In short

A damped oscillator is a harmonic oscillator with an additional velocity-proportional force that drains energy — typically drag from air or viscosity. The equation of motion is

m\ddot x + b\dot x + kx = 0,

with b > 0 the damping coefficient and \omega_0 = \sqrt{k/m} the natural (undamped) frequency. Three regimes emerge, set by the size of b:

  • Underdamped (b < 2m\omega_0): oscillates with decaying amplitude,
x(t) = A_0 \, e^{-\gamma t}\sin(\omega' t + \varphi), \quad \gamma = \frac{b}{2m}, \quad \omega' = \sqrt{\omega_0^2 - \gamma^2}.

The damped frequency \omega' is slightly less than \omega_0. The envelope A_0 e^{-\gamma t} shrinks with characteristic time \tau = 1/\gamma.

  • Critically damped (b = 2m\omega_0): returns to equilibrium fastest without overshooting — the sweet spot for shock absorbers in Indian cars and auto-rickshaws.
  • Overdamped (b > 2m\omega_0): returns without oscillating, but slowly.

Energy decays as E(t) = E_0 \, e^{-2\gamma t} (twice as fast as the amplitude, because energy \propto A^2).

Pluck a string on a sitar at a concert in Kolkata and a bright note rings out. Listen for a few seconds: the loudness drops, the vibration fades, and within a minute the string is still again. Watch a park swing in Delhi after the child has jumped off: the swing moves through a shallower and shallower arc until it settles. Drop a stone into the suspension of an autorickshaw and the vehicle bounces twice, maybe three times, then stops. Every real oscillator you have ever met does this — it loses energy to its surroundings, and the amplitude of its motion fades away.

An ideal SHM, the one treated in Equation of SHM and Phase, never loses energy. Its amplitude is constant forever. That is the textbook idealisation. Now it is time to admit what reality looks like — and it turns out that real damped oscillation is mathematically almost as clean as ideal SHM, with just one new parameter (the damping) and one new piece of mathematics (an exponential envelope).

This article asks: what happens when you add a velocity-proportional friction force F_\text{damp} = -b v to a harmonic oscillator? You will see three very different behaviours emerge — underdamped, critically damped, overdamped — separated by a clean critical condition. You will derive the underdamped solution x(t) = A_0 \, e^{-\gamma t}\sin(\omega' t + \varphi) and see exactly why the damped frequency \omega' is slightly less than the natural frequency \omega_0. You will learn why a car's shock absorber is tuned to sit right at the critical condition — any less and the car oscillates, any more and it responds sluggishly to bumps. And you will see, in closing, the central role damped oscillators play as the bridge between SHM (in the limit b \to 0) and the forced, resonant oscillators of the next chapter.

A velocity-proportional damping force

In almost every real oscillator, the dominant energy loss mechanism at moderate speeds is viscous drag — a force from the surrounding fluid (air, oil, water) that grows with the object's speed. Stokes's law for a sphere moving slowly through a fluid says F_\text{drag} = 6\pi\eta r v, where \eta is the fluid viscosity, r is the sphere's radius, and v is its speed. The drag is proportional to the velocity, and it points opposite to the velocity — if the bob moves right, drag pushes left; if the bob moves left, drag pushes right.

For an oscillator, this translates to a damping force

F_\text{damp} = -b v = -b \dot x,

where b > 0 is the damping coefficient, with SI units of kg/s (or equivalently N·s/m). The minus sign is essential: F_\text{damp} always opposes the motion, so it always does negative work on the bob, draining energy from the oscillation.

This velocity-proportional model is not the only one — for high speeds, drag grows faster (roughly as v^2 for turbulent flow); for solid friction, the damping is almost independent of speed (the "kinetic friction" model). But the linear-in-velocity model is the dominant one at low speeds, and — crucially — it is the only one that leads to a clean, closed-form solution to the equation of motion. That is why every introductory treatment (including this one) uses it.

Assumptions for this article: the restoring force is linear (Hooke's law, F = -kx); the damping is linear in velocity (F = -b\dot x); there is no external driving force. The mass, spring constant, and damping coefficient are all constant in time.

