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What is the energy loss of damped oscillator

  1. Nov 1, 2014 #1

    grandpa2390

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    1. The problem statement, all variables and given/known data
    what is the energy loss of the damped oscillator.

    2. Relevant equations
    x(t) = A*e^(-Bt)*cos(w1*t)
    T = 1/2 mv^2

    3. The attempt at a solution

    To solve for an undamped oscillator, I took the derivative of the equation of motion x(t) and plugged the amplitude into 1/2 mv^2 equation and that worked. Now a damping effect has been added. I want to do the same thing. Take the derivative of the equation above and plug its amplitude into the kinetic energy formula. Subtract to get the energy loss. I did this and I received my answer, but I can't tell if it is reasonable.
    if my mass is on a spring and I pull it back and let it go, its max velocity will be at pi/2. so when I take my derivative and I get a dx/dt = sine + cosine both multiplied by constants. the cosine term will be zero and the sine term will be 1*maxvelocity. which I can plug into my kinetic energy formula. and then subtract to get the energy loss

    Is this a reasonable approach? I couldn't find anything about this, so I am nervous.
     
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  3. Nov 1, 2014 #2

    Orodruin

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    The undamped oscillator does not lose energy, it just alternates between containing its energy as potential and kinetic. Your question is also not very precise and it is not clear what you are after, you should make sure to pose the question word by word as it was formulated in your problem. When you write your solution, it also helps a lot if you do not just describe it in words but show us your actual computations.
     
  4. Nov 1, 2014 #3

    grandpa2390

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    obviously, but my question was about damped oscillations, not undamped.

    No additional information should be necessary. I apologize if it is, but this is all I can give you.


    or how about this? derive a formula to solve for kinetic energy of a damped oscillator if you know w, B, A, and t.
    I have a formula for motion of a damped oscillator (see above). can I plug the first derivative of that equation into the kinetic energy formula and receive the correct kinetic energy at any point in time t?
     
    Last edited: Nov 1, 2014
  5. Nov 1, 2014 #4

    Orodruin

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    Yet you were using the undamped oscillator in your solution.

    For example, "Energy loss" is ambiguous. It could be taken to mean either the rate of energy loss or the total energy loss.

    Regarding the kinetic energy: Yes. The kinetic energy is mv^2/2 and v is the time derivative of the position. In your case, you have written down an under-damped oscillator, which swings back and forth. You can compute the energy loss between two passages of the equilibrium point by computing the kinetic energy there as you have done. In general, oscillators may be critically or over damped. In such a situation, it will never pass the equilibrium point.
     
  6. Nov 1, 2014 #5

    grandpa2390

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    yes because if you read my post, you see that I did in fact mention that the problem originally was undamped, and therefore required an undamped solution. then a damping effect was added, and now I am trying to solve for kinetic energy with the added damping effect

    surely you know how these problems work. you start off with an oscillator and solve it as if it were undamped, then you solve it again as if it has a damping effect.
    total energy loss because I am analyzing between two points in time.
    thankyou
     
  7. Nov 1, 2014 #6

    Orodruin

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    Yes, this is why I noted that your original problem statement is ambiguous. My interpretation of "energy loss" would be the rate of energy loss, not the total energy loss.
     
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