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Chapter 16: Problem 20

A copper wire, whose cross-sectional area is \(1.1 \times 10^{-6} \mathrm{~m}^{2},\) has a linear density of \(7.0 \times 10^{-3} \mathrm{~kg} / \mathrm{m}\) and is strung between two walls. At the ambient temperature, a transverse wave travels with a speed of \(46 \mathrm{~m} / \mathrm{s}\) on this wire. The coefficient of linear expansion for copper is \(17 \times 10^{-6}\left(\mathrm{C}^{\circ}\right)^{-1},\) and Young's modulus for copper is \(1.1 \times 10^{11} \mathrm{~N} / \mathrm{m}^{2}\). What will be the speed of the wave when the temperature is lowered by \(14 \mathrm{C}^{\circ}\) ? Ignore any change in the linear density caused by the change in temperature.

Short Answer

Expert verified
The new wave speed is approximately 50.13 m/s.

Step by step solution

01

Understand the wave speed equation

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

Calculate the original tension

Using the known wave speed and linear density, we rearrange the wave speed formula to find the tension: \( T = \mu v^2 \).Given: \( \mu = 7.0 \times 10^{-3} \mathrm{~kg/m} \) and \( v = 46 \mathrm{~m/s} \).Calculate: \( T = 7.0 \times 10^{-3} \times (46)^2 \).\[ T = 7.0 \times 10^{-3} \times 2116 = 14.812 \mathrm{~N} \].
03

Calculate the change in length due to temperature change

The change in length of the wire due to temperature change can be calculated using the formula: \( \Delta L = L_0 \alpha \Delta T \), where \( L_0 \) is the original length, \( \alpha \) is the coefficient of linear expansion, and \( \Delta T \) is the change in temperature.We know \( \alpha = 17 \times 10^{-6} \mathrm{~C}^{-1} \) and \( \Delta T = -14 \mathrm{~C} \).
04

Calculate the new tension in the wire

Young's modulus formula, \( E = \frac{\Delta T}{\epsilon} \) where \( \epsilon = \frac{\Delta L}{L_0} \), can be used to find the new tension. Here, tension exerted across the cross-sectional area is \( \frac{F}{A} \).Rearranging: \( F = E \times A \times \alpha \times (-\Delta T) \).Substitute: \( E = 1.1 \times 10^{11} \mathrm{~N/m^2} \), \( A = 1.1 \times 10^{-6} \mathrm{~m^2} \), \( \alpha = 17 \times 10^{-6} \mathrm{~C}^{-1} \), \( \Delta T = 14 \mathrm{C} \).Calculate: \( F = 1.1 \times 10^{11} \times 1.1 \times 10^{-6} \times 17 \times 10^{-6} \times 14 \).\[ F = 1.1 \times 10^{11} \times 1.1 \times 10^{-6} \times 17 \times 10^{-6} \times 14 = 2.786 \mathrm{~N} \].The new tension after the temperature drop is: \( T_{new} = 14.812 + 2.786 = 17.598 \mathrm{~N} \).
05

Calculate the new wave speed

Substitute the new tension into the wave speed formula: \( v_{new} = \sqrt{\frac{T_{new}}{\mu}} \).Given: \( T_{new} = 17.598 \mathrm{~N} \), \( \mu = 7.0 \times 10^{-3} \mathrm{~kg/m} \).Calculate: \( v_{new} = \sqrt{\frac{17.598}{7.0 \times 10^{-3}}} \).\[ v_{new} = \sqrt{2514} = 50.13 \mathrm{~m/s} \].

