Find Quartz Oscillator Disk Thickness for 88.0 MHz Frequency

[ohaithar]

Homework Statement

In a quartz oscillator, used as a stable clock in electronic devices, a transverse (shear) standing sound wave is excited across the thickness d of a quartz disk and its frequency f is detected electronically. The parallel faces of the disk are unsupported and so behave as "free ends" when the sound wave reflects from them as shown in the figure.

If the oscillator is designed to operate with the first harmonic, determine the required disk thickness if 88.0 MHz. The density and shear modulus of quartz are rho = 2650 kg/m^3 and G = 2.95 *10^10 N/m^2.

Homework Equations

Possibly:
\Deltal=(1/G)(F/A)l_{0}
f=v/2l
l=(1/2)\lambda

The Attempt at a Solution

I'm not even sure where to start with this. I tried looking in the section the question came from in the book, but I couldn't find anything about mediums other than air, let alone using a shear modulus. In the table provided, the speed of sound in quartz wasn't listed, so I assume I don't need to bother figuring that out. I'm not really certain where the shear modulus and density come into play either.

(edit:) I tried \lambda=[(3.7km/s)(1000)]/88000000Hz (3.7km/s is the speed of sound in quartz according to google) and divided that by 2 to get l, but that's apparently not the correct answer...

Please help, thanks!

 

[ideasrule]

Try looking for an equation relating the speed of sound, shear modulus, and density.

Spoiler

See http://en.wikipedia.org/wiki/Shear_modulus#Waves

 

[ohaithar]

Ahhh, thank you so much! That simplifies things a lot.