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SHM damping coefficient which envelope sign?

  1. Apr 19, 2006 #1

    dfx

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    Hi,
    I'll get right away to the point. I'm determining damping coefficients from motion sensor traces of a damped oscillator. I have gone through the standard procedure of taking natural logs of the sine wave peaks and drawing graphs etc, however what puzzles me is that the SIGN of the damping coefficient changes depending on whether u use the upper or lower envelope?

    From [tex] x = e^{-pt} x_0 Sin [/theta] t [/tex]

    Then taking logs:

    [tex] lnx = -pt + lnx_0 Sin [/theta] t [/tex]

    I asked my teacher in a hurry and he said the bottom envelope would be a reflection so the sign on the amplitude changes.... however this doesn't affect the value of the damping coefficient after taking logs.. just the value of the y-intercept! Can anyone explain, help, and show mathematically please? Any feedback appreciated... thanks very much. Cheers.

    edit: The graph I plotted was of 'lnx' against 't' to determine the value of 'p' and then further equate 'p' to 'b/2m' where 'b' is the damping coefficient. The problem is that with the upper envelope the value of 'p' - the gradient - is negative, so when you equate p = b/2m you end up with a negative value for b!
     
    Last edited: Apr 19, 2006
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  3. Apr 19, 2006 #2

    Gokul43201

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    The maxima occur at sin(wt) = 1 when wt = (4n+1)*pi/2 and the minima occur at sin(wt)=-1 at wt = (4n-1)*pi

    Plugging these in, the points on the two envelopes are :

    [tex]x_n(max) = x_0 ~ e^{-(4n+1)\pi p/2\omega} [/tex]

    and

    [tex]|x_n(min)| = x_0 ~ e^{-(4n-1)\pi p/2\omega} [/tex]

    Now take logs and plot [itex]log|x_n| [/itex] vs. n
     
    Last edited: Apr 19, 2006
  4. Apr 19, 2006 #3

    dfx

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    Hi thanks for the quick reply.

    Unfortunately this is part of a huge coursework and I have already plotted graphs of lnx against t.

    Is there no way I can get rid of the negative sign problem with my initial equation?
     
  5. Apr 19, 2006 #4

    Gokul43201

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    I"m not sure what exactly you mean by the negative sign problem, but that wouldn't be the only problem you'll have. For starters, whenever x < 0, lnx is undefined. So, for x<0, you must use ln|x|. Secondly, unless you actually use the maximal and minimal values, how are you going to get two envelopes ?
     
  6. Apr 20, 2006 #5

    dfx

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    Hi, I think you've hit the nail on the head. You said -ve x values i.e. -ve amplitudes. The problem is that the motion sensor I used does not produce a sine wave centred zero, rather it measures distances of the oscillator from the motion sensor. So what happens is that you have a maximum peak say at 40cm from the motion sensor and then a minimum at 20cm from the motion sensor if you can picture the system oscillating. So I have NO negative x values.

    And yes, I have only used minimal x values for the lower envelope and maximal x values for the upper envelope. I have extraced these maxima and minima from the motion sensor Sine wave traces.

    As for the "negative sign problem" please read the first post which I have edited to clarify as best as I can. Thank you very much for your help, I really appreciate it! :)
     
  7. Apr 20, 2006 #6

    dfx

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    I still cannot understand how you can derive a positive value for the damping coefficient.

    Say for arguments sake we had a regular sine wave centred zero, and we took the modulus of the negative amplitudes so you have ln|x|, you would essentially make the negative amplitudes positive. Thus you end up with a curve similar to the upper envelope even for the lower envelope. The problem is that this upper damping envelope curve is a NEGATIVE GRADIENT.

    And if the gradient, say, p = b/2m and p is negative, you would naturally get a negative value for 'b' the damping coefficient also!

    I'm extremely confused and frustrated as I've put in months of work and nearly 70 pages into my investigation. Please help!!
     
  8. Apr 20, 2006 #7

    dfx

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    Ok I FINALLY FIGURED IT OUT. AFTER HOURS OF THINKING.

    it's not p = b/2m, it's NEGATIVE p = b/2m. Which PERFECTLY resolves EVERYTHING!!!! thank you!!
     
  9. Apr 21, 2006 #8

    Gokul43201

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    I guess I didn't read carefully before, but I would have got p>0 (the gradient is actually given by -p, in your equation, so, if the gradient is negative, p must be positive) and using p=b/2m will give a positive value for b.
     
    Last edited: Apr 21, 2006
  10. Apr 21, 2006 #9

    dfx

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    Yes exactly. I'm glad you agree! Thanks. :)
     
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