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Find the speed of compression waves in a metal rod if the material of the rod has a Young's modulus of \(1.20 \times 10^{10} \mathrm{~N} / \mathrm{m}^{2}\) and a density of \(8920 \mathrm{~kg} / \mathrm{m}^{3}\).

Short Answer

Expert verified
The speed of compression waves is approximately 36.68 m/s.

Step by step solution

01

Understanding the Problem

We are asked to find the speed of compression waves, or sound waves, in a metal rod. This speed can be calculated using the formula for wave speed in a solid: \( v = \sqrt{\frac{E}{\rho}} \), where \( E \) is the Young's modulus and \( \rho \) is the density of the material.
02

Identifying Known Values

From the problem statement, we have: - Young's modulus \( E = 1.20 \times 10^{10} \) N/m² - Density \( \rho = 8920 \) kg/m³. We will use these values in the formula for wave speed.
03

Setting Up the Formula

Write the formula for the speed of compression waves: \[ v = \sqrt{\frac{E}{\rho}} \] Substitute the given values for \( E \) and \( \rho \) into the equation.
04

Substituting the Values

Substitute \( E = 1.20 \times 10^{10} \) and \( \rho = 8920 \) into the formula: \[ v = \sqrt{\frac{1.20 \times 10^{10}}{8920}} \]
05

Calculating the Wave Speed

Perform the calculations to determine the wave speed:1. Divide the Young's modulus by the density: \[ \frac{1.20 \times 10^{10}}{8920} \approx 1345.75 \]2. Take the square root of the result to find the speed: \[ v = \sqrt{1345.75} \approx 36.68 \, \text{m/s} \]
06

Conclusion

The speed of compression waves in the metal rod is approximately 36.68 m/s.

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Key Concepts

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

Young's Modulus
Young's modulus is an essential mechanical property that describes a material's stiffness. It is a measure of the ability of a material to withstand changes in length when under lengthwise tension or compression. In essence, Young's modulus tells us how difficult it is to stretch or compress a material.
Young's modulus is represented by the symbol \( E \), and it is expressed in units of pressure, namely pascal (N/m²). A higher Young's modulus indicates a stiffer material, which deforms less under the same applied force compared to a material with a lower Young's modulus.
Knowing the Young's modulus of a material is crucial when you're dealing with mechanical engineering and civil project applications, such as construction, where understanding material properties can impact safety and function.
In the case of the exercise, the metal rod has a Young's modulus of \( 1.20 \times 10^{10} \) N/m². This value means the metal is quite rigid, which implies that it does not easily deform under stress.
Density of Materials
Density is a fundamental property that signifies how much mass is contained in a unit volume of a substance. The formula to calculate density is \( \rho = \frac{mass}{volume} \), where \( \rho \) is the density. It is expressed in terms of kilograms per cubic meter (kg/m³).
Understanding the density of a material is important in contexts ranging from buoyancy analysis to determining material quality. For example, in construction, materials with high density might be preferred for certain load-bearing applications.
In the provided exercise, the material of the rod has a density of \( 8920 \) kg/m³. This high density suggests the material is substantial and relatively weighty, containing a lot of mass in a small volume—reflecting its robust nature.
When calculating wave speed in a solid, considering density alongside Young's modulus allows the examination of how mass distribution within the solid impacts wave propagation.
Compression Waves
Compression waves, also known as longitudinal waves, are a type of mechanical wave where particle displacement is parallel to the direction of wave propagation. These waves can travel through solids, liquids, and gases. In solids, compression waves are often referred to as sound waves.
In the context of solids, the speed of these waves is significantly influenced by the material's properties, specifically Young's modulus and density. The wave speed \( v \) in a solid is given by the formula \( v = \sqrt{\frac{E}{\rho}} \), where \( E \) is the Young's modulus and \( \rho \) is the density.
This equation tells us that the wave speed is proportional to the square root of the stiffness of the material (Young's modulus) and inversely proportional to the square root of its density. Thus, a material with high stiffness and low density allows waves to travel faster.
In the exercise, using the values \( E = 1.20 \times 10^{10} \) N/m² and \( \rho = 8920 \) kg/m³, the calculated wave speed turns out to be approximately \( 36.68 \) m/s. This speed indicates how quickly compression waves traverse through the specified metal rod.

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Most popular questions from this chapter

What is the speed of compression waves (sound waves) in water? The bulk modulus for water is \(2.2 \times 10^{9} \mathrm{~N} / \mathrm{m}^{2}\) $$ v=\sqrt{\frac{\text { Bulk modulus }}{\text { Density }}}=\sqrt{\frac{2.2 \times 10^{9} \mathrm{~N} / \mathrm{m}^{2}}{1000 \mathrm{~kg} / \mathrm{m}^{3}}}=1.5 \mathrm{~km} / \mathrm{s} $$

At S.T.P., the speed of sound in air is \(331 \mathrm{~m} / \mathrm{s}\). Determine the speed of sound in hydrogen at S.T.P. if the specific gravity of hydrogen relative to air is \(0.0690\) and if \(\gamma=1.40\) for both gases.

Find the speed of sound in a diatomic ideal gas that has a density of \(3.50 \mathrm{~kg} / \mathrm{m}^{3}\) and a pressure of \(215 \mathrm{kPa}\) We know that \(v=\sqrt{\gamma R T / M}\) and can find the temperature from the pressure. Using the gas law \(P V=(m / M) R T\), $$ \frac{R T}{M}=P \frac{V}{m} $$ Moreover, \(\rho=m / V\), and so the expression for the speed becomes $$ v=\sqrt{\frac{\gamma P}{\rho}}=\sqrt{\frac{(1.40)\left(215 \times 10^{3} \mathrm{~Pa}\right)}{3.50 \mathrm{~kg} / \mathrm{m}^{3}}}=293 \mathrm{~m} / \mathrm{s} $$ We used the fact that \(\gamma \approx 1.40\) for a diatomic ideal gas, as discussed in Chapter 20 .

Determine the speed of sound in carbon diaxide \((M-44 \mathrm{~kg} / \mathrm{kmol}, \gamma=1.30)\) at a pressure of \(0.50 \mathrm{~atm}\) and a temperature of \(400^{\circ} \mathrm{C}\).

A car moving at \(20 \mathrm{~m} / \mathrm{s}\) with its horn blowing \((f=1200 \mathrm{~Hz})\) is chasing another car going at \(15 \mathrm{~m} / \mathrm{s}\) in the same direction. What is the apparent frequency of the horn as heard by the driver being chased? Take the speed of sound to be \(340 \mathrm{~m} / \mathrm{s}\). This is a Doppler problem. Draw the observer-to-source arrow; that's the positive direction (see Fig. 23-1). Both the source and the observer are moving in the negative direction. Hence, we use \(-v_{o}\) and \(-v_{s}\) $$ f_{o}=f_{s} \frac{v \pm v_{o}}{v \mp v_{s}}=(1200 \mathrm{~Hz}) \frac{340-15}{340-20}=1.22 \mathrm{kHz} $$ Because the source is approaching the observer, the latter will measure an increase in frequency.

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