Part b of a differential equation problem

[Sparky_]

Homework Statement

Find the charge on the capacitor in an L-R-C circuit at time t = 0.001

L = 0.05H, R = 2 ohms, C = 0.01F

q(0) = 5
i(0) = 0
E(t) = 0

Homework Equations

The Attempt at a Solution

$$L \frac {di(t)}{dt} + R \frac {dq(t)}{dt} + \frac {q}{C} = 0$$

$$\frac {dq^2(t)}{dt^2} + 40 \frac {dq(t)}{dt} + 2000q = 0$$

$$m^2 = 40m + 2000 = 0$$

$$q(t) = e^{-20t} (c1 * cos(40t) + c2 * sin(40t))$$

$$q'(t) = i(t) = -20e^{-20t} (c1 * cos(40t) + c2 * sin(40t) ) + e^{-20t}(-200 sin (40t) + 40*c2*cos(40t))$$


Do you agree with my q(t) =

$$q(t) = e^{-20t} (5 cos(40t) + \frac{5} {2} sin(40t))$$


??


There is a second part to this problem -

Find the first time q is equal to 0.

Thanks to Kreizhn I have

$$0 = e^{-20t} (5cos(40t) + \frac{5} {2}sin(40t))$$

$$0 = (5cos(40t) + \frac{5} {2}sin(40t))$$

$$cos(40t) = -\frac{1} {2}sin(40t))$$

$$40t = -1.1.07$$

$$t = -0.0276$$

$$40t = -1.1.07 + pi$$

$$t = 0.0508$$

The book gets t = 0.0669.

Suggestions?

Thanks
-Sparky_

 

[dynamicsolo]

$$q(t) = e^{-20t} (5 cos(40t) + \frac{5} {2} sin(40t))$$

I am also getting this result. If you use the circuit parameters to compute the time constant for this LRC circuit, you also get a frequency of $$\omega' = 40$$ ,
so this appears to jibe.

There is a second part to this problem -

Find the first time q is equal to 0...

$$40t = -1.107$$

$$t = -0.0276$$

$$40t = -1.107 + \pi$$

$$t = 0.0508$$

The book gets t = 0.0669.

Suggestions?

The book's solution does not work in the equation q(t) = 0. The equation you used reduces to solving for 40t = arctan(-2). If you back-figure using the given solution, you find it is a result for 40t = arctan(-1/2); the simplest explanation is that the solver for this problem botched the algebra. (Yeah, like that's never happened in a textbook...)

 

[Sparky_]

Thanks,

I'll go with it.

(I'm reviewing my Diff E. from way back for the fun of it.
This is not officially for a class so I won't get the solutions.)

Thanks again
Sparky_