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Key Concepts

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

Tension and Wave Speed
Wave speed on a string primarily depends on two factors: the tension in the string and its linear density. The basic formula to calculate the speed of a wave on a string is given by \[ v = \sqrt{ \frac{T}{\mu} } \] where \( v \) is the wave speed, \( T \) is the tension in the string, and \( \mu \) is the linear mass density of the string.
  • **Tension**: This refers to the pulling force applied to the string. Greater tension leads to faster wave speeds.
  • **Linear Density**: This is the mass per unit length of the string. A heavier string (more mass in a given length) tends to slow the wave, thus decreasing speed.
In summary, increasing the tension increases the wave speed, while increasing the mass per unit length decreases it. Understanding this relationship helps us grasp why a stringed instrument sounds different when tightened or loosed.
Thermal Expansion and Wave Speed
Materials like copper expand or contract in response to temperature changes. This phenomenon is known as thermal expansion. It can significantly impact wave speed in materials such as wires or strings used in musical instruments or mechanical systems.The change in length \( \Delta L \) of a material due to temperature change is given by: \[ \Delta L = L_0 \alpha \Delta T \] where \( L_0 \) is the original length, \( \alpha \) is the coefficient of linear expansion, and \( \Delta T \) is the change in temperature.
  • **Expansion and Contraction**: As temperature rises, most materials expand, leading to a decrease in tension and slower wave speeds. Conversely, when materials are cooled, they contract, increasing tension and thus increasing wave speed.
  • **Coefficient of Expansion**: Different materials expand at different rates, characterized by their coefficient of linear expansion. For copper, the coefficient is quite small, meaning small temperature changes can still noticeably affect wave speed.
Understanding thermal expansion is crucial in applications involving precise tension and wave speed requirements.
Young's Modulus and Wave Dynamics
Young's modulus \( E \), a fundamental property of materials, measures their stiffness by quantifying the relationship between stress and strain.The formula for Young's modulus is: \[ E = \frac{\text{stress}}{\text{strain}} = \frac{F/A}{\Delta L/L_0} \] where \( F \) is the force applied, \( A \) is the cross-sectional area, \( \Delta L \) is the change in length, and \( L_0 \) is the original length.
  • **Role in Wave Dynamics**: When a material like a copper wire is stretched, Young's modulus helps determine how much it will extend or compress under a load, thereby affecting tension and wave speed.
  • **Material Stiffness**: A higher Young's modulus means the material is stiffer, contributing to higher wave speeds when tension is adjusted.
Young's modulus is a fundamental component in calculating changes in tension due to mechanical loads, and thus changes in wave speeds in strings.
Temperature Effects on Materials
Temperature fluctuations can considerably affect the mechanical properties of materials, such as tension and wave speed. As seen in thermal expansion, increasing temperature often results in expansion, while temperature decreases lead to contraction.
  • **Effect on Tension**: Lowering the temperature of a wire or string can cause it to contract, increasing tension if the wire is fixed between two points.
  • **Impact on Wave Speed**: As tension increases due to cooling, the wave speed on the string increases. Conversely, warming can lead to decreased tension and wave speed.
Adjustments for temperature fluctuations are crucial in practical applications such as musical instruments and machinery where precise wave dynamics are essential. Understanding these effects can prevent potential material failures or tuning issues in various applications.

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Most popular questions from this chapter

Argon (molecular mass \(=39.9 \mathrm{u}\) ) is a monatomic gas. As suming that it behaves like an ideal gas at \(298 \mathrm{~K}(\gamma=1.67),\) find (a) the rms speed of argon atoms and (b) the speed of sound in argon.

A sound wave travels twice as far in neon (Ne) as it does in krypton (Kr) in the same time interval. Both neon and krypton can be treated as monatomic ideal gases. The atomic mass of neon is \(20.2 \mathrm{u}\), and that of krypton is \(83.8 \mathrm{u}\). The temperature of the krypton is \(293 \mathrm{~K}\). What is the temperature of the neon?

A water-skier is moving at a speed of \(12.0 \mathrm{~m} / \mathrm{s}\). When she skis in the same direction as a traveling wave, she springs upward every \(0.600 \mathrm{~s}\) because of the wave crests. When she skis in the direction opposite to that in which the wave moves, she springs upward every \(0.500 \mathrm{~s}\) in response to the crests. The speed of the skier is greater than the speed of the wave. Determine (a) the speed and (b) the wavelength of the wave.

A typical adult ear has a surface area of \(2.1 \times 10^{-3} \mathrm{~m}^{2} .\) The sound intensity during a normal conversation is about \(3.2 \times 10^{-6} \mathrm{~W} / \mathrm{m}^{2}\) at the listener's ear. Assume that the sound strikes the surface of the ear perpendicularly. How much power is intercepted by the ear?

Concept Questions Multiple-Concept 11 provides a model for solving this type of problem. A wireless transmitting microphone is mounted on a small platform, which can roll down an incline, away from a speaker that is mounted at the top of the incline. The speaker broadcasts a fixed-frequency tone. (a) The platform is positioned in front of the speaker and released from rest. Describe how the velocity of the platform changes and why. (b) How is the changing velocity related to the acceleration of the platform? (c) Describe how the frequency detected by the microphone changes. Explain why the frequency changes as you have described. (d) Which equation given in the chapter applies to this situation? Justify your answer. Problem The speaker broadcasts a tone that has a frequency of \(1.000 \times 10^{4} \mathrm{~Hz}\), and the speed of sound is \(343 \mathrm{~m} / \mathrm{s}\). At a time of \(1.5 \mathrm{~s}\) following the release of the platform, the microphone detects a frequency of \(9939 \mathrm{~Hz}\). At a time of \(3.5 \mathrm{~s}\) following the release of the platform, the microphone detects a frequency of \(9857 \mathrm{~Hz}\). What is the acceleration (assumed constant) of the platform?
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