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

The characteristic equation is $r^2+\lambda r+k=0$, whose roots are $r_{\pm }=(-\lambda \pm \sqrt{{\lambda }^2-4k})/2$. For distinct real roots, $y=C_1{e}^{r_{+}t}+C_2{e}^{r_{-}t}$; at the repeated-root threshold ${\lambda }^2=4k$, $y=(C_1+C_{2}t)e^{-\lambda t/2}$. In the complex-root case put $\Omega =\sqrt{k-{\lambda }^2/4}$ to obtainThis is precisely the oscillatory regime of the damped harmonic oscillator. The zero solution is of course nonoscillatory; the regime describes its nontrivial motions.

The damped angular frequency is $\Omega =\sqrt{k-{\lambda }^2/4}$. The oscillation envelope is proportional to $e^{-\lambda t/2}$. A decrease by a quarter leaves three quarters of the original amplitude, so $e^{-\lambda t/2}=3/4$. ThereforeThis concerns the envelope; it does not ask when the instantaneous displacement first equals three quarters of a chosen value.

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.