Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2009 / Paper 2 / 12 / iii

Let $D=1+{\omega }^2$, with $\omega >0$. A particular solution of the forced harmonic oscillator is $y_p=e^{-t}(at+b)$. Substitution gives $Da=2$ and $Db-2a=0$, hence $a=2/D$ and $b=4/D^2$. Adding $A\cos \omega t+B\sin \omega t$ and imposing the two initial conditions gives $A=-4/D^2$ and $B=2(1-{\omega }^2)/(\omega D^2)$. ThusThe first term tends to zero. The amplitude of the surviving sinusoid isAs an independent interpretation, the residual amplitude of an oscillator after an integrable force is ${\omega }^{-1}|{\int }_0^{\infty }F(t)e^{-i\omega t}dt|$. Here that integral is $2/(1+i\omega {)}^2$, giving the same result.