Using the Angular Frequency to Solve a Differential Equation

AI Thread Summary
The discussion focuses on deriving the angular frequency for a damped harmonic oscillator from a differential equation. The derived angular frequency is expressed as ω = √(k/m - b²/4m²), which is compared to the results obtained from substituting x = e^(λt) into the equation. The participant expresses confusion about the relationship between the derived results and the meaning of λ. Suggestions include substituting a = k/m and using x = e^(ωt) instead of e^(λt) for clarity. Evaluating the sign of the argument under the square root is recommended to resolve any issues with solving for ω.
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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!
 
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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.
 
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