Question

Calculate the fundamental frequency,
when a quartz crystal having thickness
0.15 cm is vibrating at resonance .given
that youngs modulus of quartz is 7.9 x
10^10 N/m^2 and density is 2650 kg /m^3.​

Answer 1

The fundamental frequency when a quartz crystal having a thickness of 0.15 cm is vibrating at resonance is 16.38 MHz

Given

Thickness of Quartz Crystal = 0.15 cm = 15 × 10⁻⁴ m

Young's modulus of quartz = 7.9 × 10¹⁰ N/m²

Density of quartz = 2650kg/m³

To Find

Fundamental Frequency

Solution

Fundamental frequency can be found using the equation $f=\frac{1}{2t}\sqrt{\frac{Y}{p}}$ where

$f$ is the fundamental frequency to be solved

$t$ is the thickness of the quartz crystal in meters = 0.15 cm = 15 × 10⁻⁴ m

$Y$ is Young's modulus of the quartz crystal = 7.9 × 10¹⁰ N/m²

$p$ is the density of the quartz crystal = 2650 kg/m³

Substituting the values, the equation becomes

$f=\frac{1}{2 X 15 X10^{-4}} \sqrt\frac{7.9 X 10^{10} }{2650}$

  = $0.003 X 5460$

  = 16.38 MHz

Hence, the fundamental frequency when a quartz crystal having a thickness of 0.15 cm is vibrating at resonance is 16.38 MHz