Dirac Delta Function — Question 3

Question 3 – Solution

Strategy. The impulse changes velocity, not displacement. Its work depends on the velocity immediately before the kick as well as its strength.

Step 1: Jump, transform and invert. Integrating across $a$ gives $[y']_a=6$, because the integral of $4y$ tends to zero. With $[y]_a=0$, the transformed equation is $$Y=\frac{s+6e^{-as}}{s^2+4},\qquad \boxed{y(t)=\cos 2t+3H_a(t)\sin(2(t-a)).}$$ The new term vanishes at activation, while its derivative starts at $6$. Both pieces satisfy $y''+4y=0$ away from $a$ and retain the initial data.

Step 2: Evaluate the state at impact. Since $2a=\pi/2$, $$(y(a-),y'(a-))=(0,-2),\qquad (y(a+),y'(a+))=\boxed{(0,4)}.$$ For $t=a+\tau$, the pre-existing response is $-\sin 2\tau$, so the total post-impact motion is $\boxed{y(a+\tau)=2\sin 2\tau}$. Thus the impulse reverses the direction and doubles the amplitude.

Step 3: Compute the energy transfer. The energies are $E_-=2$ and $E_+=8$, giving $\boxed{\Delta E=6}$. For any impulse $J$, continuity of position and $v_+=v_-+J$ imply $$\Delta E=\tfrac 12[(v_-+J)^2-v_-^2] =Jv_-+\tfrac 12J^2.$$ At this impact time $v_-=-2$, hence $$\boxed{\Delta E=\tfrac 12J(J-4),\qquad E_+=\tfrac 12(J-2)^2.}$$

Step 4: Classify the positive kicks. For $0<J<4$ the energy decreases; at $J=4$ it is unchanged; for $J>4$ it increases. A positive impulse can remove energy when it opposes the incoming velocity. Complete rest requires both post-impact state coordinates to vanish. Position is already zero here, so the unique strength is $\boxed{J=2}$, which makes velocity zero. A zero velocity kick at a nonzero displacement would not in general leave an oscillator at rest. The plot shows the stated $J=6$ case, with continuous displacement and a corner.