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I read in Mechanics volume of Berkeley physics that $\ddot{x} + \frac{1}{\tau} \dot{x}+\omega_0^2 x = \alpha_0 \sin \omega t$ has the particular solution

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and it is easy to see that if $\mathcal{E}=\mathcal{E}_0 \cos(\omega t + \phi)$ for LCR serie circuit the differential equation that rules current is

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Let's exploit analogy to find current after transient. If $\phi=0$ we have $x \to i$, $\tau \to \frac{L}{R}$, $\omega_0 \to \frac{1}{\sqrt{LC}}$, $\alpha_0 \to \frac{-\omega \mathcal{E}_0}{L}$. By substitution we find \begin{equation} i_0 = \frac{-\mathcal{E}_0}{\sqrt{\left( \frac{1}{\omega C} - L\omega \right)^2+R^2}} \end{equation} and a phase difference between emf and current: \begin{equation} \theta = \tan^{-1} \left( \frac{R}{\frac{1}{\omega C} - L\omega} \right) \end{equation} Now, by supposing a solution like $i=i_0 \cos (\omega t)$ it is easy exploiting trigonometry to show that we must have $i_0^2 = \frac{\mathcal{E}_0^2}{{\left( \frac{1}{\omega C} - L\omega \right)^2+R^2}}$, and it's ok, but we have also,

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but how can it is be true that with a method the phase difference between emf and current is $\tan x$ and with another method is $\tan \frac{1}{x}$? These are two very different functions and solution can't be in a phase shifting: I did some trivial error in the way I developed the "analogy with mechanics" method, but what? I have long tried to figure it out why this absurd thing happens, without success. For this I ask.

Edit

Exploiting @Chemomechanics formula (I didn't know it) I solved: I can write Berkeley solution as \begin{equation} x(t) = \frac{\alpha_0}{\left[ (\omega_0^2 - \omega^2)^2 + (\omega / \tau)^2 \right]^{1/2}} \cos \left(\omega t + \tan^{-1} \left( \frac{\omega^2 - \omega_0^2}{\omega/\tau} \right) \right) \end{equation} and in this way the mechanical analogy gives \begin{equation} i(t) = \frac{- \mathcal{E}_0}{\sqrt{R^2 + \left( \omega L - \frac{1}{\omega C} \right)^2}} \cos \left( \omega t + \tan^{-1} \left( \frac{\omega L - \frac{1}{\omega C}}{R} \right) \right) \end{equation} Only a doubt is left: is it correct the minus sign? In my book (Mazzoldi) it is absent (and trigonometric considerations gives $\frac{i_0^2}{\mathcal{E}_0^2}$ so it doesn't answer to this).

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    $\begingroup$ ... $RLC$, $LCR$, what Is the difference? The differential equation Is the same if components are in serie. $\endgroup$
    – Haumea
    Nov 30, 2021 at 19:06
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    $\begingroup$ you're right I had to be more "international". in Italy we say RLC. The same is true for the electromotive force, in Italy abbreviated as fem (as I say in question), but I see that in English texts, for obvious reasons, emf Is used. $\endgroup$
    – Haumea
    Nov 30, 2021 at 19:31
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    $\begingroup$ In the first case you used a sine solution. In the second case a cossine solution. $sin(\omega t + \phi) = cos(\omega t + (\phi - \frac{\pi}{2})$ $\endgroup$ Nov 30, 2021 at 19:31
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    $\begingroup$ Why do you say "solution can't be in a phase shifting"? We have $\tan^{-1}(x)=\tan^{-1}\left(-\frac{1}{x}\right)-\frac{\pi}{2}$, where $x=\frac{\omega L-1/\omega C}{R}$. If you excite something with a sine wave, the solution in terms of the cosine function is $\frac{\pi}{2}$ out of phase. $\endgroup$ Nov 30, 2021 at 20:18
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    $\begingroup$ In fact, doing the math starting from the $i=i_0 \sin (\omega t)$, there is a different phase difference from the one I photographed, with inverted numerator and denominator (and changed sign) as my $\theta$. This upsets me because I find it counterintuitive. Thanks for the valuable comments on which I will reflect. $\endgroup$
    – Haumea
    Nov 30, 2021 at 20:26

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Note that in some definitions of electrical-to-mechanical analogies, the capacitive terms in each are reciprocals of one another. Might this be the source of your confusion?

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    $\begingroup$ Thank you. I solved with @Chemomechanics formula. I edited and only a doubt is left. $\endgroup$
    – Haumea
    Nov 30, 2021 at 23:30

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