Highest frequency of sound waves in a thin aluminium rod

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SUMMARY

The discussion centers on estimating the highest frequency and smallest wavelength of sound waves in aluminium, utilizing its mass density (2.7 x 103 kg/m3), Young's modulus (6 x 1010 N/m2), and atomic weight (27). The derived wave speed (v) is calculated as v = √(Y/ρ), resulting in a value of approximately 4714.045 m/s. The maximum frequency is determined by substituting the minimum wavelength, which corresponds to the interatomic spacing of aluminium atoms, into the wave equation. The discussion emphasizes the importance of calculating the number of aluminium atoms per cubic meter to ascertain atomic spacing.

PREREQUISITES
  • Understanding of wave equations and their applications in solid materials.
  • Familiarity with Young's modulus and its significance in material science.
  • Knowledge of mass density and its role in wave propagation.
  • Basic concepts of atomic structure and interatomic spacing in metals.
NEXT STEPS
  • Calculate the interatomic spacing in aluminium using its atomic weight and density.
  • Explore the relationship between frequency, wavelength, and wave speed in different materials.
  • Investigate boundary conditions in wave equations for solid materials.
  • Learn about the implications of atomic structure on sound wave propagation in metals.
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Students in physics and materials science, engineers working with acoustic properties of metals, and anyone interested in the fundamental principles of wave mechanics in solid materials.

aednat
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Homework Statement


Estimate the highest possible frequency (in Hertz) and the smallest possible wavelength, of a sound wave in aluminium due to the discrete atomic structure of this material. The mass density, Young's modulus, and atomic weight of aluminium are 2.7x103kg m-3, 6x1010 N m-2, and 27 respectively.

Homework Equations


Second partial of Ψ(x,t) WRT t = second partial of Ψ(x,t) WRT x multiplied by (Young's modulus / mass density)

The Attempt at a Solution


I've obtained the expression v = sqrt(Y/p), but v=frequency x wavelength
so v = 4714.045 = frequency x wavelength.
I understand that the minimum wavelength occurs at wavelength = the interatomic spacing between the aluminium atoms, but I am unsure as to how to obtain an expression for it's value.

The maximum frequency occurs at the minimum wavelength so I can just algebraically sub in the min wavelength to find the max frequency.

I'm assuming I have to use the atomic weight of aluminium somewhere in the question, but I can't seem to find the next step.

I have a feeling I may have strayed in the wrong direction ; perhaps I could use some boundary conditions and solve the wave equation? I don't know if I'm oversimplifying the problem.

Any help is massively appreciated.
 
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aednat said:
The maximum frequency occurs at the minimum wavelength so I can just algebraically sub in the min wavelength to find the max frequency.
Right.

You have 2700kg/m3 and every atom has a weight of 27 u. How many aluminium atoms do you expect per m3? What does that tell you about the distances between those atoms?
 

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