times 0.86 = 200 \times 0.86 = 172 \text{ Hz} \].

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Key Concepts

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

Linear Density
Linear density is a concept that refers to how much mass is distributed along a length of the string or wire. It is expressed as mass per unit length and is denoted by the symbol \( \mu \). The formula used to calculate linear density is \( \mu = \frac{m}{L} \), where \( m \) is the mass and \( L \) is the length of the wire. Understanding linear density is crucial because it plays a significant role in determining the frequency of a vibrating string.
In problems involving wires of the same diameter and length, linear density can be directly compared by using volume density (\( \rho \)), because cross-sectional areas (\( A \)) cancel out. Thus, \( \mu = \rho \cdot A \). For wires or strings that are identical in dimensions but differ in material, linear density helps assess how their material properties affect their behavior under tension.
Tension in Wires
Tension is the force that is exerted along a string or wire and is essential in understanding how vibrations occur. When a wire is under tension, it is stretched tightly, which is crucial for vibration and frequency considerations. The tension force is usually denoted by \( T \) in physics problems.
The frequency of vibration is affected by the tension in the wire. The greater the tension, the higher the frequency, assuming other factors like linear density remain constant. Mathematically, frequency \( f \) is related to tension via the square root term \( \sqrt{\frac{T}{\mu}} \) in the formula \( f = \frac{1}{2L} \sqrt{\frac{T}{\mu}} \). This highlights that an increase in tension would lead to an increase in vibrational frequency, making understanding tension pivotal in problems dealing with vibrating strings.
Density and Material Properties
Density is a fundamental material property that is defined as mass per unit volume, denoted by \( \rho \). For two wires made of different materials but with the same dimensions, the difference in density directly affects the linear density and subsequently the frequency of vibration.
Different materials have different densities due to their atomic structures. For example, steel has a density of \( 7.80 \text{ g/cm}^3 \) while silver has a density of \( 10.6 \text{ g/cm}^3 \). These values affect how mass is distributed in the wire, thus impacting the wire's behavior under tension. By understanding a material's density, you can infer its resistance to vibration or tension, which is why knowing the density of different materials is valuable when calculating linear density and frequency.
Fundamental Frequency Calculation
The fundamental frequency is the lowest frequency at which a system resonates. For a vibrating wire, it is the frequency of the simplest mode of vibration, with the wire length corresponding to half the wavelength of the sound wave produced.
To calculate the fundamental frequency of a wire, you can use the formula \( f = \frac{1}{2L} \sqrt{\frac{T}{\mu}} \). Knowing the length \( L \), tension \( T \), and linear density \( \mu \) allows you to compute \( f \). In scenarios where one wire's frequency is known, as with the steel wire here, you can determine another wire's fundamental frequency by considering proportions of linear densities. For silver wires in this example, the formula accounts for changes in material density, resulting in a calculated frequency of \( 172 \text{ Hz} \), given a steel wire frequency of \( 200 \text{ Hz} \). This illustrates the interconnection between material properties and vibrational outcomes.

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Most popular questions from this chapter

What must be the length of an iron rod that has the fundamental frequency \(320 \mathrm{~Hz}\) when clamped at its center? Assume longitudinal vibration at a speed of \(5.00 \mathrm{~km} / \mathrm{s}\).

(a) Determine the shortest length of pipe closed at one end that will resonate in air when driven by a sound source of frequency \(160 \mathrm{~Hz}\). Take the speed of sound in air to be \(340 \mathrm{~m} / \mathrm{s} .(b)\) Repeat the analysis for a pipe open at both ends. (a) Figure \(22-4(a)\) applies in this case. The shortest pipe will be \(\lambda / 4\) long. Therefore, $$ L=\frac{1}{4} \lambda=\frac{1}{4}\left(\frac{v}{f}\right)=\frac{340 \mathrm{~m} / \mathrm{s}}{4\left(160 \mathrm{~s}^{-1}\right)}=0.531 \mathrm{~m} $$ (b) In this case the pipe will have antinodes at both ends and a node at its center. Then, $$ L=2\left(\frac{1}{4} \lambda\right)=\frac{1}{2}\left(\frac{v}{f}\right)=\frac{340 \mathrm{~m} / \mathrm{s}}{2\left(160 \mathrm{~s}^{-1}\right)}=1.06 \mathrm{~m} $$

A uniform flexible cable is \(20 \mathrm{~m}\) long and has a mass of \(5.0\) 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? (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 $$ \lambda=\frac{v}{f}=\frac{9.9 \mathrm{~m} / \mathrm{s}}{7.0 \mathrm{~Hz}}=1.4 \mathrm{~m} $$

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?

A string \(180-\mathrm{cm}\) -long resonates in a standing wave that has three segments when driven by a 270 - \(\mathrm{Hz}\) vibrator. What is the speed of the waves on the string?

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