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Natural frequency calculation

  1. Mar 9, 2014 #1
    A Cantilever beam of SiO2 of length 100μm, width 20μm and thickness of 5μm. How to calculate the resonant frequency? I know f=1/2∏√(k/m) where k=3EI/L^3 and mass = l*w*h *ρ (density) and I = w*t^3/12, but i don't get the desired answer. pl correct me.

    What changes are to be considered for the natural frequency if the cantilever is coated by a gold layer of 0.5μm uniform thickness forming a composite beam.

    thanks in advance.
     
  2. jcsd
  3. Mar 9, 2014 #2

    Simon Bridge

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    Welcome to PF;
    Please show how you did the calculation.
    What was the desired answer?

    Depends on how you model the coating - would it act a bit like putting two springs in parallel?
    Perhaps you course notes include some example of the kind of model you are expected to use?
     
  4. Mar 11, 2014 #3
    By substituting the values, I got mass = 2.2*10^-8 gm, I=2.08*10^-22 m^4, K=4.56*10^4g/s^2, f= 229KHz . But the expected freq is 82KHz. Pl Check. Thank you for the response.

    Yes! The gold coating will act like two springs in parallel.
     
  5. Mar 11, 2014 #4

    Simon Bridge

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    Since you basically just plugged numbers into a formula, and got the wrong answer, then you either got bad arithmetic, or the wrong equation.

    Off your notes post #1:$$\omega = \sqrt{\frac{k}{m}}$$... is correct - so you want to look closely at the substitutions.
    Work through the problem carefully, documenting each step, try to be consistent with your notation.
     
  6. Mar 11, 2014 #5
    sorry for the trouble, i am working for the theoretical frequency calculation of a SiO2 cantilever with gold coating. I came across a similar paper which has the result given for the resonant frequency of a polysilicon cantilever with gold coating.

    RESULT:
    The frequency of polysilicon cantilever is 9.35KHz,
    and the frequency with gold coating is 19.7KHz

    Kindly check the attachment for my calculations for the same dimensions given in that paper. My result is 8.69KHz instead of 9.35KHz. Kindly help.
     

    Attached Files:

  7. Mar 11, 2014 #6

    Simon Bridge

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    I'm not in the habit of checking people's arithmetic - that often results in me doing their work for them ;)
    If you get a different figure from someone else, i.e. in the paper, then go through their calculation carefully and see where they do something different to you. If you don't understand part of the calculation, I can help you with that.

    It can also hep to check your course notes against another source.
    http://iitg.vlab.co.in/?sub=62&brch=175&sim=1080&cnt=1
     
  8. Apr 27, 2014 #7
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