Damped oscillations

What just happened?

You increased the damping and the oscillation amplitude dropped faster. The envelope of the displacement shrinks exponentially — the larger $\beta$, the sooner the motion dies out.

Damped oscillations:

— the displacement oscillates with frequency $\omega' = \sqrt{\omega_0^2 - \beta^2}$, but is confined within a decaying envelope $A\,e^{-\beta t}$. A — initial amplitude, β — damping coefficient (1/s), $\omega_0$ — natural frequency.

What is damping?

It is the dissipation of oscillation energy — medium resistance, friction, viscosity. Mechanical energy turns into heat, so it decreases monotonically: $E \sim e^{-2\beta t}$. The motion never "gains" energy on its own.

Three regimes — and what is critical damping?

We compare β with the natural frequency $\omega_0$:

With weak damping the system oscillates, slowly losing amplitude. Critical damping is the fastest return to equilibrium without a single oscillation. Above it — overdamping — the system also does not oscillate, but returns sluggishly.

Where does this apply?

Car shock absorbers are tuned close to critical damping: after a bump the wheel should return at once, without bouncing. The same goes for door closers, balance scales and vibration dampers in buildings.

Why you need this for the exam:

Read the regime off the graph: oscillations in a shrinking envelope = weak damping; a return without crossing zero = critical or overdamped. Remember that β increases the decay rate, and $\omega' < \omega_0$ — damped oscillations are slower than free ones.

Remember: damping dissipates energy, so the amplitude decays exponentially — $x(t)=A e^{-\beta t}\cos(\omega' t)$. Larger β → faster decay; at β = ω₀ the return is fastest and without oscillations.