The equation of motion

Add the restoring force -kx and the damping force -b\dot x, and apply Newton's second law:

m\ddot x = -kx - b\dot x.

Rearrange to put everything on the left:

\boxed{\; m\ddot x + b\dot x + kx = 0. \;} \tag{1}

This is a second-order linear ordinary differential equation with constant coefficients. Every term is proportional to x or one of its derivatives; nothing involves x^2 or \sin x or any other non-linear mixing. That linearity is what makes the equation solvable, and it is what makes the solutions behave in the remarkably clean way you are about to see.

It is convenient to rewrite equation (1) in terms of two standard parameters:

Dividing equation (1) by m gives

\ddot x + 2\gamma \dot x + \omega_0^2 x = 0. \tag{1'}

Equation (1') has only two physical parameters, \gamma and \omega_0, and everything about the behaviour depends on their ratio.

Solving equation (1') — try an exponential

The standard move for a linear ODE with constant coefficients is to guess that the solution is an exponential, x(t) = e^{\lambda t} for some (possibly complex) number \lambda, and then plug in and see what \lambda must be.

Step 1. Compute the derivatives of the trial solution.

If x = e^{\lambda t}, then \dot x = \lambda e^{\lambda t} and \ddot x = \lambda^2 e^{\lambda t}.

Step 2. Substitute into equation (1').

\lambda^2 e^{\lambda t} + 2\gamma \lambda e^{\lambda t} + \omega_0^2 e^{\lambda t} = 0.

Factor out e^{\lambda t} (which is never zero):

\lambda^2 + 2\gamma\lambda + \omega_0^2 = 0. \tag{2}

Why: the differential equation \ddot x + 2\gamma\dot x + \omega_0^2 x = 0 has been converted, for an exponential trial, into an algebraic equation in \lambda. This is the characteristic equation of the ODE.

Step 3. Solve the quadratic for \lambda.

Using the quadratic formula:

\lambda = \frac{-2\gamma \pm \sqrt{4\gamma^2 - 4\omega_0^2}}{2} = -\gamma \pm \sqrt{\gamma^2 - \omega_0^2}. \tag{3}

The quantity under the square root is \gamma^2 - \omega_0^2, and its sign changes everything.

Three regimes — the sign of \gamma^2 - \omega_0^2

Three cases, set by how \gamma compares to \omega_0:

Regime Condition Physical meaning
Underdamped \gamma < \omega_0 \gamma^2 - \omega_0^2 < 0: the discriminant is negative, roots are complex. Solutions oscillate with decaying amplitude.
Critically damped \gamma = \omega_0 Discriminant is zero, roots are real and equal. Fastest non-oscillatory return to equilibrium.
Overdamped \gamma > \omega_0 Discriminant is positive, roots are real and negative. Slow, non-oscillatory return to equilibrium.

Equivalently, since \gamma = b/(2m) and \omega_0 = \sqrt{k/m}, the condition \gamma < \omega_0 becomes b < 2\sqrt{mk} or equivalently b < 2m\omega_0. The critical damping coefficient is

b_\text{crit} = 2m\omega_0 = 2\sqrt{mk}.

Below b_\text{crit} the system rings. Right at b_\text{crit} the system snaps back to equilibrium as fast as possible without oscillating. Above b_\text{crit} the system returns slowly and sluggishly. Engineers design shock absorbers to sit right at critical damping — or very slightly below — for the best ride.

The underdamped solution — the one you should memorise

Take \gamma < \omega_0. Write the square root as

\sqrt{\gamma^2 - \omega_0^2} = \sqrt{-(\omega_0^2 - \gamma^2)} = i\sqrt{\omega_0^2 - \gamma^2} = i\omega',

where

\omega' = \sqrt{\omega_0^2 - \gamma^2}

is the damped angular frequency — a real number, because \omega_0 > \gamma in this regime.

So the two roots of the characteristic equation are

\lambda_\pm = -\gamma \pm i\omega'.

