# Frequency of sound wave

1. Oct 5, 2009

### BringBackF77

1. The problem statement, all variables and given/known data

"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.

2. Relevant equations

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

3. 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: Oct 5, 2009
2. Oct 5, 2009

### kuruman

Highest frequency is shortest wavelength. Can the wavelength be shorter than the interatomic distance?

3. Oct 15, 2009

### BringBackF77

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!