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Damped Oscillator equation - Energy

  1. Dec 8, 2012 #1
    the damped oscillator equation:

    (m)y''(t) + (v)y'(t) +(k)y(t)=0

    Show that the energy of the system given by

    E=(1/2)mx'² + (1/2)kx²

    satisfies:

    dE/dt = -mvx'


    i have gone through this several time simply differentiating the expression for E wrt and i end up with

    dE/dt = x'(-vx')

    im at a brick wall. Am i doing something wrong? Any help is much appreciated! Thanks
     
  2. jcsd
  3. Dec 8, 2012 #2

    Dick

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    There is some sort of problem with the equation you have been asked to prove. A damped oscillator is always losing energy. Your solution shows that is true. The given solution would say the oscillator is sometimes gaining energy if the sign of x' is correct. I don't think that's correct.
     
  4. Dec 9, 2012 #3

    rude man

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    You can't just differentiate E the way you did and prove the theorem. You need to incorporate the basic diff eq representing a damped spring-mass system. The expression for E represents ANY spring-mass system, damped or not, linear or not, etc.

    The only thing I can think of is to solve the diff eq (it's a simple 2nd order one with constant coeff). Apply an initial condition to x = x0. Derive the solution x(t) and then substitute in E, take dE/dt and there you are.

    (BTW why is y used in the diff eq and x in E?)

    Dick's comment is well taken! Not only his, but I noticed the dimensions don't make sense. dE/dt has dimensions of FLT-1 whereas -mvx' has dimensions of MF where
    M = mass
    F = force = MLT-2
    L = length
    T = time.

    Thus -mvx' has the wrong dimensions to be dE/dt.
     
    Last edited: Dec 9, 2012
  5. Dec 9, 2012 #4

    Dick

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    Paddyod1509 did substitute the differential equation into dE/dt to get his answer. The given answer can't be right. -vx'^2 has the correct units of J/s. -mvx' doesn't.
     
  6. Dec 9, 2012 #5

    rude man

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    OK, I was misled by his wording.
     
  7. Dec 9, 2012 #6

    Dick

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    Well, he didn't actually say that that's what he did. But once you form dE/dt it's the obvious way to get to -vx'^2.
     
  8. Dec 9, 2012 #7

    rude man

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    So did he solve the d.e. for x(t) and then substitute in E, or what?
     
  9. Dec 9, 2012 #8

    Dick

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    No. He found dE/dt=mx'x''+kxx'=x'(mx''+kx). The DE then tells you mx''+kx=(-vx'). You don't need to actually solve the DE.
     
  10. Dec 9, 2012 #9

    rude man

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    Thanks. Good job.
     
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