Unraveling the Enigma: Understanding the Equation

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The discussion focuses on calculating the charge on a capacitor after five complete cycles using a specific equation from the textbook. To find the charge, one must first calculate the angular frequency (ω') using the values for resistance (R), inductance (L), and capacitance (C). Initial conditions, including charge (q(0) = 25μC) and current (I(0) = 0), are used to determine q0 and φ. The duration of one cycle (T') is derived from ω', and then the charge at five cycles (q') is calculated by substituting t = 5T' into the initial equation for q(t). Finally, the energy stored in the capacitor is computed using the formula E = (1/2)(q'^2/C).
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Homework Statement
An RLC circuit has a resistance of 200 Ω, an inductance of 15 mH, and a capacitance of 34nF. At time t = 0, the charge on the capacitor is 25
μC and there is no current flowing. After five complete cycles the energy stored in the capacitor is:

A) 0.64 μJ
B) 77 μJ
C) 0.49 mJ
D) 1.8 mJ
E) 9.2 mJ

The answer is A.
Relevant Equations
(see better expression below)
The equation is the only equation discussed in the textbook.
Is there a hint of how I could start this?
 

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Yes well, you should find the charge on the capacitor after five complete cycles. To find it you will use the first equation you give in the attached picture.

You can calculate ##\omega'## from the given values for R L C, and also ##q_0## and ##\phi## from the initial conditions of ##q(0)=25\mu C,I(0)=0##.

Also once you find ##\omega'## find ##T'=\frac{2\pi}{\omega'}## which is the duration of one cycle. Finally put ##t=5T'## in the initial equation for ##q(t)## and find ##q'=q(5T')##. The energy stored will be $$E=\frac{1}{2}\frac{q'^2}{C}$$.
 
Thanks for the hint. I got it.
 
Thread 'Correct statement about size of wire to produce larger extension'
The answer is (B) but I don't really understand why. Based on formula of Young Modulus: $$x=\frac{FL}{AE}$$ The second wire made of the same material so it means they have same Young Modulus. Larger extension means larger value of ##x## so to get larger value of ##x## we can increase ##F## and ##L## and decrease ##A## I am not sure whether there is change in ##F## for first and second wire so I will just assume ##F## does not change. It leaves (B) and (C) as possible options so why is (C)...

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