Each of e^{\lambda_+ t} and e^{\lambda_- t} is a solution to the ODE. By linearity, any linear combination is also a solution. The general solution is

x(t) = c_+ e^{(-\gamma + i\omega')t} + c_- e^{(-\gamma - i\omega')t} = e^{-\gamma t}\bigl(c_+ e^{i\omega' t} + c_- e^{-i\omega' t}\bigr),

where c_+ and c_- are complex constants.

For the physical position x(t) to be real, c_+ and c_- must be complex conjugates. Using Euler's formula e^{i\omega't} = \cos\omega't + i\sin\omega't, the combination inside the parentheses reduces to a real sinusoid with some amplitude A_0 and phase \varphi:

\boxed{\; x(t) = A_0 \, e^{-\gamma t} \sin(\omega' t + \varphi), \qquad \omega' = \sqrt{\omega_0^2 - \gamma^2}. \;} \tag{4}

Why: two real constants (A_0 and \varphi) are needed to match the two initial conditions (initial position and initial velocity) of a second-order ODE. The exponentially-decaying sine form captures both of them cleanly.

This is the equation of the underdamped oscillator. It says that the motion is a sinusoid (oscillation at frequency \omega') multiplied by an exponential envelope e^{-\gamma t} that shrinks toward zero with a characteristic time \tau = 1/\gamma.

The envelope and the oscillation

Plot x(t) versus t and you see two things happening simultaneously. At every instant, x(t) is somewhere between -A_0 e^{-\gamma t} and +A_0 e^{-\gamma t}. The two exponential curves form an envelope — an imaginary upper and lower bound that the oscillation slides between, itself shrinking with time. Within the envelope, the oscillation sweeps back and forth at the damped frequency \omega'. Every time the sinusoidal part hits a peak, the motion touches the envelope; between peaks, it is inside.

The peaks of consecutive oscillations (separated by a period T' = 2\pi/\omega') have magnitudes that differ by a ratio:

\frac{x_n}{x_{n+1}} = \frac{A_0 e^{-\gamma t_n}}{A_0 e^{-\gamma (t_n + T')}} = e^{\gamma T'}.

The logarithmic decrement \Lambda = \ln(x_n/x_{n+1}) = \gamma T' is a dimensionless measure of how fast the oscillation dies. Experimentally, you measure \Lambda by timing a few successive peak amplitudes in a damped oscillator; that gives \gamma T', and (if you know \omega_0 separately) \gamma itself.

Animated: underdamped oscillator position versus time A red dot traces out the function x(t) = exp(minus 0.3 times t) times sine of 2 t, starting at amplitude 1 and decaying over 10 seconds. Two dashed curves mark the plus and minus exponential envelope exp(minus 0.3 t). The oscillation touches the envelope at each peak and moves inside it between peaks, with a characteristic time of about 3.3 seconds. t (s) x 0 +1 −1 1 5 10 +e^(−γ t) −e^(−γ t)
Position of an underdamped oscillator with $\omega_0^2 = \omega'^2 + \gamma^2$, $\omega' = 2$ rad/s, $\gamma = 0.3$ /s. The red dot traces $x(t) = e^{-0.3 t}\sin(2t)$; the dashed curves are the envelope $\pm e^{-0.3 t}$. Every peak of the oscillation just touches the envelope. After one characteristic time $\tau = 1/\gamma \approx 3.3$ s, the envelope has shrunk to $1/e \approx 37\%$ of its initial height. Click replay to watch again.

Why \omega' < \omega_0 — the damping slows the ringing

The damped angular frequency \omega' = \sqrt{\omega_0^2 - \gamma^2} is always less than the natural frequency \omega_0. This is a subtle but real effect: adding damping makes the oscillator ring at a slightly slower tempo.

For light damping, expand the square root:

\omega' = \omega_0\sqrt{1 - \gamma^2/\omega_0^2} \approx \omega_0\left(1 - \frac{1}{2}\frac{\gamma^2}{\omega_0^2}\right) = \omega_0 - \frac{\gamma^2}{2\omega_0}.

The shift is quadratic in \gamma, so for small damping the effect is small. An oscillator with \gamma/\omega_0 = 0.1 (a reasonably well-damped but still clearly ringing system) has \omega'/\omega_0 = \sqrt{1 - 0.01} \approx 0.995 — a 0.5% shift. For \gamma/\omega_0 = 0.5 the shift becomes \omega'/\omega_0 \approx 0.866, about 13% slower than the undamped natural frequency.

