Real mechanical oscillators lose energy through friction, fluid resistance, internal material losses, and other mechanisms. A common linear model represents those losses with a viscous damping force proportional to velocity:

$$\boxed{F_d=-c\dot x},$$

where $c>0$ is the damping coefficient. The minus sign makes the force oppose the motion.

For a mass $m$ attached to an ideal spring of stiffness $k$, free motion obeys

$$\boxed{m\ddot x+c\dot x+kx=0}.$$

Define

$$\omega_0=\sqrt{\frac{k}{m}},\qquad \gamma=\frac{c}{2m}.$$

Then

$$\ddot x+2\gamma\dot x+\omega_0^2x=0.$$

The competition between the restoring tendency $\omega_0$ and the damping rate $\gamma$ determines the character of the motion.

Underdamped motion

When

$$\gamma<\omega_0,$$

the oscillator still crosses equilibrium repeatedly, but its amplitude decays. Try a damped sinusoid

$$x(t)=Ae^{-\gamma t}\cos(\omega_dt+\phi).$$

Differentiating twice and substituting into the equation shows that it is a solution when

$$\boxed{\omega_d=\sqrt{\omega_0^2-\gamma^2}}.$$

Thus

$$\boxed{x(t)=Ae^{-\gamma t}\cos(\omega_dt+\phi)}.$$

The exponential envelope $Ae^{-\gamma t}$ shrinks with time, and the damped angular frequency $\omega_d$ is slightly smaller than the undamped value $\omega_0$.

Overdamped motion

Nonoscillatory solutions can be found by trying

$$x=e^{rt}.$$

Substitution gives

$$r^2+2\gamma r+\omega_0^2=0,$$

so

$$r=-\gamma\pm\sqrt{\gamma^2-\omega_0^2}.$$

If

$$\gamma>\omega_0,$$

the two roots $r_1$ and $r_2$ are distinct, real and negative. Each exponential $e^{r_1t}$ and $e^{r_2t}$ satisfies the equation, so linearity gives

$$\boxed{x(t)=A e^{r_1t}+B e^{r_2t}}.$$

The two constants can match the two initial conditions. The displacement approaches equilibrium without sustained oscillation, so the motion is overdamped.

Critical damping

At the boundary

$$\gamma=\omega_0,$$

the two exponential rates coincide at

$$r=-\omega_0.$$

One solution is therefore $e^{-\omega_0t}$. Direct differentiation also shows that

$$t e^{-\omega_0t}$$

satisfies the critically damped equation

$$\ddot x+2\omega_0\dot x+\omega_0^2x=0.$$

The two independent solutions can therefore be combined as

$$\boxed{x(t)=(A+Bt)e^{-\omega_0t}}.$$

Critical damping separates oscillatory decay from overdamped decay and gives the fastest nonoscillatory return in this ideal second-order model.

Damping ratio

The same regimes are often summarized by the dimensionless damping ratio

$$\boxed{\zeta=\frac{c}{2\sqrt{km}}=\frac{\gamma}{\omega_0}}.$$

Then

• $0<\zeta<1$: underdamped;
• $\zeta=1$: critically damped;
• $\zeta>1$: overdamped.

Damping removes mechanical energy

For the spring-mass system,

$$E=\frac12m\dot x^2+\frac12kx^2.$$

Differentiating,

$$\frac{dE}{dt}=\dot x(m\ddot x+kx).$$

Using the equation of motion,

$$m\ddot x+kx=-c\dot x,$$

so

$$\boxed{\frac{dE}{dt}=-c\dot x^2\le0}.$$

Viscous damping therefore continuously removes mechanical energy except at instants when the velocity is zero.

Example

Let

$$m=1.0,\mathrm{kg},\qquad k=25,\mathrm{N/m},\qquad c=4.0,\mathrm{kg/s}.$$

Then

$$\omega_0=5.0,\mathrm{rad/s},\qquad \gamma=2.0,\mathrm{s^{-1}}.$$

Because $\gamma<\omega_0$, the motion is underdamped, with

$$\omega_d=\sqrt{25-4}=\sqrt{21}\approx4.58,\mathrm{rad/s}.$$

The displacement therefore oscillates while its amplitude decays with envelope $e^{-2t}$.

Viscous damping is an idealized force law, not a universal description of every loss mechanism. Its value is that it captures, with one tractable model, how dissipation changes oscillatory motion and creates underdamped, critical, and overdamped regimes.