/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 86 A metal wire, with density \(\rh... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A metal wire, with density \(\rho\) and Young's modulus \(Y\) is stretched between rigid supports. At temperature \(T,\) the speed of a transverse wave is found to be \(v_{1}\) . When the temperature is increased to \(T+\Delta T,\) the speed decreases to \(v_{2} < v_{1} .\) Determine the coefficient of linear expansion of the wire.

Short Answer

Expert verified
\( \alpha = \frac{1}{\Delta T} \left( 1 - \frac{v_2^2}{v_1^2} \right) \)

Step by step solution

01

Understand the Relationship Between Wave Speed and Tension

The speed of a transverse wave on a wire is given by \( v = \sqrt{\frac{T}{\mu}} \), where \( T \) is the tension, and \( \mu \) is the linear mass density \( \mu = \rho A \) (with \( A \) being the cross-sectional area). Since the ends are fixed, a thermal expansion occurs which changes the tension in the wire.
02

Express Tension in Terms of Modulus of Elasticity

The tension in the wire, which affects the wave speed, can be expressed using Young's Modulus \( Y \). This is given by \( T = Y \times A \times \frac{\Delta L}{L} \), where \( \Delta L \) is the elongation due to thermal expansion.
03

Define the Relationship of Wave Speed with Temperature Change

When the temperature increases, the length change due to the coefficient of linear expansion \( \alpha \) is \( \Delta L = \alpha L \Delta T \). Substitute into the expression for tension: \( T = Y \times A \times \alpha \Delta T \). Consequently, as the temperature increases the speed of the wave changes as \( v_2 = \sqrt{\frac{T_2}{\mu}} \), and originally \( v_1 = \sqrt{\frac{T_1}{\mu}} \).
04

Solve for the Coefficient of Linear Expansion

Using the wave speeds, we know \( v_1^2 = \frac{Y A}{\rho A} \) initially and modifies to \( v_2^2 = \frac{Y A (1 - \alpha \Delta T)}{\rho A} \). Simplifying gives \( \alpha \Delta T = 1 - \frac{v_2^2}{v_1^2} \), thus \( \alpha = \frac{1}{\Delta T}\left(1 - \frac{v_2^2}{v_1^2}\right) \).
05

