Underdamping

For $0<\zeta <1$, define the damped angular frequency ${\omega }_{\mathrm{d}}={\omega }_0\sqrt{1-{\zeta }^2}$. The initial value problem has solutionEvery nonzero underdamped response crosses equilibrium repeatedly. Its oscillation period is $T_{\mathrm{d}}=2\pi /{\omega }_{\mathrm{d}}$, not $2\pi /{\omega }_0$. The amplitude envelope decays as $e^{-\zeta {\omega }_{0}t}$, but the actual displacement also changes sign. The envelope coefficient need not equal $|x_0|$: initial velocity fixes the oscillation phase as well as that coefficient.

For $\ddot{x}+2\beta \dot{x}+{\omega }_0^{2}x=0$ with $0<\beta <{\omega }_0$, put ${\omega }_d=\sqrt{{\omega }_0^2-{\beta }^2}$. The initial conditions $x(0)=x_0$ and $\dot{x}(0)=0$ giveThus the turning points have the times and displacements in the title. For release from $x_0<0$, the first positive maximum is $|x_0|e^{-\beta \pi /{\omega }_d}$, sufficient to test whether the body reaches a fixed positive stop before that first maximum. Every later positive maximum is smaller.