Damped oscillator- graphical interpretation

[rsaad]

Hi!
The damped oscillator equation is as follows:
x(t)= A exp(γt/2) cos(ωt)

where ω= √( (w0)^2 + (γ^2)/4 )

I have attached a graph of a damped oscillator.
The question is if I use graph to measure angular frequency, will it be w0 or ω?

It should be w0 because if I put γ=0, I should be getting the normal undamped system. The enveloped curve would disappear since exp(γt/2) is 1. BUT then where is ω on the graph!

 

[PhysicoRaj]

How can u expect ω in an x-t graph? You will have to calculate.

 

[rsaad]

of course I know that. Let m rephrase. The time T between successive maxima is constant. So I consider the complete oscillations, k, for a given time, t. To get angular frequency, W= k* 2pi/t
The question is, what is this this W? is it w0 or is it the angular frequency ω of the damped oscillator

 

[andrien]

it is the angular frequency of damped oscillator.

 

[Philip Wood]

The answer to your question is ω. Peaks in the damped sinusoid $e^{-kx} cos(\omega t)$occur a little before the peaks in the pure sinusoid $cos(\omega t)$, but by the same amount each time, so the time between peaks is the same as that between the peaks in $cos(\omega t)$.

 

[AlephZero]

e^{-kx} cos(\omega t)[/itex]occur a little before the peaks in the pure sinusoid $cos(\omega t)$, but by the same amount each time, so the time between peaks is the same as that between the peaks in $cos(\omega t)$.

An easier way to see the anwer is $\omega$ is to think about the times when x(t) = 0. They are the roots of $\cos \omega t = 0$.

 

[Philip Wood]

AlephZero. Agree, but thought rsaad (in post 3) was worried about maxima.