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In a liquid with density 1300 kg/m3, longitudinal waves with frequency 400 Hz are found to have wavelength 8.00 m. Calculate the bulk modulus of the liquid. (b) A metal bar with a length of 1.50 m has density 6400 kg/m3. Longitudinal sound waves take 3.90 \(\times\) 10\(^{-4}\) s to travel from one end of the bar to the other. What is Young's modulus for this metal?

Short Answer

Expert verified
Bulk modulus of the liquid: \(1.3312 \times 10^7 \) Pa. Young's modulus of the metal: \(9.436 \times 10^{10} \) Pa.

Step by step solution

01

Determine the velocity of sound in the liquid

The velocity of sound in a medium is given by the formula \( v = f \cdot \lambda \), where \( f \) is the frequency and \( \lambda \) is the wavelength. For the liquid, the frequency \( f = 400 \) Hz and the wavelength \( \lambda = 8.00 \) m. Substitute these values into the equation:\[v = 400 \text{ Hz} \times 8.00 \text{ m} = 3200 \text{ m/s}\]
02

Calculate the bulk modulus of the liquid

The bulk modulus \( B \) can be expressed using the formula:\[ B = \rho \cdot v^2 \]where \( \rho \) is the density of the liquid and \( v \) is the velocity of sound in the liquid. We've calculated \( v = 3200 \) m/s and the given density \( \rho = 1300 \) kg/m\(^3\). Now, substitute these values:\[B = 1300 \text{ kg/m}^3 \times (3200 \text{ m/s})^2 = 1.3312 \times 10^7 \text{ Pa}\]
03

Determine the velocity of sound in the metal bar

The velocity of sound in an object is given by the formula \( v = \frac{d}{t} \), where \( d \) is the distance the sound travels (the length of the bar) and \( t \) is the time taken. Here, \( d = 1.50 \) m and \( t = 3.90 \times 10^{-4} \) s. Substitute these values:\[v = \frac{1.50 \text{ m}}{3.90 \times 10^{-4} \text{ s}} = 3846.15 \text{ m/s}\]
04

Calculate Young's modulus for the metal

Young's modulus \( E \) is related to the velocity of sound in a solid by the formula:\[ E = \rho \cdot v^2 \]where \( \rho \) is the density of the metal and \( v \) is the velocity of the sound in the metal. We've calculated \( v = 3846.15 \) m/s and the given density \( \rho = 6400 \) kg/m\(^3\). Substitute these values:\[E = 6400 \text{ kg/m}^3 \times (3846.15 \text{ m/s})^2 = 9.436 \times 10^{10} \text{ Pa}\]

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Key Concepts

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

Longitudinal Waves
Longitudinal waves are a type of wave where the particle displacement is parallel to the direction of wave propagation. Such waves consist of compressions and rarefactions, much like a slinky moving back and forth as it expands and contracts. A common example of longitudinal waves is sound traveling through air, liquids, or solids.
This type of wave is crucial for understanding the behavior of sound in different media, like how sound travels through a metal bar or a liquid in the context of the given problem. When you know the frequency and the wavelength of a longitudinal wave, you can determine the velocity of sound in that medium using the formula: \( v = f \cdot \lambda \), where \( v \) is the velocity, \( f \) is the frequency, and \( \lambda \) is the wavelength. This relationship helps in evaluating the characteristics of the medium, such as density and compressibility, further connecting to concepts like the Bulk Modulus.
Young's Modulus
Young's Modulus is a measure of the stiffness of a solid material. This elastic modulus is crucial as it quantifies the ability of a material to resist deformation under tensile stress. In the context of the problem, calculating Young's Modulus involves understanding how sound speed and material density relate in a solid object, such as the metal bar.
To find Young's Modulus \( E \), we use the formula:\[ E = \rho \cdot v^2 \]Here, \( \rho \) is the density of the material, and \( v \) is the velocity of sound in the material. Young's Modulus not only reveals the inherent material rigidity but also connects to other physical properties like Poisson’s ratio, which further describes the behavior of materials under various stresses.
  • It helps engineers and designers choose materials with appropriate stiffness.
  • It is fundamental for calculations in mechanical and structural engineering.
Sound Velocity in Medium
The velocity of sound in a medium plays a critical role in understanding how quickly sound energy is transmitted through different materials. This velocity depends on both the medium's elastic properties and its density.
In the problem, sound velocity was calculated using known parameters of frequency and wavelength for a liquid and considering the length and time it takes for sound to travel through a metal bar. This reinforces the formula for wave speed, \(v = f \cdot \lambda \) for liquid, and \( v = \frac{d}{t} \) for solids.
  • The velocity of sound is higher in solids than in liquids and gases, due to tighter molecule packing.
  • Understanding sound velocity aids in determining crucial properties like Bulk Modulus and Young's Modulus.
  • It is essential for applications in acoustics and materials science, contributing to advances in technology and industry.

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Most popular questions from this chapter

Many opera singers (and some pop singers) have a range of about 2\(\frac{1}{2}\) octaves or even greater. Suppose a soprano's range extends from A below middle C (frequency 220 Hz) up to E-flat above high C (frequency 1244 Hz). Although the vocal tract is quite complicated, we can model it as a resonating air column, like an organ pipe, that is open at the top and closed at the bottom. The column extends from the mouth down to the diaphragm in the chest cavity, and we can also assume that the lowest note is the fundamental. How long is this column of air if \(v =\) 354 m/s? Does your result seem reasonable, on the basis of observations of your own body?

(a) A sound source producing 1.00-kHz waves moves toward a stationary listener at one-half the speed of sound. What frequency will the listener hear? (b) Suppose instead that the source is stationary and the listener moves toward the source at one-half the speed of sound. What frequency does the listener hear? How does your answer compare to that in part (a)? Explain on physical grounds why the two answers differ.

A turntable 1.50 m in diameter rotates at 75 rpm. Two speakers, each giving off sound of wavelength 31.3 cm, are attached to the rim of the table at opposite ends of a diameter. A listener stands in front of the turntable. (a) What is the greatest beat frequency the listener will receive from this system? (b) Will the listener be able to distinguish individual beats?

You have a stopped pipe of adjustable length close to a taut 62.0-cm, 7.25-g wire under a tension of 4110 N. You want to adjust the length of the pipe so that, when it produces sound at its fundamental frequency, this sound causes the wire to vibrate in its second \(overtone\) with very large amplitude. How long should the pipe be?

You live on a busy street, but as a music lover, you want to reduce the traffic noise. (a) If you install special soundreflecting windows that reduce the sound intensity level (in dB) by 30 dB, by what fraction have you lowered the sound intensity (in W\(/\)m\(^2\))? (b) If, instead, you reduce the intensity by half, what change (in dB) do you make in the sound intensity level?

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