Using the Angular Frequency to Solve a Differential Equation

[imagemania]

Homework Statement

Im somewhat unsure of what the result i have derived is exactly. I know the angular frequency should be
$$\omega = \sqrt{\frac{k}{m} - \frac{{b}^{2}}{4{m}^{2}}}$$

The Attempt at a Solution

$$m\frac{{d}^{2}x}{d{t}^{2}} = -kx -b\frac{dx}{dt}$$
Sub in $$\omega = \sqrt{\frac{k}{m}}$$
Do $$x = {e}^{\lambda t}$$
x' = ...
x''=...

$${e}^{\lambda t}({\lambda}^{2} + \lambda \frac{b}{m} + {\omega}^{2}) = 0$$
$$\lambda = -\frac{b}{2m} \pm \sqrt{\frac{{b}^{2}}{{4m}^{2}} - 4\frac{{\omega}^{2}}{4}}$$

However, i thought
$$\omega = \sqrt{\frac{k}{m} - \frac{{b}^{2}}{4{m}^{2}}}$$

Are these results related as i cannot quite put my tongue on how to get this result from the $$\lambda$$ result (And I am not entirely sure what $$\lambda$$ representes in real terms)

Thanks!

 

[obafgkmrns]

Substitute something like a = k/m (k/m isn't the angular frequency in a damped system.), and let x = exp(omega t) instead of exp(lambda t).

If you get stumped when you solve for omega, evaluate the sign of the argument under the square root.