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Relationship between size and frequency of an oscillator

  1. Apr 3, 2015 #1
    I've always just accepted that as you scale down a mechanical system the frequency and Q factor both increase. For example, this Wikipedia page simply says that "Small bells ring at higher frequencies than large bells".

    But for a driven damped harmonic oscillator, what is the exact relationship between size, resonant frequency, and Q factor?

    http://www.ieee-uffc.org/frequency-control/learning/pdf/Kaajakari-MEMS_Resonators_v2b.pdf [Broken]states that a tuning fork scales:

    ##\omega_r \propto \frac{1}{size}##

    But how about a mass-spring-damper? Will this also scale the same as the tuning fork?

    I'm trying to find out how the frequency and Q factor scale as you scale down MEMS oscillators, and the PDF above states that a MEMS oscillator can be modelled as a mass-spring-damper.

    ##m \ddot{x} + c \dot{x} + kx = F_{input}##

    If the oscillator can be modelled as above, can we use the same scaling law as the tuning fork? If we can, how can this be proved mathematically? If not, how will it scale? I'm guessing it all depends on how ##m##, ##c##, and ##k## scale with size. If this is complex, what is a fair assumption to simplify the problem? If the maths is still too complex, are there any rules of thumb of how mass-spring-damper frequency scales with size? Is there anything I need to watch out for as the MEMS resonator can be scaled down to micro and nano scale sizes?
     
    Last edited by a moderator: May 7, 2017
  2. jcsd
  3. Apr 3, 2015 #2
    I've almost found the answer. http://www.ieee-uffc.org/frequency-control/learning/pdf/Piazza-MEMS_Resonators_for_Frequency_v1.pdf [Broken]says the answer is:

    ##f_0 \propto \frac{T}{L^2} \sqrt{\frac{E}{\rho}}##

    I now just need to know what the symbols are. Any ideas?
     
    Last edited by a moderator: May 7, 2017
  4. Apr 8, 2015 #3
    from normal bell, frequency tuner, and other oscillations it is all f = sqrt (k/m), as the resonator mass decreases for a constant stiffness the frequency increases. or as stiffness increases the frequency increases.

    the equation above takes into account the geometry and young's modulus of the material - I think its material flexural vibration modes. take beam of material and you trying to calculate the stiffness and mass. the not an equals sign is to indicate that there a multiple modes possible.
    so basically a L length of beam, with a E young's modulus and a density rho, T is the beam's thickness.
    the first mode is going to be the greatest usually and the lowest frequency - simple bending.
     
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