# Solving and manipulating the damped oscillator differential equation

the differential equation that describes a damped Harmonic oscillator is:
$$\ddot x + 2\gamma \dot x + {\omega}^2x = 0$$ where ##\gamma## and ##\omega## are constants.
we can solve this homogeneous linear differential equation by guessing ##x(t) = Ae^{\alpha t}##
from which we get the condition:
$$\alpha = -\gamma \pm ({\gamma}^2 - {\omega}^2)^{1/2}$$

we can then say that the most general solution for ##x(t)## is:
$$x(t) = e^{-\gamma t}(Ae^{\Omega t} + Be^{-\Omega t})$$ where ##\Omega = ({\gamma}^2 - {\omega}^2)^{1/2} ##

since I am trying to derive ##x(t)## for under damping i.e. ##\gamma < \omega## or ##{\Omega}^2 < 0##
say ##w' = ({\omega}^2 - {\gamma }^2 )^{1/2}##
then we have ##\Omega = (-1)^{1/2}w' = Iw'##

$$x(t) = e^{-\gamma t}(Ae^{Iw't} + Be^{-Iw't})$$

##x(t)## has to be real as it represents displacement.
My book thus concludes that $$x(t) = Ce^{-\gamma t}cos(w't + \phi) ->(1)$$
it suggests we can get this by using ##e^{i\theta} = cos\theta + isin\theta## and it also states that if ##x(t)## is to be real then ##A* = B## where ##A*## is the complex conjugate of ##A##(I dont see why this has to be true). I need help arriving at eqn(1).

anuttarasammyak
Gold Member
As you assume ##A=B^*=a e^{i\phi}##. Your result goes
$$x(t)=e^{-\gamma t} 2a [e^{i(\omega'+\phi)t}+e^{-i(\omega'+\phi)t}]$$
You see ##C=4a##

EDIT Correction
$$x(t)=e^{-\gamma t} 2a [e^{i(\omega't+\phi)}+e^{-i(\omega't+\phi)}]=e^{-\gamma t} 4a \cos(\omega't+\phi)$$

Last edited:
As you assume ##A=B^*=a e^{i\phi}##. Your result goes
$$x(t)=e^{-\gamma t} 2a [e^{i(\omega'+\phi)t}+e^{-i(\omega'+\phi)t}]$$
You see ##C=4a##
i dont understand how they are converting the exponential form(i.e. ##x(t) = e^{-\gamma t}(Ae^{\Omega t} + Be^{-\Omega t})##) to the nice trigonometric form by applying Euler's formula.

Last edited:
As you assume ##A=B^*=a e^{i\phi}##. Your result goes
$$x(t)=e^{-\gamma t} 2a [e^{i(\omega'+\phi)t}+e^{-i(\omega'+\phi)t}]$$
You see ##C=4a##

EDIT Correction
$$x(t)=e^{-\gamma t} 2a [e^{i(\omega't+\phi)}+e^{-i(\omega't+\phi)}]=e^{-\gamma t} 4a \cos(\omega't+\phi)$$
$$x(t)=e^{-\gamma t} a [e^{i(\omega'+\phi)t}+e^{-i(\omega'+\phi)t}]$$
so, should'nt ##C = 2a##?

I tried doing the same manipulation for the differential equation describing simple harmonic motion:
$$\ddot x + \omega^2 x = 0$$
after guessing the solution to be of form ##x(t) = Ae^{\alpha t}## we get
$$x(t) = Ae^{i\omega t} + Be^{-i\omega t}$$
now again from here I want to write ##x(t)## in terms of a trigonometric function:
since ##x(t)## has to be real ##A* = B = ae^{i\phi}##
$$x = ae^{i(wt - \phi)} + ae^{-i(wt - \phi)}$$
$$x = Ccos(wt - \phi)$$ so ##C = 2a##
is this correct?

anuttarasammyak
Gold Member
You may remember
$$cos \phi = \frac{e^{i\phi}+e^{-i\phi}}{2}$$

• Hamiltonian299792458
vela
Staff Emeritus
Homework Helper
$$x(t) = A e^{i\omega t}+B e^{-i\omega t} = (A+B) \cos\omega t + i(A-B)\sin\omega t.$$ Let ##A = a+bi## and ##B = c+di##. For ##x(t)## to be real, ##(A+B)## has to be real and ##(A-B)## has to be imaginary, so you get