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Solution of damped oscillation D.E.

  1. Jul 29, 2012 #1
    d2x + 2Kdx + w0 2 x(t) = 0
    dt dt


    while solving this we assume the solution to be of form x = f(t)e-kt

    why is this exponential taken???
     
  2. jcsd
  3. Aug 1, 2012 #2

    Philip Wood

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    Gold Member

    Because the solution to the equation is known to be of this form! You are, if you like, assuming the form of the answer in advance! This is quite a common procedure in 'solving' differential equations.

    If your physical intuition is good, you could identify the [itex]2k\frac{dx}{dt}[/itex] term as due to a resistive force, and guess that there'd be an exponential decay factor in the solution.

    Once you've made the substitution [itex]x = e^{-kt} f(t)[/itex], you have a differential equation for [itex]f(t)[/itex], which is just the ordinary shm equation with angular frequency [itex]\omega[/itex] given by [itex]\omega^2 = \omega_0^2 - k^2[/itex], provided that [itex]\omega > k[/itex]. So the solution can be seen immediately, by elementary methods, to be the product of sinusoidally oscillating, and exponential damping, factors. [If [itex]\omega < k[/itex] we have a non-oscillatory fall-off in x with time.]

    Making the substitution [itex]x = e^{-kt} f(t)[/itex] involves a bit of drudgery and, imo, one might as well go the whole hog and (if [itex]\omega > k[/itex]) make the substitution [itex]x=Ae^{-kt}sin ({\omega t + \epsilon})[/itex] at the outset. You'll find that this expression does fit, provided that [itex]\omega^2 = \omega_0^2 - k^2[/itex].

    If you're happy with complex numbers, and the idea of linear combinations, there is a very slick method which simply requires you to substitute [itex]x=Ae^{-\alpha t}[/itex]. Alpha turns out to be complex if [itex]\omega > k[/itex], and you need to form a linear combinations of two solutions. The mathematics is a bit more advanced than that needed for the substitution [itex]x = e^{-kt} f(t)[/itex] or [itex]x=Ae^{-kt}sin ({\omega t + \epsilon})[/itex].
     
    Last edited: Aug 1, 2012
  4. Aug 4, 2012 #3

    Philip Wood

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    In previous post [itex]\omega > k[/itex] should read [itex]\omega_0 > k[/itex], and [itex]\omega < k[/itex] should read [itex]\omega_0 < k[/itex]. Sorry.
     
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