Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2017 / Paper 1 / 3

A causal Green function vanishes before the impulse. For $t>\tau$ it solves the homogeneous linear ordinary differential equation, whose characteristic roots are $-q\pm 2i$. Thus it has the formContinuity at $t=\tau$ is necessary to avoid a delta derivative in $G_{tt}$, so $A=0$. Integrating its forced equation across the impulse gives the jump condition for an impulse, $G_t({\tau }^{+},\tau )-G_t({\tau }^{-},\tau )=1$. The derivative from below is zero and the derivative from above is $2B$, givingHere $\Theta$ is the Heaviside step function. The zero initial data exclude any additional homogeneous response. By the superposition principle, the output for a general locally integrable input isThe upper endpoint causes no unwanted term in the first derivative because $G(t,t)=0$; in the second derivative its unit derivative jump reproduces the input.

The pulse height is the constant $t_0$, not the time variable. In the undamped case the convolution therefore reduces toIntegration gives the continuous displacement and velocityThe post-pulse expression is equivalently $x(t)=(t_0/2)\sin t_0\sin (2t-t_0)$. Since $t_0>0$, it vanishes for every subsequent time precisely when $\sin t_0=0$. ThusThese are integer multiples of the oscillator's natural period $\pi$. At such switch-off times both $x(t_0)$ and $x^{\prime }(t_0)$ are zero. Vanishing displacement alone would not suffice: a nonzero switch-off velocity would restart an oscillation. This is rectangular-pulse cancellation of an undamped oscillator.