Solution of damped oscillation D.E.

[uday1236]

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?

 

[Philip Wood]

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 2k\frac{dx}{dt} 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 x = e^{-kt} f(t), you have a differential equation for f(t), which is just the ordinary shm equation with angular frequency \omega given by \omega^2 = \omega_0^2 - k^2, provided that \omega > k. So the solution can be seen immediately, by elementary methods, to be the product of sinusoidally oscillating, and exponential damping, factors. [If \omega < k we have a non-oscillatory fall-off in x with time.]

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

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 x=Ae^{-\alpha t}. Alpha turns out to be complex if \omega > k, and you need to form a linear combinations of two solutions. The mathematics is a bit more advanced than that needed for the substitution x = e^{-kt} f(t) or x=Ae^{-kt}sin ({\omega t + \epsilon}).

 

[Philip Wood]

In previous post \omega > k should read \omega_0 > k, and \omega < k should read \omega_0 < k. Sorry.