Dirac Delta Function — Question 5

Question 5 – Solution

Strategy. The second kick must occur at a zero of displacement, then cancel the incoming velocity. Timing errors leave a measurable residual oscillation.

Step 1: Transform and invert. For $s>0$, $$Y=\frac{Ie^{-as}+Ke^{-bs}}{s^2+\omega^2},\qquad \boxed{y=\frac I\omega H_a(t)\sin\omega(t-a) +\frac K\omega H_b(t)\sin\omega(t-b).}$$ Each activated sine is zero initially, with derivative equal to its impulse strength. Thus $y$ is continuous and $[y']_a=I$, $[y']_b=K$.

Step 2: Impose both state conditions. For $d=b-a$, the pre-impact state is $$y(b-)=\frac I\omega\sin\omega d,\qquad y'(b-)=I\cos\omega d.$$ Displacement cannot jump. Therefore complete rest requires and is ensured by $$\boxed{d=\frac{n\pi}{\omega},\qquad K=(-1)^{n+1}I,\qquad n=1,2,\ldots.}$$ After these jumps the homogeneous oscillator has zero data, so it remains zero. The corresponding output has compact support and a transform for every real $s$.

Step 3: Restrict the second impulse to be positive. Positive $K$ requires odd $n$. The earliest choice is $\boxed{b=a+\pi/\omega,\ K=I}$. The net applied impulse is $2I$, yet the oscillator stops: during the intervening motion the restoring force also changes momentum. Indeed $\int_a^b\omega^2y\,dt=2I$, balancing the applied impulses in the integrated equation. Ignoring this force would give an incorrect momentum test.

Step 4: Quantify a timing error. At the actual second kick the age is $\pi/\omega+\eta>0$. The state immediately afterward is $$y_+=-\frac I\omega\sin\omega\eta,\qquad v_+=I(1-\cos\omega\eta).$$ The subsequent homogeneous amplitude is $$R=\sqrt{y_+^2+(v_+/\omega)^2} =\boxed{\frac{2I}{\omega}|\sin(\omega\eta/2)|} =I|\eta|+O(|\eta|^3).$$ For every nonzero $\eta$ in the stated interval this amplitude is positive, so the tail’s exact real transform domain is $s>0$. The plot compares perfect timing and $\eta=0.25$ for $I=\omega=a=1$.