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JamesJames
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A square wave pulse (generated using an oscilloscope) is used to induce damped oscillations in a circuit that consits of an inductance L and a capacitance C connected in series. A resistance is present even though no resistor is present in the circuit.
a) Find the differential equation for the capacitor charge.
b) Find the underdamped solution. Hint: Understand why application of a square-wave corresponds to kicking a damped harmonic oscillator:
q(0)=0,q'(0)=0
Here are my attempts:
a)
LI(dI/dt) + (q/C) dq/dt = -I^2 / R and then using I = dq/dt,
[tex]L \frac{d^{2}q}{dt^{2}} + R \frac{dq}{dt} + \frac{q}{C}[/tex] = 0
I am quite sure about this part.
b)
Here is where I am getting really confused. How do they know that q(0) = 0, q'(0) = 0 are the necessary initial conditions?
These initial conditions imply that the capacitor is not charged initially. Taking these as given (even though I don't understand why) the solution to the differential equation HAS to be
q(t) = [tex]e^{-Rt/2L}sin(\omega t)[/tex]
This is the only way that I can ensure that q(0) = 0 because for a cosine solution, q(0) will not be zero.
Finally regarding the undercritical damping, the solution above is infact the undercritical case. By definition of underctirical damping, the frequency [tex]\omega[/tex] is essentially equal to the undamped frequency.
i.e. [tex]\omega = \sqrt \frac{1}{LC}[/tex].
Is my part b solution correct?
James
a) Find the differential equation for the capacitor charge.
b) Find the underdamped solution. Hint: Understand why application of a square-wave corresponds to kicking a damped harmonic oscillator:
q(0)=0,q'(0)=0
Here are my attempts:
a)
LI(dI/dt) + (q/C) dq/dt = -I^2 / R and then using I = dq/dt,
[tex]L \frac{d^{2}q}{dt^{2}} + R \frac{dq}{dt} + \frac{q}{C}[/tex] = 0
I am quite sure about this part.
b)
Here is where I am getting really confused. How do they know that q(0) = 0, q'(0) = 0 are the necessary initial conditions?
These initial conditions imply that the capacitor is not charged initially. Taking these as given (even though I don't understand why) the solution to the differential equation HAS to be
q(t) = [tex]e^{-Rt/2L}sin(\omega t)[/tex]
This is the only way that I can ensure that q(0) = 0 because for a cosine solution, q(0) will not be zero.
Finally regarding the undercritical damping, the solution above is infact the undercritical case. By definition of underctirical damping, the frequency [tex]\omega[/tex] is essentially equal to the undamped frequency.
i.e. [tex]\omega = \sqrt \frac{1}{LC}[/tex].
Is my part b solution correct?
James
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