= Solution
Let the initial conditions be $x(0)=0$ and $\dot x(0^+)=v_0>0$. The <velocity-impulse response of a damped oscillator> is
$$
\boxed{x(t)=\begin{cases}
\dfrac{v_0}{\omega_d}e^{-\gamma t}\sin(\omega_dt),&\gamma<\omega_0,\\
v_0t e^{-\omega_0t},&\gamma=\omega_0,\\
\dfrac{v_0}{s}e^{-\gamma t}\sinh(st),&\gamma>\omega_0,\quad s=\sqrt{\gamma^2-\omega_0^2}.
\end{cases}}
$$
All curves start at the origin with the same positive slope $v_0$. <Underdamping> gives a decaying sequence of positive and negative lobes. <Critical damping> gives one positive lobe followed by return to zero. <Overdamping> also remains positive for these initial data, but has a longer tail. The initial impulse changes <velocity> abruptly, not displacement.
\Image[/past-exam-of-the-physics-a-course-of-the-university-of-cambridge/2019/paper-2-b7-impulse-responses.png]
{title=Calculated displacement after the same initial velocity impulse for underdamped, critically damped and overdamped oscillators}
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