Suspension system, linear differential equation

[bobred]

Homework Statement

In the limit as t→∞, the solution approaches x(t) =K \sin[ω(t − t_0)]
where K and t₀ depend on ω. A>0 and ω≥0. Show that

K(ω) = \frac{A}{\sqrt{ω^4 + 2ω^2 + 1}}
.

Homework Equations

Here is the differential equation

\frac{\textrm{d}^{2}x}{\textrm{d}t^{2}}+2\frac{\textrm{d}x}{\textrm{d}t}+x=A\sin(\omega t)

The Attempt at a Solution

Here is the general solution

x=\left(C+Dt\right)e^{-t}-\frac{A\left(2\omega\cos\left(\omega t\right)+\omega^{2}\sin\left(\omega t\right)-\sin\left(\omega t\right)\right)}{\omega^{4}+2\omega^{2}+1}

As t get large the exponential term vanishes but cannot see how the solution approaches x(t) =K \sin[ω(t − t_0)]

Any pointers?

 

[pasmith]

Use the identity K\sin(\theta - \phi) = K\sin \theta \cos\phi - K \cos \theta \sin \phi and set <br /> K \sin (\omega t - \omega t_0) = -\frac{A(2 \omega \cos(\omega t) + (\omega^2 - 1)\sin (\omega t))}{\omega^4 + 2\omega^2 + 1}. Compare the coefficients of \sin(\omega t) and \cos(\omega t) of each side. This gives you two simultaneous equations of the form <br /> C = K \cos (\omega t_0) \\<br /> D = K \sin (\omega t_0) to be solved for K \geq 0 and - \frac \pi 2 \leq \omega t_0 \leq \frac \pi 2.

 

[bobred]

Thanks pasmith, need to try and remember those trig identities!