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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 $$x(t)=Ae^{-\gamma t}e^{i\omega_d t}+Be^{-\gamma t}e^{-i\omega_d t}$$ where $\gamma = \frac{c}{2m}$ and $ \omega_d=\sqrt{\omega^2-\gamma ^2}$.Using Eulers equation, we can expand this as follows: $$Ae^{-\gamma t}(\cos(\omega _dt)+i\sin(\omega_d t))+Be^{-\gamma t}(\cos(\omega _dt)-i\sin(\omega_d t))$$ $$\Rightarrow e^{-\gamma t}(A+B)\cos(\omega_d t) +e^{-\gamma t}(Ai-Bi)\sin(\omega_d t)$$ 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!

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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 $x(t) = e^{-\gamma t} C\cos(\omega_d t + \phi)$.

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  • $\begingroup$ Thanks for the response! Dont you mean that $x(t)=e^{-\gamma t}2 Re(A)cos(\omega_d t)-e^{-\gamma t}2 Im(A)sin(\omega_d t)$ ? Also, are you saying that if we want a physical solution we simply cant have a non zero imaginary part ? Since the requirement that $x^*(t)=x(t)$ requires that the entire solution solution be real. I was under the impression that if $x(t)=a+bi$ where both a and b are non zero, then we can simply take $x(t)=a$ as the physical solution discarding the non zero (or it could even happen to be zero, it doesn't matter since we are concerned with the real part) imaginary part? $\endgroup$ – SalahTheGoat Sep 3 at 9:57
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    $\begingroup$ Yes sorry for the mistake, I've made a correction. Let me be a little clearer with what I meant regarding the physical solution. You're trying to solve a second order differential equation involving a real variable, $x(t)$. You can solve the equation on $\mathbb{R}$ directly and it will give you $x(t) = e^{-\gamma t} C \cos(\omega_d t + \phi)$.Or, if you want to be more general, you can solve the equation on a larger set ($\mathbb{C}$), because it makes it easier to solve it using complex exponentials. $\endgroup$ – QuantumApple Sep 3 at 12:17
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    $\begingroup$ However, because you solved the equation in a larger frame, you need to put a constraint on the solution to ensure that it corresponds to a physical one (i.e, here, that $x(t)$ is real). You can also just "disregard" the imaginary part as you've suggested, but I don't think this is necessarily good practice. In any case, I think your mistake came from the fact that you assumed that $A$ and $B$ were real and thus $A i - B i$ must be purely imaginary and can be discarded (which is not the case if you consider $A$ and $B$ to be non real). $\endgroup$ – QuantumApple Sep 3 at 12:18
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    $\begingroup$ Another way to phrase it (because now I see what might have confused you) is that "discarding the imaginary part of the solution" isn't the same as "discarding the imaginary part of $A$ and $B$". You should do the former and not the latter. $\endgroup$ – QuantumApple Sep 3 at 12:21
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    $\begingroup$ This looks correct to me, yes. You can also use the quicker method by enforcing $\overline{x(t)} = x(t)$, which simplifies the calculation a bit earlier but in the end that's the same. I've skipped the last part where you transform the sum of a cos and a sin into a single cos with a phase but you can check @Eli answer for details. $\endgroup$ – QuantumApple Sep 4 at 12:32
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The general solution for your ODE is:

$$x(t)=(a+i\,b)\,e^{-\gamma t}e^{i\omega_d t}+(a-i\,b)e^{-\gamma t}e^{-i\omega_d t}\tag 1$$

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:

$$\tan(\phi)=\frac{b}{a}~,C=2\,\sqrt{a^2+b^2}$$

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