Final Expression and Verification

So, the coefficient of linear expansion is calculated as \( \alpha = \frac{1 - \frac{v_2^2}{v_1^2}}{\Delta T} \). This equation is derived based on how linear expansion impacts tension and results in a change in wave speed.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Wave Speed
Understanding wave speed is essential when studying physics and materials science. Wave speed, or the velocity of a wave, is calculated by dividing the distance that the wave travels by the time it takes to travel that distance. For a transverse wave on a wire, the formula is given by:
  • \( v = \sqrt{\frac{T}{\mu}} \)
where \( v \) is the wave speed, \( T \) is the tension in the wire, and \( \mu \) is the linear mass density.
The tension in the wire pulls it taut, while the linear mass density relates to the mass per unit length of the wire.
Wave speed can change due to factors like tension adjustments and thermal expansion, as observed when temperature varies. This is because a rise in temperature causes the wire to expand, thereby altering the tension that directly affects the wave speed.
Thus, observing the change in wave speed at different temperatures helps to deduce the coefficient of linear expansion.
Young's Modulus
Young's modulus is a fundamental concept in elasticity, providing insight into a material's ability to resist changes in length when under tension or compression. It quantifies a material's stiffness or rigidity, calculated as:
  • \( Y = \frac{T}{A \cdot \frac{\Delta L}{L}} \)
where \( Y \) is Young's modulus, \( T \) is the tension force, \( A \) is the cross-sectional area, \( \Delta L \) is the change in length, and \( L \) is the original length.
In the context of wave speed in a wire, Young's modulus allows for predicting how a wire will stretch given a certain force.
For a fixed tension and area, it determines the proportion of elongation due to thermal effects. Since mechanical properties and temperature are linked, changes in Young's modulus provide clues about thermal behavior and coefficient of linear expansion.
Thermal Expansion
Thermal expansion is the tendency of matter to change in volume in response to a change in temperature. When a material heats up, its particles begin to move faster and typically expand. The coefficient of linear expansion, \( \alpha \), quantifies this expansion:
  • \( \Delta L = \alpha L \Delta T \)
where \( \Delta L \) is the change in length, \( \alpha \) is the coefficient of linear expansion, \( L \) is the original length, and \( \Delta T \) is the change in temperature.
This principle affects the wave speed in a metal wire because as the temperature increases, the length and tension of the wire change, leading to a different wave speed.
Understanding thermal expansion is crucial for determining the coefficient of linear expansion, particularly in scenarios where precise measurements and adjustments are needed, for instance, in engineering and construction.
Linear Mass Density
Linear mass density (\( \mu \)) describes how much mass is distributed along a unit length of the object, typically expressed as:
  • \( \mu = \frac{\text{mass}}{\text{length}} \)
In the formula of wave speed, \( \mu \) represents the distribution of mass along the wire (\( \mu = \rho A \)). Here, \( \rho \) is the density of the material and \( A \) is the cross-sectional area.
The linear mass density influences the tension needed to achieve a particular wave speed. This is because a higher mass density requires more tension to achieve the same wave speed compared to a lesser mass density.
Calculating \( \mu \) is integral for solving problems relating to vibrations and wave propagation, ensuring that appropriate properties are maintained for desired outcomes.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A Styrofoam bucket of negligible mass contains 1.75 \(\mathrm{kg}\) of water and 0.450 \(\mathrm{kg}\) of ice. More ice, from a refrigerator at \(-15.0^{\circ} \mathrm{C},\) is added to the mixture in the bucket, and when thermal equilibrium has been reached, the total mass of ice in the bucket is 0.868 \(\mathrm{kg} .\) Assuming no heat exchange with the surroundings, what mass of ice was added?

A wood ceiling with thermal resistance \(R_{1}\) is covered with a layer of insulation with thermal resistance \(R_{2} .\) Prove that the effective thermal resistance of the combination is \(R=R_{1}+R_{2}\) .

"The Ship of the Desert." Camels require very little water because they are able to tolerate relatively large changes in their body temperature. While humans keep their body temperatures constant to within one or two Celsius degrees, a dehydrated camel permits its body temperature to drop to \(34.0^{\circ} \mathrm{C}\) overnight and rise to \(40.0^{\circ} \mathrm{C}\) during the day. To see how effective this mechanism is for saving water, calculate how many liters of water a \(400-\mathrm{kg}\) camel would have to drink if it attempted to keep its body temperature at a constant \(34.0^{\circ} \mathrm{C}\) by evaporation of sweat during the day (12 hours) instead of letting it rise to \(40.0^{\circ} \mathrm{C} .\) (Note: The specific heat of a camel or other mammal is about the same as that of a typical human, 3480 \(\mathrm{J} / \mathrm{kg} \cdot \mathrm{K}\) . The heat of vaporization of water at \(34^{\circ} \mathrm{C}\) is \(2.42 \times 10^{6} \mathrm{J} / \mathrm{kg}.\))

Like the Kelvin scale, the Rankine scale is an absolute temperature scale: Absolute zero is zero degrees Rankine \(\left(0^{\circ} \mathrm{R}\right)\) However, the units of this scale are the same size as those of the Fahrenheit scale rather than the Celsius scale. What is the numerical value of the triple-point temperature of water on the Rankine scale?

While painting the top of an antenna 225 \(\mathrm{m}\) in height, a worker accidentally lets a \(1.00-\mathrm{L}\) wattle fall from his lunchbox. The bottle lands in some bushes at ground level and does not break. If a quantity of heat equal to the magnitude of the change in mechanical energy of the water goes into the water, what is its increase in temperature?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.