Find Quartz Oscillator Disk Thickness for 88.0 MHz Frequency

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In summary, the conversation discusses the use of a quartz oscillator as a stable clock in electronic devices. It explains how a transverse standing sound wave is excited across the thickness of a quartz disk and its frequency is detected electronically. The conversation then poses a problem of determining the required disk thickness for the oscillator to operate at a frequency of 88.0 MHz using the first harmonic. The solution involves using equations that relate the speed of sound, shear modulus, and density of quartz.
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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:
[tex]\Delta[/tex][tex]l=(1/G)(F/A)l_{0}[/tex]
[tex]f=v/2l[/tex]
[tex]l=(1/2)\lambda[/tex]

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

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

Related to Find Quartz Oscillator Disk Thickness for 88.0 MHz Frequency

1. How do I calculate the thickness of a quartz oscillator disk for a specific frequency?

To calculate the thickness of a quartz oscillator disk for a specific frequency, you can use the formula: d = c / (2 * f * N), where d is the thickness, c is the speed of light, f is the frequency, and N is the overtone number (typically 3 or 5 for quartz oscillators).

2. Why is the thickness of a quartz oscillator disk important for a specific frequency?

The thickness of a quartz oscillator disk is important because it determines the frequency at which the oscillator will vibrate. This is due to the piezoelectric effect of quartz, where an electric field applied to the crystal causes it to vibrate at a specific frequency.

3. How do I know if a quartz oscillator disk is suitable for a specific frequency?

You can determine if a quartz oscillator disk is suitable for a specific frequency by calculating the required thickness using the formula mentioned in question 1. Then, you can compare this thickness to the actual thickness of the disk. If they are similar, then the disk is suitable for the frequency you are looking for.

4. What happens if the quartz oscillator disk thickness is not accurate for the desired frequency?

If the quartz oscillator disk thickness is not accurate for the desired frequency, the oscillator will not vibrate at the desired frequency. This can result in inaccurate measurements or malfunction of electronic devices that rely on the oscillator for timing.

5. Can the thickness of a quartz oscillator disk be adjusted for a specific frequency?

Yes, the thickness of a quartz oscillator disk can be adjusted by either physically grinding the disk to the desired thickness or by using a trimmer capacitor to fine-tune the frequency. However, this process can be time-consuming and may require the expertise of a skilled technician.

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