In a quartz oscillator, used as a stable clock in electronic devices, a transverse (shear) standing sound wave is excited across the thickness
圖片參考：http://session.masteringphysics.com/render?tex=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 90.0
圖片參考：http://imgcld.yimg.com/8/n/HA00021001/o/700912100025313873383930.jpg
sound wave
2009-12-10 12:00 pm
回答 (2)
2009-12-10 5:38 pm
✔ 最佳答案
Velocity of sound in quartz = square-root[2.95x10^10/2650] = 3336 m/sWavelength of sound in quartz at frequency of 90 MHz = 3336/90x10^6 m = 3.71x10^-5 m
For the fundamental frequency, d =(1/2) x wavelength = 3.71x10^-5/2 m = 1.85 x10^-5 m
2009-12-10 5:27 pm
Sound velocity in the quartz = G/p = 1.113 x 109 m/s
Wavelength = 1.113 x 109/(9 x 107) = 0.1237 m
In fundamental mode, thickness should be half of the wavelength = 0.0618 m or 6.18 cm
Wavelength = 1.113 x 109/(9 x 107) = 0.1237 m
In fundamental mode, thickness should be half of the wavelength = 0.0618 m or 6.18 cm
參考: Myself
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