How to begin oscillation in steady state?

[bullet_ballet]

I need to find the initial conditions such than an underdamped harmonic oscillator will immediately begin steady-state motion under the time dependent force F = m f cosωt.

For the underdamped case:
x(t) = ae^{-\gamma t}cos(\Omega t+\alpha)+\frac{f}{r}cos(\omega t-\theta)

and if it matter, r^2 = (\omega^2_0-\omega^2)^2+4\gamma^2\omega^2
and \theta = Tan^{-1}\frac{2\gamma\omega}{\omega^2_0-\omega^2}<br />

I thought I would just have to find x0 and v0 such that the transient was 0, but that doesn't seem to be leading down the right track. What direction should my solution be heading?

 

[genxhis]

Why not just let x₀=0 and v₀=0? This should zero out the transient portion of the solution and leave the driving force intact.

 

[bullet_ballet]

Why not just let x₀=0 and v₀=0? This should zero out the transient portion of the solution and leave the driving force intact.

Makes sense to me, but the back of the book doesn't seem to agree. It has x_0=\frac{f (\omega^2_0-\omega^2)}{r^2} and v_0=\frac{2\gamma\omega^2f}{r^2}.

 

[genxhis]

oh. take x(0) and x'(0) and let a = 0. If a = 0 then the transient solution is immediate null, but you'll see x₀ and v₀ are not. You'll have to subtitute for theta as well.

 

[ehild]

I need to find the initial conditions such than an underdamped harmonic oscillator will immediately begin steady-state motion under the time dependent force F = m f cosωt.

For the underdamped case:
x(t) = ae^{-\gamma t}cos(\Omega t+\alpha)+\frac{f}{r}cos(\omega t-\theta)

and if it matter, r^2 = (\omega^2_0-\omega^2)^2+4\gamma^2\omega^2
and \theta = Tan^{-1}\frac{2\gamma\omega}{\omega^2_0-\omega^2} <br />

I thought I would just have to find x0 and v0 such that the transient was 0, but that doesn't seem to be leading down the right track. What direction should my solution be heading?

It was a good start. Let a=0. Find x(0) and v(0). You have
x(0)=\frac{f}{r}cos(\theta ) \mbox{ and }v(0)=\frac{f\omega}{r}\sin(\theta ), use that
cos(\theta ) = \frac{1}{\sqrt{1+tan^2(\theta )}}\mbox{, }sin(\theta )=\frac{tan(\theta )}{\sqrt{1+tan^2(\theta )}} \mbox{ and } tan(tan^{-1}(\theta))=\theta.

ehild