Solutions to damped harmonic oscillator?

Question

For the damped harmonic oscillator equation
$$frac{d^2x}{dt^2}+frac{c}{m}frac{dx}{dt}+frac{k}{m}x=0$$
we get that the general solution is
$$x(t)=Ae^{-gamma t}e^{iomega_d t}+Be^{-gamma t}e^{-iomega_d t}$$
where $gamma = frac{c}{2m}$  and $ omega_d=sqrt{omega^2-gamma ^2}$.Using Eulers equation, we can expand this as follows:
$$Ae^{-gamma t}(cos(omega _dt)+isin(omega_d t))+Be^{-gamma t}(cos(omega _dt)-isin(omega_d t))$$
$$Rightarrow e^{-gamma t}(A+B)cos(omega_d t) +e^{-gamma t}(Ai-Bi)sin(omega_d t)$$
But now we are dealing with a physical problem so we only examine the real part which is $e^{-gamma t}(A+B)cos(omega_d t)$. But this does not have any phase difference. Yet textbooks always make the claim that the real part of the solution is
$$e^{-gamma t}(C)cos(omega_d t+phi)$$
where $phi$ is some arbitrary initial phase. But where does that initial phase come from if the real part of the solution does not have a phase change in it? I understand that $A$ and $B$ themselves need not be real however I do not understand how this fact could ever lead to a non zero initial phase in the real part of the solution.
This issue has bothered me for quite some time now so any help would be immensely appreciated!

QuantumApple · Accepted Answer

As you've said, $A$ and $B$ are not necessarily real, they can be complex. In fact, if you want a physical solution, you don't need the $sin$ term to be $0$, but rather you want to impose $overline{x(t)} = x(t)$, which is equivalent to $overline{A} = B$.
From here, you can rewrite the last part of the solution as:
$$x(t) = e^{-gamma t} 2, mathrm{Re}(A)cos(omega_d t) - e^{-gamma t} 2, mathrm{Im}(A)sin(omega_d t),$$
where $mathrm{Re}(A)$ and $mathrm{Im}(A)$ are respectively the real and imaginary part of the complex number $A$. With some trigonometry, you can rewrite it  as $x(t) =  e^{-gamma t}  Ccos(omega_d t + phi)$.

Eli · Answer

The general solution for your ODE is:
$$x(t)=(a+i,b),e^{-gamma t}e^{iomega_d t}+(a-i,b)e^{-gamma t}e^{-iomega_d t}tag 1$$
Where $a=x(0)$ and $b=D(x)(0)$ are the initial conditions
expand equation (1) you obtain:
$$x(t)=2,e^{-gamma t}(a,cos(omega_d t)-b,sin(omega_d t))tag 2$$
thus the solution is real
you can also write the solution equation (2) with two new constants $~C$ and $phi$ instead of $~a$ and b  :
$$x(t)=C,,e^{-gamma t},cos(omega_d t+phi)=C,,e^{-gamma t},[cos(omega_d t),sin(phi)-sin(omega_d t),cos(phi)]tag 3$$
comparing equation(3) with (2) you obtain:
$$tan(phi)=frac{b}{a}~,C=2,sqrt{a^2+b^2}$$

Solutions to damped harmonic oscillator?

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