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Homework Statement
It is the driven series RLC circuit. It is given in the following images.
It is from the section 12.3 in this note.
Homework Equations
The differential equation, as given by 12.3.3, is ##L \frac{d^2 Q}{d t^2} + R \frac{d Q}{d t} + \frac{Q}{C} = V_0 \sin{(\omega t)}##.
It says that one possible solution is ##Q(t) = Q_0 \cos{(\omega t - \phi)}##, where ##Q_0 = \frac{V_0}{\omega \sqrt{R^2 + (\omega L - 1 / \omega C)^2}}## and ##\tan \phi = \frac{1}{R} \left( \omega L - \frac{1}{\omega C} \right)##.
The Attempt at a Solution
So, assuming ##Q(t) = Q_0 \cos{(\omega t - \phi)}##, I get$$\frac{d Q}{d t} = -Q_0 \omega \sin{(\omega t - \phi)}\\\frac{d^2 Q}{d t^2} = -Q_0 \omega^2 \cos{(\omega t - \phi)}$$
Substituting,$$-L Q_0 \omega^2 \cos{(\omega t - \phi)} - R Q_0 \omega \sin{(\omega t - \phi)} + \frac{Q_0}{C} \cos{(\omega t - \phi)} = V_0 \sin{(\omega t - \phi)}$$
I cannot figure out show to eliminate ##t## and ##\phi## and solve this for ##Q_0##.
I try applying$$\cos{(\omega t - \phi)} = \cos{(\omega t)} \cos \phi + \sin{(\omega t)} \sin \phi\\\sin{(\omega t - \phi)} = \sin{(\omega t)} \cos \phi - \cos{(\omega t)} \sin \phi$$
But still there's no way cancel ##t## and ##\phi## and solve for ##Q_0##, or cancel ##Q_0## and ##t## and solve for ##\phi##.
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