Question

A quartz crystal of thickness 't' with Young's modulus Y' and density $\rho$ " the fundamental frequency $\nu$ can be expressed as

Answer 1

$\nu = \frac{1}{2t} \sqrt{\frac{Y}{\rho}}$

Answer 2

For a quartz crystal vibrating in its fundamental thickness mode, the thickness 't' corresponds to half a wavelength ($\lambda/2$).

So, $\lambda = 2t$.

The speed of a longitudinal wave (sound) in a solid material is given by: $v = \sqrt{\frac{Y}{\rho}}$ where Y is Young's modulus and $\rho$ is the density of the material.

The fundamental frequency ($\nu$) is related to the wave speed and wavelength by: $\nu = \frac{v}{\lambda}$

Substitute the expressions for 'v' and '$\lambda$' into the frequency equation: $\nu = \frac{\sqrt{\frac{Y}{\rho}}}{2t}$ $\nu = \frac{1}{2t} \sqrt{\frac{Y}{\rho}}$