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.