Physical intuition: damping removes kinetic energy from the oscillator most effectively when the speed is largest (since F_\text{damp} \cdot v is the power dissipated, and F_\text{damp} \propto v, so power \propto v^2). Each swing is slowed by the drag it experiences at high speed, lengthening the time between maximum excursions. The slightly lower frequency is the macroscopic consequence.

Critically damped and overdamped — the non-oscillatory regimes

Return to the characteristic equation \lambda^2 + 2\gamma\lambda + \omega_0^2 = 0 and consider the other two cases.

Critical damping, \gamma = \omega_0

The discriminant \gamma^2 - \omega_0^2 = 0, so both roots coincide: \lambda = -\gamma. One solution is therefore e^{-\gamma t}. But a second-order ODE has two linearly independent solutions. The second one is t\, e^{-\gamma t} (a standard result from ODE theory — when the characteristic equation has a repeated root, multiplying by t gives you the second solution).

The general critically-damped solution is

x(t) = (A + Bt)\, e^{-\gamma t}, \tag{5}

with A and B two real constants set by the initial conditions. The motion is an exponential decay modified by a linear-in-t factor. It does not oscillate; depending on the initial velocity, it may overshoot zero at most once before returning monotonically to equilibrium.

Why critical damping matters engineeringly. Among all values of b, b = b_\text{crit} = 2m\omega_0 gives the fastest return to equilibrium without any oscillation. A car's shock absorber is tuned close to this condition: when you drive over a pothole, the spring-mass-damper system in the suspension should return to its ride height as quickly as possible, but not oscillate up and down (which would be uncomfortable). An over-damped car would wallow; an underdamped car would bounce. A critically-damped car settles cleanly. The same design logic shows up in door-closing hinges (critically damped so the door closes firmly but doesn't slam), in analog meter movements (so the needle stops at the reading quickly), and in camera shutters.

Overdamping, \gamma > \omega_0

The discriminant \gamma^2 - \omega_0^2 > 0, and both roots

\lambda_\pm = -\gamma \pm \sqrt{\gamma^2 - \omega_0^2}

are real. Since \sqrt{\gamma^2 - \omega_0^2} < \gamma, both \lambda_+ and \lambda_- are negative. The general solution is

x(t) = A e^{\lambda_+ t} + B e^{\lambda_- t}, \tag{6}

a combination of two decaying exponentials with different time constants. The slower of the two (\lambda_+, the less-negative root) dominates at long times, and the system creeps back to equilibrium over a time \sim 1/|\lambda_+|.

An overdamped system takes longer to return to equilibrium than a critically-damped one. It is sometimes unavoidable — a very viscous fluid (think of dropping a marble into honey) — but it is rarely the design goal.

Exploring the three regimes — the interactive

Drag the damping coefficient b below (with m = 1 kg and k = 1 N/m fixed, so \omega_0 = 1 rad/s and b_\text{crit} = 2) to see the three regimes. The curve shows the response x(t) to an initial unit displacement with zero velocity.

Interactive: x(t) for different damping coefficients b A curve shows the displacement x(t) of a unit-mass unit-spring oscillator released from rest at x equals 1, for varying damping coefficient b between 0.1 and 4. The curve reshapes as b changes. For small b the motion oscillates and decays slowly; at b equal 2 (critical) the motion drops monotonically to zero; above b equal 2 the motion drops slowly without oscillating. Readouts display b, the damped frequency omega prime, and the regime. t (s) x(t) 0 1 −0.5 5 10 15 2 1 0 b (kg/s) b_crit = 2 b_crit = 2 (below = underdamped) drag red dot vertically to change b
The response $x(t)$ of a unit-mass, unit-spring oscillator (so $\omega_0 = 1$ rad/s, $b_\text{crit} = 2$) released from $x = 1$ at rest, as a function of the damping coefficient $b$. Drag the red point vertically from 0.1 up toward the critical value 2. Small $b$ gives many cycles of slowly-decaying ringing; as $b$ grows, the ringing fades faster and the damped frequency $\omega' = \sqrt{1 - b^2/4}$ shrinks; approaching $b = 2$ the motion stops oscillating and drops monotonically to zero — the critical damping limit. (Beyond $b = 2$, into the overdamped regime, the solution becomes a sum of two real decaying exponentials rather than a decaying sinusoid; the plot here stays inside the underdamped regime.)

