Solving an Oscillatory Circuit

[1MileCrash]

Homework Statement

attached

Homework Equations

The Attempt at a Solution

Is the correct answer a choice?

I get Q= .17e-6cos(4000t), every single time.

 

[Dick]

Homework Statement

attached

Homework Equations

The Attempt at a Solution

Is the correct answer a choice?

I get Q= .17e-6cos(4000t), every single time.

Well, how do you get that?

 

[1MileCrash]

Q'' + 4e6Q = 0 =>

r^2 + 4e6 = 0 (characteristic equation) => r = 0 +/- 4000i

=> Q = c1cos(4000t) + c2sin(4000t)
=> Q' = -4000c1sin(4000t) + 4000c2cos(4000t)

=> Q(0) = c1 = .17e-6
=> Q'(0) = 4000c2 = 0 => c2 = 0

So

Q = .17e-6cos(4000t)

 

[Dick]

Q'' + 4e6Q = 0 =>

r^2 + 4e6 = 0 (characteristic equation) => r = 0 +/- 4000i

=> Q = c1cos(4000t) + c2sin(4000t)
=> Q' = -4000c1sin(4000t) + 4000c2cos(4000t)

=> Q(0) = c1 = .17e-6
=> Q'(0) = 4000c2 = 0 => c2 = 0

So

Q = .17e-6cos(4000t)

The square root of 4e6 is 2000, not 4000.

 

[1MileCrash]

The square root of 4e6 is 2000, not 4000.

I'm confused. Why would I be taking the square root of 4e6?

Shouldn't I be taking the square root of b^2 - 4(a)(c)?

Which is 0 - 4(1)(4e6) = 16000000, and the square root of that is 4000.


Thanks

 

[Dick]

I'm confused. Why would I be taking the square root of 4e6?

Shouldn't I be taking the square root of b^2 - 4(a)(c)?

Which is 0 - 4(1)(4e6) = 16000000, and the square root of that is 4000.


Thanks

You don't NEED to use the quadratic formula to solve r^2+4e6=0, but if you do you shouldn't forget there is a 2a in the denominator.

 

[1MileCrash]

You don't NEED to use the quadratic formula to solve r^2+4e6=0, but if you do you shouldn't forget there is a 2a in the denominator.

Wow, I was thinking a = 1 and ignored it.

And I know I don't need to, it just like to for complex since it spits it out in the right form already.

That's what I get for rushing.

Thanks again