What Is the Maximum Frequency of Sound in Aluminium?

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SUMMARY

The maximum frequency of sound in aluminium can be calculated using the formula f = (k / 2π) * √(Y/ρ), where Y is Young's modulus (6x1010 N m-2) and ρ is the mass density (2.7x103 kg m-3). The discussion emphasizes that the highest frequency corresponds to the shortest wavelength, which raises the question of whether the wavelength can be shorter than the interatomic distance. The user initially struggled with calculating the wave number k but later recognized the simplicity of the approach needed to solve the problem.

PREREQUISITES
  • Understanding of wave equations in physics
  • Familiarity with Young's modulus and mass density concepts
  • Knowledge of sound wave properties
  • Basic proficiency in mathematical manipulation of equations
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  • Research the calculation of wave number k in solid materials
  • Explore the relationship between frequency and wavelength in sound waves
  • Study the implications of atomic structure on sound propagation in materials
  • Learn about the limitations of sound wave frequencies in different media
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Students studying physics, particularly those focusing on wave mechanics, material science enthusiasts, and anyone interested in the acoustic properties of metals like aluminium.

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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



Assuming the mode will follow the form

Ψ(x,t) = Acos(kx)cos(ωt - φ)

then the second partial WRT t will be

Ψ''(x,t) = -ω2Acos(kx)cos(ωt - φ)

and the second partial WRT x will be

Ψ''(x,t) = -k2Acos(kx)cos(ωt - φ)

Plugging into wave equation I get

2Acos(kx)cos(ωt - φ) = -c2k2Acos(kx)cos(ωt - φ)

--> ω2 = κ2(Y/ρ)

--> ω = k(Y/ρ)1/2

--> 2\pi f = k(Y/ρ)1/2

--> f = \frac{k}{2\pi} \sqrt{\frac{Y}{\rho}}

Have no clue where to go from here. This may not even be the way to go about doing it. I guess I technically have the Young's modulus and mass density for the problem but I do not know how to calculate k, and don't understand how this system could vary in frequency to find the highest possible one. Any help would be appreciated, thanks. (Sorry, that I suck a latex btw)
 
Last edited:
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Highest frequency is shortest wavelength. Can the wavelength be shorter than the interatomic distance?
 
kuruman said:
Highest frequency is shortest wavelength. Can the wavelength be shorter than the interatomic distance?

Whoops, forgot to reply. Yes, thank you, I understand where I was going wrong now, it was much simpler than I tried to make out to be. Thanks again!
 

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