Energy decay — amplitude² means double the rate

From Energy in SHM, the total energy of a harmonic oscillator is E = \tfrac{1}{2} k A^2, proportional to the amplitude squared. For an underdamped damped oscillator, the envelope is A(t) = A_0 e^{-\gamma t}, so the total energy — averaged over one oscillation (to smooth out the rapid trade between K and U) — is

\boxed{\; E(t) = \tfrac{1}{2} k [A_0 e^{-\gamma t}]^2 = E_0 \, e^{-2\gamma t}, \qquad E_0 = \tfrac{1}{2} k A_0^2. \;} \tag{7}

The energy decays with a time constant \tau_E = 1/(2\gamma) — half the time constant of the amplitude. After one amplitude time constant 1/\gamma, the amplitude has dropped to 1/e \approx 37\% but the energy has dropped to 1/e^2 \approx 13.5\%.

This is directly verifiable. Compute dE/dt from the equation of motion (using E = \tfrac{1}{2} m\dot x^2 + \tfrac{1}{2} k x^2 and Newton's second law m\ddot x = -kx - b\dot x):

\frac{dE}{dt} = m\dot x \ddot x + k x \dot x = \dot x(m\ddot x + kx) = \dot x \cdot (-b\dot x) = -b\dot x^2.

Why: the damping force does work on the bob at rate F_\text{damp} \cdot v = -b\dot x \cdot \dot x = -b\dot x^2. That power leaves the oscillator — it becomes heat in the fluid. The rate of energy loss is always negative (damping never adds energy), and proportional to the instantaneous kinetic energy (since \dot x^2 is twice the kinetic energy per unit mass).

Averaging \dot x^2 over a cycle for the underdamped solution (using \langle \cos^2 \rangle = 1/2), you get \langle dE/dt \rangle = -2\gamma \langle K \rangle = -\gamma \langle E \rangle... which, if you be careful, works out to the cycle-averaged E(t) \propto e^{-2\gamma t} of equation (7).

The quality factor Q

Engineers package the damping behaviour into a single dimensionless number, the quality factor:

Q = \frac{\omega_0}{2\gamma} = \frac{m\omega_0}{b}.

More precisely, Q/(2\pi) is approximately the number of oscillations required for the amplitude to fall to e^{-\pi} \approx 4\% of its initial value. A very-high-Q system is one that "rings" cleanly — a crystal oscillator in a quartz wristwatch has Q of order 10^5, a guitar string has Q of order a few hundred, the suspension of a Maruti WagonR has Q of order 1 (designed to be close to critical).

The quality factor reappears centrally in Forced Oscillations and Resonance, where it determines the sharpness of the resonance peak.

Worked examples

Example 1: The guitar string that rings for 3 seconds

A guitar string oscillates at f_0 = 440 Hz (musical A) after being plucked. You observe that the sound's amplitude has fallen to 1/100 of its initial value after 3 seconds. Find the damping parameter \gamma, the quality factor Q, and the number of oscillations in that 3-second window.

Envelope decay of a guitar string by a factor of 100 in 3 seconds A graph of exponential envelope decay. At t = 0 the amplitude is at A0. At t = 3 s the amplitude has dropped to A0 over 100, about 1 percent. A dashed horizontal line marks the 1/100 level; a vertical line at t = 3 s marks where the envelope crosses that level. t (s) amplitude A₀/100 3 s A₀ A(t) = A₀ e^(−γ t)
The envelope decays by a factor of 100 over 3 seconds — so $e^{-\gamma \cdot 3} = 1/100$.

