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carlj0nes

  • one year ago

Whoever can answer this the best I will send you 10 owlbucks. Can someone please explain how we could use this differential equation to model an un-damped oscillator. (d^2 x)/(dt^2 )-10 dx/dt+34x=0 Then if we were to extend this differential equation into a forced damped oscillator, explain how you could model such an oscillator using a forced second order differential equation. The description must be commensurate to a level 5 qualification.

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  1. anonymous
    • one year ago
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    Need help @carlj0nes

  2. phi
    • one year ago
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    I believe the damping term is associated with the dx/dt term find the roots to the "characteristic equation", and examine the discriminant to classify the behavior of the system

  3. phi
    • one year ago
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    Your question is not clear to me. You can start with the physics of , for example, a spring that you pull and then release. Making some simple assumptions, you can derive a differential equation that models the behavior of the spring. You can introduce damping to model how the spring slows down. You will get an equation that looks like the one you posted.

  4. carlj0nes
    • one year ago
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    The original question was. A Damped oscillation with no external forces can be modelled by the equation I previously posted. Where x(in mm) is the amplitude of the oscillation at time t seconds. The initial amplitude of the oscillation is 2mm (i.e. when t=0) and the velocity is 0mm/s. By using the information provided above, determine the particular solution of the differential equation. Just to confirm I have already completed this part of the question but I need to do the question posted for a higher grade

  5. phi
    • one year ago
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    an undamped harmonic oscillator would have this equation \[ \frac{ d^2x }{ dt^2 }+ \cancel{-10\frac{ dx }{ dt }}+34x = 0 \]

  6. phi
    • one year ago
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    btw, are you sure it is -10 dx/dt ? that -10 leads to e^5t term which grows (as opposed to being damped) with time.

  7. carlj0nes
    • one year ago
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    This is what I have so far (d^2x)/(dt^2)-10 dx/dt +34x=0 r ^2 −10r+34=0 (r−5)^ 2 −25+34=0 (r−5) ^2 =−9 (r−5)=±√−9 =±√9 √ −1 r−5=±3i r=5±3i so x(t)=ae ^(5+3i)t +be ^(5−3i)t x(t)=e ^5t (asin(3t)+bcos(3t))= e^ 5t (ksin(3t+θ)) −e 5t ≤e 5t sin(3t+θ)≤e 5t

  8. carlj0nes
    • one year ago
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    not sure if it is correct but I am after a explanation as well

  9. phi
    • one year ago
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    I would stop at x(t)=e ^5t (asin(3t)+bcos(3t)) and use the initial conditions to find a and b at t=0, x= 2 so \[ 2 = e^{5 \cdot 0} ( a \sin(3\cdot 0) + b \cos(3\cdot 0)\\ 2= 1 \cdot(0 + b\cdot 1) \\ b=2 \]

  10. phi
    • one year ago
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    now find dx/dt (which is the velocity), so we can use the second initial condition v=0 at t=0 \[ \frac{d}{dt} x(t)= \frac{d}{dt} \left(e^{5t} (a \sin(3t)+b\cos(3t))\right) \]

  11. phi
    • one year ago
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    should replace b with 2 and dx/dt with 0 (i.e. v=0) in that, and solve for a

  12. phi
    • one year ago
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    you should get \[ \frac{d}{dt} \left(e^{5t} (a \sin(3t)+b\cos(3t))\right) = 0 \\ \left(\frac{d}{dt} e^{5t}\right) (a \sin(3t)+b\cos(3t)) + e^{5t} \frac{d}{dt} (a \sin(3t)+b\cos(3t)) = 0 \]

  13. phi
    • one year ago
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    or \[ 5 e^{5t} (a \sin(3t) + 2 \cos(3t)) + e^{5t} ( 3a \cos(3t) -6\sin(3t) ) = 0 \] now plug in t=0

  14. phi
    • one year ago
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    to "force" the oscillator, set the equation equal to some function of t \[ \frac{ d^2x }{ dt^2 }+ -10\frac{ dx }{ dt }+34x = f(t) \] for example, let f(t)= sin(t)

  15. phi
    • one year ago
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    ***A Damped oscillation with no external forces can be modelled by the equation I previously posted.*** Again, I would double check the -10 coefficient. Real oscillations grow smaller, not grow larger with time.

  16. carlj0nes
    • one year ago
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    @phi thanks for your help but can someone explain how we could use this differential equation to model an un-damped oscillator.

  17. amoodarya
    • one year ago
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    http://openstudy.com/study#/updates/558c1099e4b0c2c04901546b

  18. amoodarya
    • one year ago
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    if it was +10dx/dt was damping \[x(t)=e^{-5t}(asin(3t)+bcos(3t))\] |dw:1435827570430:dw|

  19. carlj0nes
    • one year ago
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    The question I was given said -10. But it does make sense to be +10 Maybe it was a typo. Out of interest we know what is b as that = 2 but what does a=

  20. carlj0nes
    • one year ago
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    I have seen my tutor today and he said the original is correct for damped but for undamped it does change to +10

  21. anonymous
    • one year ago
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    can I help sir?

  22. phi
    • one year ago
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    **I have seen my tutor today and he said the original is correct for damped but for undamped it does change to +10*** Can you post a screen snapshot of the original material? This make sense only if the original equation was written e.g as \[ \frac{d^2 x}{dt^2 } = -10 \frac{d x}{dt } -34x \] or the leading coefficient of the d^2/dt^2 term is negative.

  23. phi
    • one year ago
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    **Out of interest we know what is b as that = 2 but what does a= *** See the above posts where b is found, and (most!) of the work is done to find "a" Can you give it a try to find a?

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