/*! 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 73 Two solid rods have the same bul... [FREE SOLUTION] | 91Ó°ÊÓ

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Two solid rods have the same bulk modulus but one is 2.5 times as dense as the other. In which rod will the speed of longitudinal waves be greater, and by what factor?

Short Answer

Expert verified
The speed in the first rod is greater by a factor of approximately 1.58.

Step by step solution

01

Understanding the Relationship

The speed of longitudinal waves in a rod is determined by the formula \( v = \sqrt{\frac{K}{\rho}} \), where \( v \) is the speed of sound, \( K \) is the bulk modulus, and \( \rho \) is the density of the material. In this problem, both rods have the same bulk modulus, \( K \), but different densities.
02

Calculate the Speed for Each Rod

Let \( \rho_1 \) be the density of the first rod and \( \rho_2 = 2.5 \times \rho_1 \) be the density of the second rod. The speed of sound in the first rod is defined as \( v_1 = \sqrt{\frac{K}{\rho_1}} \) and in the second rod as \( v_2 = \sqrt{\frac{K}{\rho_2}} \).
03

Substitute and Compare Speeds

Substitute the given density relation: \( v_2 = \sqrt{\frac{K}{2.5 \rho_1}} = \frac{1}{\sqrt{2.5}} \sqrt{\frac{K}{\rho_1}} = \frac{1}{\sqrt{2.5}} v_1 \). Therefore, \( v_2 = \frac{1}{\sqrt{2.5}} v_1 \).
04

Determine the Factor

Simplify \( \sqrt{2.5} \) to find the factor difference: \( \sqrt{2.5} \approx 1.58 \). Therefore, the speed of sound in the first rod is approximately 1.58 times greater than in the second rod.

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Key Concepts

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

Bulk Modulus
The bulk modulus is a measure of a material's resistance to uniform compression. It is a critical parameter when studying longitudinal waves because it defines how much a material will compress under pressure.
In simpler terms, the bulk modulus tells us how hard it is to squeeze a material into a smaller volume. If a material has a high bulk modulus, it will not easily change its volume when pressure is applied.
  • Unit of bulk modulus: Pascal (Pa)
  • High bulk modulus: Difficult to compress
  • Low bulk modulus: Easier to compress
In the context of longitudinal waves, the bulk modulus helps to determine the speed at which waves travel through a material. Since the speed of longitudinal waves is proportional to the square root of the bulk modulus to density ratio, the bulk modulus, alongside density, is crucial for calculating wave speed.
Density and Wave Speed
Density is another vital property that impacts the speed of longitudinal waves in a material. It refers to how much mass is packed into a given volume of the material. Understanding density's influence on wave speed helps explain why sound travels differently in various mediums.
In the exercise, the density of one rod is mentioned as 2.5 times the density of another. This difference in density affects wave speed. According to the formula for wave speed, \( v = \sqrt{\frac{K}{\rho}} \), as density \( \rho \) increases, wave speed \( v \) decreases if the bulk modulus \( K \) remains constant.
Thus, a denser medium results in slower wave speeds because it has more mass packed into each unit volume. Conversely, in a less dense medium, waves can travel faster because there is less mass to push through.
Speed of Sound in Solids
The speed of sound in solids depends largely on the material properties such as bulk modulus and density. Solids typically conduct sound faster than liquids and gases because their particles are closely packed, allowing sound waves to travel more efficiently.
Formula: The speed of sound \( v \) in a solid is expressed as \( v = \sqrt{\frac{K}{\rho}} \), where \( K \) is the bulk modulus and \( \rho \) is the density. The interplay of these values dictates how quickly vibrations can propagate through the material.
  • Higher bulk modulus: Increase in speed
  • Higher density: Decrease in speed
  • Same bulk modulus, different densities: Varying speeds
In the exercise, we see that though both rods have the same bulk modulus, their differing densities result in different propagation speeds. This reflects how understanding the speed of sound in solids is essential for predicting wave behavior in different materials.

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Most popular questions from this chapter

(II) A cord of length \(1.0 \mathrm{~m}\) has two equal-length sections with linear densities of \(0.50 \mathrm{~kg} / \mathrm{m}\) and \(1.00 \mathrm{~kg} / \mathrm{m}\). The tension in the entire cord is constant. The ends of the cord are oscillated so that a standing wave is set up in the cord with a single node where the two sections meet. What is the ratio of the oscillatory frequencies?

A highway overpass was observed to resonate as one full loop \(\left(\frac{1}{2} \lambda\right)\) when a small earthquake shook the ground vertically at \(3.0 \mathrm{~Hz}\). The highway department put a support at the center of the overpass, anchoring it to the ground as shown in Fig. \(15-41 .\) What resonant frequency would you now expect for the overpass? It is noted that earthquakes rarely do significant shaking above 5 or \(6 \mathrm{~Hz}\). Did the modifications do any good? Explain.

(II) A transverse wave pulse travels to the right along a string with a speed \(v=2.0 \mathrm{~m} / \mathrm{s}\). At \(t=0\) the shape of the pulse is given by the function $$ D=0.45 \cos (2.6 x+1.2) $$ where \(D\) and \(x\) are in meters. (a) Plot \(D\) vs. \(x\) at \(t=0\). (b) Determine a formula for the wave pulse at any time \(t\) assuming there are no frictional losses. (c) Plot \(D(x, t)\) vs. \(x\) at \(t=1.0 \mathrm{~s} .\) (d) Repeat parts \((b)\) and \((c)\) assuming the pulse is traveling to the left. Plot all 3 graphs on the same axes for easy comparison.

(I) A violin string vibrates at \(441 \mathrm{~Hz}\) when unfingered. At what frequency will it vibrate if it is fingered one-third of the way down from the end? (That is, only two-thirds of the string vibrates as a standing wave.)

(II) A ski gondola is connected to the top of a hill by a steel cable of length \(660 \mathrm{~m}\) and diameter \(1.5 \mathrm{~cm} .\) As the gondola comes to the end of its run, it bumps into the terminal and sends a wave pulse along the cable. It is observed that it took \(17 \mathrm{~s}\) for the pulse to return, \((a)\) What is the speed of the pulse? (b) What is the tension in the cable?

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