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Oscillation damped oscillations ? how to calculate energy after t

  1. Feb 28, 2012 #1
    1. The problem statement, all variables and given/known data

    3. A damped oscillator's amplitude dec¡eases from 8 cm to 4 cm in 20 seconds, If the intial energy of the oscillator is 64 J, what is the energy âfter 40 seconds? (Recall: E: (l/2)kA2)

    2. Relevant equations
    not sure how to approach the problem


    3. The attempt at a solution

    intia E which is 64J = k*A^2 solve for k since 64 = K * 0.08^2. k = 9375 N

    problem how do I take the damped oscillator into account for after 40 seconds

    so after 40 seconds E = 9375*A^2 ??? or am i doing it wrong totally
     
  2. jcsd
  3. Feb 29, 2012 #2

    ehild

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    Do you know the mathematical expression as the amplitude of a damped oscillator changes with time?


    ehild
     
  4. Feb 29, 2012 #3
    is it A=Ae^-t/2T where T= M/b ?? i'm not sure what the time constant is in this case ? or would i apply the forumla A = sqrt( xo^2 + (vo/w)^2) ?? in which case i have no idea what the v or the w and the xo is since distance is constantly changing
     
  5. Feb 29, 2012 #4

    ehild

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    Read the problem: The amplitude decreases from 8 cm to 4 cm in 20 seconds. The amplitude decreases exponentially, as you have written, A=Ainitiale^(-t/2T). Plug in the data: what is the time constant?

    ehild
     
  6. Feb 29, 2012 #5
    would the time constant be 20 seconds since amplitude is decreased at 20 seconds, since T = m/b how do i calculate for T if the time constant isn't 20 seconds, but I think the time constant is indeed 20 seconds since the amplitude changes at 20 seconds
     
  7. Feb 29, 2012 #6

    ehild

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    The amplitude decreased to half in 20 seconds. The energy is proportional to the amplitude. So the energy would decrease to 1/4 of the initial value in 20 seconds. In 40 seconds, the amplitude halves again. What about energy?

    If general case, when the amplitude changes from A1 to A2 in t seconds,

    [tex]A_2=A_1 e^{-\frac{t}{2T}}[/tex]

    [tex]\frac{t}{2T}=\ln(\frac{A_1}{A_2 })[/tex]

    ehild
     
  8. Feb 29, 2012 #7
    ahh i see so using the ln equation i can use the ln equation using A1 = 8 cm and A2= 4 cm, then I can figure out T ?

    then using that constant I can figure out A = (0.08m)*e^(-t/2T) ??? with that I can plug into E= KA^s ???
     
  9. Feb 29, 2012 #8
    but would the K apply since i found K using intial energy and initial amplitutde
     
  10. Feb 29, 2012 #9

    ehild

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    Yes.

    Yes, but you can use the energy directly. As the energy is proportional to the square of the amplitude, the energy also decreases exponentially
    [tex]A_2^2=A_1^2(e^{-\frac{t}{2T}})^2=A_1^2e^{-\frac{t}{T}}[/tex]
    [tex]E_2=E_1e^{-t/T}[/tex].

    E1=64J, and you know T.


    ehild
     
  11. Feb 29, 2012 #10
    so the k will still work plugging into second E=K*A^2 with the new A, and thank you did not know most of the equations you posted my prof sucks!!!! got all the hard equations
     
  12. Feb 29, 2012 #11

    ehild

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    I am glad if it cleared ...

    ehild
     
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