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Energy of a Forced, Damped Oscilator

  1. Apr 4, 2015 #1
    1. The problem statement, all variables and given/known data
    A forced damped oscilator of mass ##m## has a displacement varying with time of ##x=Asin(\omega t) ## The restive force is ## -bv##. For a driving frequency ##\omega## that is less than the natural frequency ## \omega_{0}##, sketch graphs of potential energy, kinetic energy and total energy of the oscillator over the cycle.

    2. Relevant equations


    3. The attempt at a solution
    From the equation given, and the definitions of kinetic and potential energy for an oscillator:
    (1) Kinetic Energy:
    $$ E_{K}=\frac{1}{2}m\dot{x}^{2} = \frac{1}{2}mA^{2}\omega^{2}cos(\omega t) $$

    (2) Potential Energy,
    $$ E_{P}=\frac{1}{2}kx^{2}=\frac{1}{2}mA^{2}sin^{2}(\omega t) $$

    These are simple and easy to plot. My issue comes when thinking about the total energy:

    (3) Total Energy:
    $$ E_{T}= E_{K}+E_{P} +E_{D} $$
    ##E_{D} ## is the energy lost due to the damping force, ##F_{D}##, where ##F_{D}=-b\dot{x}##.

    At some time ##t##, then :

    $$ E_{D} = \int_{0}^{x(t)} F_{D} \ dx = \int^_{0}^{x(t)} \omega \sqrt{1-x^{2}} \ dx $$
    This can be evaluated by making the substitution ##x=sin(u) \implies dx=cos(u) du ##.
    So:
    $$ E_{D} = \int_{0}^{sin(x(t))} \omega cos^{2}u \ du = \bigg|_{0}^{sin(x(t)} \frac{sin(2u)}{4} +\frac{u}{2} $$

    So:

    $$E_{T}=\frac{1}{2}mA^{2}\bigg(sin^{2}(\omega t)+\omega^{2}cos(\omega t)\bigg)-bAsin(\omega t) $$

    Is this the right approach though? The reason I am confused is that there will be points in the motion where the damping force will do work with the system - whereas it should be at all times against it?

    Thanks!
     
    Last edited: Apr 4, 2015
  2. jcsd
  3. Apr 4, 2015 #2
    PLEASE IGNORE ABOVE - OLDER VERSION WAS COPIED IN BY MISTAKE! APOLOGIES.
     
  4. Apr 5, 2015 #3

    vela

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    The units don't work out in your expression for potential energy. Also, don't you also need to take into account the work done by the applied force? It's continually adding energy to the system as the frictional force dissipates it.
     
  5. Apr 6, 2015 #4
    Ah, I think I understand better now - can we say that if it's in a steady state that the loss of energy due to friction is equal to the energy which is input into the system by the driver?

    In that case, we just get back to the total energy being ## \frac{1}{2}m\omega^{2}A^{2} ##. ?

    By the way, can you tell me what the significance of ## \omega ## being less than ## \omega_{0} ## is here? I don't really understand why it matters. Thanks!
     
  6. Apr 6, 2015 #5

    vela

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    On average, it's true that the energy input is equal to the energy lost, but the total energy varies with time. I think the problem is asking you to calculate the energy as a function of time.
     
  7. Apr 6, 2015 #6
    Oh, oops .The expression for the potential energy is rubbish. Sorry. It should be:

    $$ E_{U} = \frac{1}{2}kA^{2}sin^{2}\omega t =\frac{1}{2}m\omega_{0}^{2}A^{2}sin^{2}\omega t $$

    After that, I went through it again:

    I get that the kinetic energy plus the potential energy: ## E_{K}+E_{U} ## is:
    $$ E_{K}+E_{U}= \frac{1}{2}mA^{2}(\omega^{2}sin^{2}\omega t+\omega_{0}^{2}cos^{2}\omega t) =\frac{1}{2}A^{2}m[\omega^{2}+(\omega_{0}^{2}-\omega^{2})cos^{2}\omega t)]$$

    Using the argument that ## P=F\dot{x} ##, I then say that the net energy due to the driving force ##F## and the damping force ##F_{D}## is:
    $$ E_{N} = \int_{0}^{t} F_{D}\dot{x} - F\dot{x} \ \text{dt'} \ $$
    We know that ##F ## is going to be of the form:
    $$ F=F_{0}sin(\omega t)$$
    ##F_{D}## is given by ##F_{D}=-b\dot{x}=-b\omega A cos\omega t \implies F_{D}\dot{x} = -b \omega^{2}A^{2}cos^{2}\omega t ## So:
    $$ E_{N} = \int_{0}^{t} -b\omega^{2}A^{2}\bigg(\frac{cos2\omega t' +1}{2} \bigg) + \omega F_{0}cos2\omega t' \ \text{dt'} $$
    $$ \ \therefore E_{N}= \frac{-b\omega^{2}A^{2}}{2}t+\frac{sin(2\omega t)\omega}{2}\bigg(\frac{F_{0}-b\omega A^{2}}{2} \bigg) $$

    So I suppose the total energy ##E_{T} ## is then :

    $$ E_{T}=E_{N}+E_{K}+E_{U} =\frac{-b\omega^{2}A^{2}}{2}t+\frac{sin(2\omega t)\omega}{2}\bigg(\frac{F_{0}-b\omega A^{2}}{2} \bigg)+\frac{1}{2}A^{2}m[\omega^{2}+(\omega_{0}^{2}-\omega^{2})cos^{2}\omega t)] $$

    But this expression seems very unhelpful in terms of sketching the graph - what should I do? :P
     
  8. Apr 6, 2015 #7

    vela

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    There's a phase shift between the forcing term F and the displacement x.
     
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