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Mean Input Power & Q value , Damped Harmonic Motion

  1. Feb 16, 2016 #1

    RJLiberator

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    1. The problem statement, all variables and given/known data
    33333.JPG

    2. Relevant equations


    3. The attempt at a solution

    I'm working on part a.

    The numerical value of Q.

    I have an equation stating that Q = ω_0/ϒ.

    I dont really know what ϒ is, in other places (http://farside.ph.utexas.edu/teaching/315/Waves/node13.html) it seems like the frequency.

    But, I also do not have w_0. So perhaps this is the wrong equation or perhaps I am looking at it wrong.

    I also have: P_max = (1/2) * (Q*F_0^2)/(m*w_0)
    but I do not think this helps directly.
     
  2. jcsd
  3. Feb 16, 2016 #2

    BvU

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    Correct. What would help is an expression for P as a function of ##\omega##. Because from the picture you can see that P halves when you are ##\omega/50## away from ##\omega_0## ....

    Time to fill in the blank under "2. Relevant equations" :rolleyes:
     
  4. Feb 16, 2016 #3

    RJLiberator

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    Sometimes with foreign topics it is hard to know what is a relevant equation and what is irrelevant :p.

    P(w) = F_0^2*w_0/(2*K*Q) * [1/((w_0/w-w/w_0)^2 + 1/Q^2)]

    At w = w_0, we get it to be maximized.

    P(w) = F_0^2*w_0*Q/2K

    i'm not entirely sure how this helps, tho.
    As we do not know F_0, other then that it is held constant.
    w_0 is unknown, but you are saying that P halves when we are w/50 away from w_0. Why 50?

    If we take (1/2)*P(w) = F_0^2*w_0*Q/4K

    That doesn't seem to help.

    I am guessing that I am not making a connecting between F_0, K, and w_0?
     
  5. Feb 16, 2016 #4

    BvU

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    From the picture we see ##P(\omega_0 + \omega_0/50) = {1\over 2} P(\omega_0)##. A lot of the factors that bother you divide out when you work out this equation....and who knows, you end up with something containing ##Q## ...
     
  6. Feb 16, 2016 #5

    gneill

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    It would be instructive to look up the definition of the Q factor. There is a definition that involves bandwidth (-3 dB, or half-power points) that will be of particular interest.
     
  7. Feb 16, 2016 #6

    RJLiberator

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    Definition of Q factor : https://en.wikipedia.org/wiki/Q_factor

    So, reformatting my previous equation we see:

    2*K*P(w_0)/w_0*F_0^2 = Q

    How would the definition of Q help me here? I would think that we are looking for Q so we do not want to replace it with anything.

    Maybe this is it: P(max) = Q*F_0^2/(2m*w_0)
    If I input this in for P(w_0) then we see

    k*Q/(m*w_0^2) = Q
    But then I cancel out Q and i'm stuck.

    k/(m*w_0^2) = 1
     
  8. Feb 16, 2016 #7

    gneill

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    Sure you would! Q has a definition in terms of the bandwidth. The bandwidth is defined in terms of the half-power points. You are given the half power points on the diagram in terms of the resonant frequency ##\omega_o##. So, what's the bandwidth on your diagram?
     
  9. Feb 16, 2016 #8

    RJLiberator

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    0.04w_0 = bandwidth = Q ?
     
  10. Feb 16, 2016 #9

    gneill

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    Go back to the Wikipedia page and look at the definition of Q in terms of the bandwidth!
     
  11. Feb 16, 2016 #10

    RJLiberator

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    Aha.,

    f_0/The change in frequency = Q

    So, w_0/0.04*w_0 = 1/0.04 = 25 is the value for Q!
     
  12. Feb 16, 2016 #11

    gneill

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    Huzzah! Yes!
     
  13. Feb 16, 2016 #12

    RJLiberator

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    Solved ! Thank you kindly.
     
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