# Thread: Angular displacement from its equilibrium position

1. ## Angular displacement from its equilibrium position

Hi all. I'm currently stuck on the following question and I was wondering what I need to do. Any guidance and a few pointers in the right direction will be deeply appreciated!

2. Originally Posted by Benji123
Hi all. I'm currently stuck on the following question and I was wondering what I need to do. Any guidance and a few pointers in the right direction will be deeply appreciated!

This is a perfectly normal second order linear constant coefficient ODE initial value problem. So what do your notes say?

Try a trial solution of the form $e^{kt}$, and solve the resulting quadratic for $k$ etc

CB

3. Originally Posted by CaptainBlack
This is a perfectly normal second order linear constant coefficient ODE initial value problem. So what do your notes say?

Try a trial solution of the form $e^{kt}$, and solve the resulting quadratic for $k$ etc

CB
Thanks for the advice! Using the auxiliary equation I obtained: k^2 + 4k + 13 = 0
Therefore, k = (-4 ±√16+52)/2
k1 = 2.213
k2 = -6.123
Θ = Ae^2.123t + Be^-6.123t

I dont think this is correct because I haven't done anything with t=0, Θ=2 and dΘ/dt = 8...

4. Use your initial conditions to find your "constants of integration" A and B. You'll get two equations with two unknowns. Just plug and chug. Make sense?

5. Originally Posted by Benji123
Thanks for the advice! Using the auxiliary equation I obtained: k^2 + 4k + 13 = 0
Therefore, k = (-4 ±√16+52)/2
k1 = 2.213
k2 = -6.123
Θ = Ae^2.123t + Be^-6.123t

I dont think this is correct because I haven't done anything with t=0, Θ=2 and dΘ/dt = 8...
No the roots of the characteristic equation has roots $-2 \pm 3 \text{i}$

CB

6. Originally Posted by CaptainBlack
No the roots of the characteristic equation has roots $-2 \pm 3 \text{i}$

CB
So this would be...

m^2 + 4m + 13 = 0
m= -2±3i
^
y = Ae^(-2+3i)t + Be^(-2-3i)t

y = e^3t (4cos-2t + 13sin-2t)?

7. No, your last line is confusing the real and imaginary parts. You should have

y = e^(2t) (A cos(3t) + B sin(3t)).

8. Ohh, thanks alot for the help buddy! I knew with a bit of help I'd eventually get there

9. Actually, I wrote it wrong. It's

y = e^(-2t) (A cos(3t) + B sin(3t)).

Make sure to thank CB. He did more for you than I did!

10. Thanks for the help everyone! I really appreciate it!