Amplitude of harmonic oscillator

[longrob]

Homework Statement

I have a simple harmonic oscillator system with the driving force a sinusoidal term. The question is to find the general solution and the amplitude of the steady state solution

Homework Equations

I found the steady state part of the solution. It is of the form:
(sin(wt)*(1-w^2)-cos(wt)*w)/f(w)

The Attempt at a Solution

I'm told that the amplitude of the system in steady state should be 1/sqrt(f(w)) but I can't derive it. Can anyone help ?

Thanks

 

[payumooli]

use trig identity and reduce the equation to sin(A-B), amplitude is the factor by which the stated term is raised

 

[longrob]

But the coefficients of each sinusoidal term are different...

 

[payumooli]

is your steady state solution correct? it gets crazy if i reduce it to some trig identity
and what is f(w) equal to?

 

[longrob]

Yes I'm sure it's correct. This is homework so I'm loathed to post the whole details. However, take a look here:
http://nuweb.neu.edu/dheiman/U600/DHO.pdf
Look at the section "Driven Harmonic Oscillator" on page 4. The solution given is exactly what I want (with lambda, w_0 and F_0/m all equal to 1). I just can't derive it from my own solution. By hand I just used a trial solution of p.cos(wt) + q.sin(wt) and equated coefficients, which lead to the form I posted above. I also used Maple to check this, and it agrees.

 

[vela]

The (1-w^2) factor isn't under a square root?

 

[longrob]

no, it's not...

 

[longrob]

Another look at this in Maple:
[Sorry, fiddling with latex...]

 

[longrob]

In Maple:
$${\it s2s}\, := \,{\frac { \left( -{\omega_{{0}}}^{2}+{\omega}^{2} \right) \sin \left( \omega\,t \right) +\cos \left( \omega\,t \right) \omega\,\lambda}{{\omega}^{4}+ \left( {\lambda}^{2}-2\,{\omega_{{0}}}^{2} \right) {\omega}^{2}\\<br /> \mbox{}+{\omega_{{0}}}^{4}}}\]}$$

which simplifies to:

$${\frac {\cos \left( \omega\,t-\arctan \left( -{\omega_{{0}}}^{2}+{\omega}^{2},\omega\,\lambda \right) \right) }{ \sqrt{{\lambda}^{2}{\omega}^{2}+{\omega_{{0}}}^{4}-2\,{\omega_{{0}}}^{2}{\omega}^{2}\\<br /> \mbox{}+{\omega}^{4}}}}\]}$$

And that's what I'm trying to achieve by hand...

Maple code is
s2s:=((-omega[0]^2+omega^2)*sin(omega*t)+cos(omega*t)*omega*lambda)/(omega^4+(lambda^2-2*omega[0]^2)*omega^2+omega[0]^4);
simplify(convert(s2s,phaseamp,t));

Thanks

 

[vela]

Oh, that helps. Note that the denominator is $(\omega^2-\omega_0^2)^2+(\lambda\omega)^2$. Consider a right triangle with one leg of length $\omega^2-\omega_0^2$ and the other of length $\lambda\omega$. Does that help?

 

[longrob]

Indeed, yes. So easy now. Thanks !