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Chapter 16: Problem 5

(a) In a liquid with density 1300 \(\mathrm{kg} / \mathrm{m}^{3}\) , longitudinal waves with frequency 400 \(\mathrm{Hz}\) are found to have wavelength 8.00 \(\mathrm{m}\) . Calculate the bulk modulus of the liquid. (b) A metal bar with a length of 1.50 \(\mathrm{m}\) has density 6400 \(\mathrm{kg} / \mathrm{m}^{3}\) . Longitudinal sound waves take \(3.90 \times 10^{-4} \mathrm{s}\) to travel from one end of the bar to the other. What is Young's modulus for this metal?

Short Answer

Expert verified
(a) Bulk modulus: \(1.33 \times 10^{10} \, \text{Pa}\); (b) Young's modulus: \(9.45 \times 10^{10} \, \text{Pa}\).

Step by step solution

01

Determine Wave Speed in Liquid

To find the speed of the longitudinal wave, use the wave equation: \[ v = f \lambda \]where \( v \) is the wave speed, \( f \) is the frequency, and \( \lambda \) is the wavelength. Given that \( f = 400 \) Hz and \( \lambda = 8.00 \) m, calculate \( v \):\[ v = 400 \times 8.00 = 3200 \, \text{m/s} \]
02

Calculate Bulk Modulus of Liquid

The bulk modulus \( B \) of a liquid is given by the formula:\[ B = \rho v^2 \]where \( \rho = 1300 \, \text{kg/m}^3 \) is the density. Substituting the known values:\[ B = 1300 \times (3200)^2 = 1300 \times 10240000 = 1.33 \times 10^{10} \, \text{Pa} \]
03

Determine Wave Speed in Metal Bar

Use the relation \( v = \frac{L}{t} \) for the speed of the wave in the bar, where \( L = 1.50 \, \text{m} \) is the length and \( t = 3.90 \times 10^{-4} \, \,\text{s} \) is the time taken. Calculate \( v \):\[ v = \frac{1.50}{3.90 \times 10^{-4}} = 3846.15 \, \text{m/s} \]
04

Calculate Young's Modulus of Metal

Young's modulus \( E \) is related to the speed of sound in a solid by the formula:\[ E = \rho v^2 \]where \( \rho = 6400 \, \text{kg/m}^3 \) is the density. Using the speed from Step 3:\[ E = 6400 \times (3846.15)^2 = 6400 \times 14771779.22 = 9.45 \times 10^{10} \, \text{Pa} \]
05

Conclusion

The bulk modulus of the liquid is \( 1.33 \times 10^{10} \, \text{Pa} \) and the Young's modulus of the metal bar is \( 9.45 \times 10^{10} \, \text{Pa} \).

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Key Concepts

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

Bulk Modulus
Bulk modulus is a fascinating concept in wave mechanics. It measures how incompressible a material is when under uniform pressure. Think of it as the stiffness of a fluid. The higher the bulk modulus, the less compressible the fluid. You can imagine trying to squeeze a balloon filled with water versus one filled with air. The balloon with water is tougher to compress, indicating water's high bulk modulus.

To calculate the bulk modulus, we use the formula:
  • \( B = \rho v^2 \)
Here, \( \rho \) stands for the fluid's density, and \( v \) is the speed of sound in the fluid. The formula highlights the strong relationship between density, wave speed, and stiffness. It ultimately helps us understand how sound travels through different mediums. Knowing the bulk modulus enables engineers and scientists to predict how materials behave under pressure, aiding in designing better structures and systems for a variety of applications.
Young's Modulus
Young's modulus is a critical concept when studying the mechanical properties of solids. It defines the relationship between stress and strain in a material undergoing elongation or compression. In simpler terms, it describes how much a material will stretch or compress under a certain amount of force. This concept is crucial for engineers and builders to ensure materials can withstand the forces they face without breaking.

The formula for calculating Young's modulus \( E \) is:
  • \( E = \rho v^2 \)
This equation shows that Young's modulus depends on the material's density \( \rho \) and the speed of longitudinal sound waves \( v \) through it. These properties are indicative of how rigid a material is. High Young's modulus means the material is stiff and resilient. Low Young's modulus indicates that the material is more flexible. Testing for Young's modulus helps in choosing the right materials for specific applications. Ensuring safety and efficiency in structures, such as buildings and bridges, relies heavily on understanding this property.
Longitudinal Waves
Longitudinal waves are a crucial aspect of wave mechanics. These waves involve particle motion along the direction of wave propagation. Imagine a slinky, when you push and pull it back and forth, it creates waves that move in a direction parallel to the movement.

In the context of materials like solids, liquids, or gases, these waves transport energy and information. Sound waves moving through air are a common example of longitudinal waves. When calculating their properties, we often use the relationship:
  • \( v = f \lambda \)
where \( v \) is the wave's speed, \( f \) is the frequency, and \( \lambda \) is the wavelength. Understanding how these components interact provides insights into the behavior of sound and other types of waves. This knowledge is essential for various fields, from designing acoustically efficient spaces to developing new communication technologies.

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Most popular questions from this chapter

The Sacramento City Council recently adopted a law to reduce the allowed sound intensity level of the much despised leaf blowers from their current level of about 95 \(\mathrm{dB}\) to 70 \(\mathrm{dB}\) . With the new law, what is the ratio of the new allowed intensity to the previously allowed intensity?

A long tube contains air at a pressure of 1.00 atm and a temperature of \(77.0^{\circ} \mathrm{C}\) . The tube is open at one end and closed at the other by a movable piston. A tuning fork near the open end is vibrating with a frequency of 500 \(\mathrm{Hz}\) . Resonance is produced when the piston is at distances \(18.0,55.5,\) and 93.0 \(\mathrm{cm}\) from the open end. (a) From these measurements, what is the speed of sound in alr at \(77.0^{\circ} \mathrm{C} ?\) (b) From the result of part (a), what is the value of \(\gamma ?\) (c) These data show that a displacement antinode is slightly outside of the open end of the tube. How far outside is it?

A woman stands at rest in front of a large, smooth wall. She holds a vibrating tuning fork of frequency \(f_{0}\) directly in front of her (between her and the wall). (a) The woman now runs toward the wall with speed \(v_{\mathrm{W}}\) . She detects beats due to the interference between the sound waves reaching her directly from the fork and those reaching her after being reflected from the wall. How many beats per second will she detect? (Note: If the beat frequency is too large, the woman may have to use some instrumentation other than her ears to detect and count the beats.) (b) If the woman instead runs away from the wall, holding the tuning fork at her back so it is between her and the wall, how many beats per second will she detect?

Singing in the Showen. A pipe closed at both ends can have standing waves inside of it, but you normally don't hear them because little of the sound can get out. But you can hear them if you are inside the pipe, such as someone singing in the shower. (a) Show that the wavelengths of standing waves in a pipe of length \(L\) that is closed at both ends are \(\lambda_{n}=2 L / n\) and the frequencies are given by \(f_{n}=n v / 2 L=n f_{1},\) where \(n=1,2,3, \ldots\) (b) Modeling it as a pipe, find the frequency of the fundamental and the first two overtones for a shower 2.50 \(\mathrm{m}\) tall. Are these frequencies audible?

A railroad train is traveling at 30.0 \(\mathrm{m} / \mathrm{s}\) in still air. The frequency of the note cmitted by the train whistle is 262 \(\mathrm{Hz}\) . What frequency is heard by a passenger on a train moving in the opposite direction to the first at 18.0 \(\mathrm{m} / \mathrm{s}\) and (a) approaching the first; and (b) receding from the first?
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