Question bank — Damped oscillations — underdamped, critically damped, overdamped

Partly true and partly a trap: an underdamped system oscillates at ${\omega }_d<{\omega }_0$, so the frequency is lower — but once $\gamma \ge {\omega }_0$ there is no oscillation frequency at all, so "lowers the frequency" is meaningless there.

False. Overdamped means two real negative roots, so both terms are pure decaying exponentials — the mass approaches equilibrium monotonically (at most crossing zero once) with zero wobble.

False. It gives the fastest possible return without overshoot, but it still takes finite time — the solution $(A+Bt)e^{-\gamma t}$ only reaches zero as $t\to \infty$.

False. Increasing $b$ speeds up settling only until critical damping; beyond that (overdamped) the slow root ${\lambda }_1=-\gamma +\sqrt{{\gamma }^2-{\omega }_0^2}$ gets closer to zero and the system becomes slower again.

False. Amplitude decays as $e^{-\gamma t}$, but energy $\propto$ amplitude${}^2$, so energy decays as $e^{-2\gamma t}$ — twice as fast in the exponent.

True. Since $Q={\omega }_0/2\gamma$, setting $Q=\frac{1}{2}$ gives $\gamma ={\omega }_0$, which is exactly critical damping — the borderline of the discriminant.

False. Because it is proportional to $-\dot{x}$ with $b>0$, it always opposes the current velocity, so it can never add energy.

False. A repeated root gives only one exponential; the required second independent solution is $t{e}^{-\gamma t}$, so the general form is $(A+Bt)e^{-\gamma t}$.

Wrong to drop it: for underdamped the root is imaginary, $\sqrt{{\gamma }^2-{\omega }_0^2}=i{\omega }_d$, and that imaginary part is precisely what produces the oscillation — you keep it as $\cos ({\omega }_{d}t+\varphi )$.

Error: ${\omega }_d=0$ means no oscillation, not no motion. The mass still moves back to equilibrium via $(A+Bt)e^{-\gamma t}$; only the periodic part vanishes.

Sign flipped: overdamped requires ${\gamma }^2-{\omega }_0^2>0$ (discriminant positive) to make the square root real; negative would give the imaginary roots of the underdamped case.

Overkill: overdamping stops the bounce but returns sluggishly, feeling mushy over bumps. Engineers aim near critical damping for the fastest bounce-free recovery.

Only for zero damping: with damping present you actually measure ${\omega }_d=\sqrt{{\omega }_0^2-{\gamma }^2}<{\omega }_0$; ${\omega }_0$ is the idealised undamped value.

Backwards: high $Q={\omega }_0/2\gamma$ means small $\gamma$, i.e. light damping — the fork rings for many cycles and loses energy slowly.

It's not just cosmetic: dividing by $m$ exposes the physically meaningful $\gamma =b/2m$ and ${\omega }_0=\sqrt{k/m}$, which are the quantities the whole classification depends on.

Because viscous fluid/air drag at low speed genuinely scales with speed — the faster you push through the medium, the harder it resists — making $-b\dot{x}$ the simplest physically realistic loss term.

Because differentiating $e^{\lambda t}$ just multiplies it by $\lambda$, turning the differential equation into an ordinary algebraic (quadratic) equation in $\lambda$ that we can solve directly.

By Euler's relation $e^{i{\omega }_{d}t}=\cos {\omega }_{d}t+i\sin {\omega }_{d}t$, so an imaginary exponent generates sines and cosines — the mathematical fingerprint of back-and-forth motion.

Pure convenience: it makes the quadratic roots come out as $-\gamma \pm \sqrt{{\gamma }^2-{\omega }_0^2}$ with clean coefficients rather than trailing factors of $\frac{1}{2}$.

Heavier (overdamped) damping introduces a slow root ${\lambda }_1$ near zero that lingers; at exactly critical, the decay exponent $-\gamma$ is as large in magnitude as possible while still avoiding overshoot.

Amplitude falls like $e^{-\gamma t}$, and stored energy $\sim \frac{1}{2}k{A}^2$ depends on amplitude squared, so $E\propto (e^{-\gamma t}{)}^2=e^{-2\gamma t}$.

Because both roots are real, the solution is a sum of two decaying exponentials with different rates — there is no periodic term, hence no frequency to quote.

The roots become $\lambda =\pm i{\omega }_0$, giving pure Simple Harmonic Motion at ${\omega }_0$ with a constant-amplitude, never-ending oscillation.

The discriminant is zero, so there is one repeated root $\lambda =-\gamma$ and the solution is $(A+Bt)e^{-\gamma t}$ — a single, possibly-once-crossing, non-oscillatory return.

The fast root ${\lambda }_2\approx -2\gamma$ dies almost instantly, while the slow root ${\lambda }_1\approx -{\omega }_0^2/2\gamma \to 0^{-}$ becomes ever slower, so the system creeps back extremely sluggishly.

Yes — it oscillates within the shrinking envelope $\pm A{e}^{-\gamma t}$, crossing equilibrium many times, each swing smaller than the last.

At most once: depending on initial velocity it may overshoot the equilibrium a single time, but it can never return through it periodically because there is no oscillatory term.

Very small and real, so the system oscillates at a very low frequency with a long period, while the envelope $e^{-\gamma t}$ decays comparatively fast — you see barely one or two faint wobbles.

We've idealised the system as loss-free; the real oscillator would still lose energy, so a genuine $b=0$ (undamped SHM) is a mathematical idealisation, connecting to Forced Oscillations & Resonance where energy is instead resupplied.

Identically: $R$ plays the role of $b$, so $R^2<4L/C$ is underdamped, $R^2=4L/C$ critical, and $R^2>4L/C$ overdamped — the same discriminant governs both via Second-order linear ODEs.