3Free damped vibration
Add the damper: $m\ddot x + c\dot x + kx=0$. With $x=e^{st}$, the characteristic equation $s^2+2\zeta\omega_n s+\omega_n^2=0$ has roots
$$ s_{1,2} = -\zeta\omega_n \pm \omega_n\sqrt{\zeta^2-1} $$Where does the characteristic equation come from?
Try a solution of the form $x=Ae^{\lambda t}$. Then $\dot x=\lambda Ae^{\lambda t}$ and $\ddot x=\lambda^2Ae^{\lambda t}$, and substituting into $M\ddot x+C\dot x+Kx=0$ gives
$$\left(M\lambda^2+C\lambda+K\right)Ae^{\lambda t}=0.$$The exponential is never zero, and $A=0$ is the trivial "nothing moves" solution, so any actual motion requires
$$M\lambda^2+C\lambda+K=0\quad\Longrightarrow\quad \lambda=\frac{-C\pm\sqrt{C^2-4MK}}{2M}=\left(-\zeta\pm\sqrt{\zeta^2-1}\right)\omega_n,$$with $\omega_n=\sqrt{K/M}$ and $\zeta=C/2\sqrt{KM}$. Dividing by $m$ is what turns $s$ (or $\lambda$) into the form shown above.
The sign under the square root splits the behavior into three cases:
Underdamped, $\zeta<1$. Complex roots. The mass oscillates inside a decaying envelope, at the damped natural frequency $\omega_d=\omega_n\sqrt{1-\zeta^2}$:
$$ x(t)=e^{-\zeta\omega_n t}\left[x_0\cos\omega_d t + \frac{v_0+\zeta\omega_n x_0}{\omega_d}\sin\omega_d t\right] $$Critically damped, $\zeta=1$. Repeated root. Returns to equilibrium as fast as possible without oscillating: $x(t)=\left[x_0+(v_0+\omega_n x_0)t\right]e^{-\omega_n t}$.
Overdamped, $\zeta>1$. Two real negative roots. A slow, sluggish creep back: $x(t)=C_1e^{s_1 t}+C_2e^{s_2 t}$.
From complex roots to cosines and sines
With two distinct roots the general solution is $x=A_1e^{\lambda_1t}+A_2e^{\lambda_2t}$. In the underdamped case $\lambda_{1,2}=-\zeta\omega_n\pm i\omega_d$, so
$$x=e^{-\zeta\omega_nt}\left(A_1e^{i\omega_dt}+A_2e^{-i\omega_dt}\right).$$Euler's formula $e^{\pm i\omega_dt}=\cos\omega_dt\pm i\sin\omega_dt$ turns the bracket into a combination of cosine and sine:
$$x=e^{-\zeta\omega_nt}\left(C\cos\omega_dt+S\sin\omega_dt\right),\qquad C=A_1+A_2,\quad S=i(A_1-A_2).$$For a real motion $A_1$ and $A_2$ are complex conjugates, which makes $C$ and $S$ real. The initial conditions give $C=x_0$ and $S=(v_0+\zeta\omega_nx_0)/\omega_d$. With $\zeta=0$ this reduces to the undamped $x=C\cos\omega_nt+S\sin\omega_nt$, which never decays.
The roots $s_{1,2}$ are the whole story. Their real part sets how fast the motion decays, and their imaginary part sets how fast it oscillates. Watch them move as the damping changes:
Roots in the complex plane (dashed circle: radius ωₙ). The shaded right half means growth.
The response for those roots, released from x₀ with v₀ = 0.
Underdamped roots slide around the circle toward the real axis; at ζ = 1 they meet; beyond that they split along the real axis. One root then creeps toward zero, which is why heavy damping is slow.
Measuring damping: the logarithmic decrement. For an underdamped system, the ratio of successive peaks is constant. If $x_1$ and $x_{n+1}$ are peaks $n$ cycles apart,
$$ \delta = \frac1n\ln\frac{x_1}{x_{n+1}} = \frac{2\pi\zeta}{\sqrt{1-\zeta^2}} \qquad\Longleftrightarrow\qquad \zeta=\frac{\delta}{\sqrt{4\pi^2+\delta^2}} $$This is how damping is measured in the lab: pluck the structure and read the peaks. The demo below does exactly that on the simulated response.
All three regimes side by side ($v_0=0$)
Try this
- Use the presets. Which returns to rest fastest: $\zeta=0.5$, $1$, or $2$?
- With $\zeta=0.05$, compare $\omega_d$ and $\omega_n$. Light damping barely changes the frequency, but strongly affects decay.
- Check that "ζ from δ" agrees with the slider value. This is your virtual lab measurement.
- Overdamped, $x_0>0$, large negative $v_0$: can it overshoot? How many times can it cross zero at most?
Different systems, same motion
Divide the equation of motion by $M$ and it becomes the standard form $\ddot x+2\zeta\omega_n\dot x+\omega_n^2x=f(t)/M$. The free response therefore depends on only two numbers, $\zeta$ and $\omega_n$, plus the initial conditions, not on $M$, $C$ and $K$ individually. Two systems that look nothing alike, with different masses, geometry, springs and dampers, will vibrate identically if their $\zeta$ and $\omega_n$ match. The standard form makes that visible at a glance, and it also tells you immediately whether a system is under-, critically or overdamped.
Each system has different m, c and k (listed on the left), chosen so that its ζ and ωₙ match the sliders. Released from the same x₀, all three move in lockstep.
Where does the energy go? In free vibration nothing adds energy, and the damper removes it as heat. Watch the energy shift from the spring and the mass into the damper:
The total never changes. Energy only moves from kinetic and potential into heat, fastest when the mass moves fastest (the damper force is cẋ).
Old car-mechanic test: push down hard on one corner of a parked car and let go. A healthy suspension settles after about one rebound. If it keeps bouncing, the shock absorbers (the dampers) are worn and ζ has dropped too low.