- #1

Hamiltonian

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$$\ddot x + 2\gamma \dot x + {\omega}^2x = 0$$ 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).