Where does the phi come from in the charge of a capacitor in an LC circuit?

AI Thread Summary
The discussion centers on the derivation of the charge equation for a capacitor in an LC circuit, starting from the equation -L(di/dt) - q/C = 0. The transformation of current to charge leads to a second-order differential equation, q'' + (1/LC)q = 0, which is solved using standard methods for homogeneous systems. The solution involves trigonometric functions, resulting in q = k1*cos(at) + k2*sin(at). The discrepancy with the book's formula, q = Q*cos(at + phi), is clarified, indicating the relationship between the sine and cosine functions in phase-shifted forms. The discussion concludes with a reference to a helpful YouTube video that provided additional insight.
thed0ctor
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I'm reading University Physics 13e by Young and Freedman and we're given this equation:

-L(di/dt)-q/C=0

So we know that i, current, is, i=dq/dt. So the first equation can be read, after some simplification, as

q''+(1/LC)q=0

where q prime means the derivative of q (charge) with respect to time.

This is a homogeneous system so I figure solve it like anything else. So

s^2 = -(1/LC)
=>
s=i*(1/LC)^1/2... the i here is imaginary

let: a=(1/LC)^1/2

thus q=k1*cos(at) + k2*sin(at)

Yet in the formula in the book we end up with the formula:
q=Q*cos(at + phi)

with no sine...

Any help?
 
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Figured it out:

s48fb6.jpg


This was inspired by the youtube video I found here.
 
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