Step 1. Extract \gamma from the envelope decay.

e^{-\gamma \cdot 3} = \frac{1}{100} \quad\Longrightarrow\quad \gamma \cdot 3 = \ln 100 \approx 4.605.
\gamma \approx \frac{4.605}{3} \approx 1.535 \text{ s}^{-1}.

Why: take natural log of both sides. \ln 100 = \ln 10^2 = 2\ln 10 \approx 2 \times 2.303 = 4.606.

Step 2. Compute the angular frequency and quality factor.

\omega_0 = 2\pi f_0 = 2\pi \times 440 \approx 2764 \text{ rad/s}.

(Since damping is so small compared to \omega_0, we can use \omega_0 \approx \omega' to excellent accuracy: \omega'/\omega_0 = \sqrt{1 - \gamma^2/\omega_0^2} = \sqrt{1 - (1.535/2764)^2} \approx 1.)

Q = \frac{\omega_0}{2\gamma} = \frac{2764}{2 \times 1.535} = \frac{2764}{3.07} \approx 900.

Step 3. Number of oscillations in 3 seconds.

N = f_0 \times t = 440 \times 3 = 1320 \text{ oscillations}.

Why: the string goes through 440 full cycles per second, over 3 seconds. Because the damping is so light, each period is essentially 1/440 s; the damped period correction is negligible (a 10^{-7} effect).

Step 4. Consistency check via the number-of-cycles-per-e-folding formula.

The number of cycles for the amplitude to fall by a factor of e is N_e = f_0/\gamma = 440/1.535 \approx 287, which agrees with N_e = Q/\pi \approx 900/\pi \approx 286. And to drop by a factor of 100 (i.e. e^{\ln 100} = e^{4.605}) takes 4.605 \times N_e \approx 1320 cycles — matching the 1320 cycles observed in the 3-second window.

Why: the amplitude ratio A_0/A(t) = e^{\gamma t}, and the number of full oscillations in that time is N = f_0 t. So \gamma t = \ln(A_0/A) becomes N = f_0 \ln(A_0/A)/\gamma = (f_0/\gamma)\ln(A_0/A) = N_e \ln(A_0/A). With N_e \approx 287 and \ln 100 \approx 4.605, the product is \approx 1321, a near-exact match.

Result: \gamma \approx 1.54 /s, Q \approx 900, and the string completes about 1320 oscillations in the 3 seconds of observable ringing. The envelope has dropped to 1% by then.

What this shows: A well-constructed stringed instrument is a high-Q oscillator. A Q of 900 means the damping per cycle is very small — only a 0.35% amplitude decrease per oscillation. This is why a plucked sitar string rings for tens of seconds: the energy leaks out slowly, and the instrument's listeners hear a clean, sustained tone.

Example 2: Designing a critically-damped car suspension

An Indian hatchback has a mass of 1200 kg (shared equally among four wheels, so 300 kg per wheel). Each wheel has a spring with k = 30{,}000 N/m. Find the critical damping coefficient per wheel and the time for the car to return to its ride height after a bump.

Car suspension modelled as spring-damper system under each wheel A schematic of one corner of a car suspension. A car body of 300 kg sits on top of a spring (coil drawn zig-zag) and a damper (cylinder with piston drawn alongside the spring). The spring has stiffness 30 kN per metre. The damping coefficient b is to be found. car body (m = 300 kg per wheel) spring k = 30 kN/m damper b = ? road (wheel not shown)
A quarter-car model of a suspension: a body mass $m$ on a spring of stiffness $k$ in parallel with a damper of coefficient $b$. The designer's job is to pick $b$ so the car returns to ride height promptly without wallowing.

Step 1. Compute the natural frequency.

\omega_0 = \sqrt{\frac{k}{m}} = \sqrt{\frac{30{,}000}{300}} = \sqrt{100} = 10 \text{ rad/s}.

The corresponding time period is T_0 = 2\pi/\omega_0 \approx 0.63 s, and the frequency f_0 \approx 1.59 Hz — close to the 1–2 Hz range typical of passenger-car suspensions.

Step 2. Compute the critical damping coefficient.

b_\text{crit} = 2m\omega_0 = 2 \times 300 \times 10 = 6000 \text{ N·s/m}.

A real-world damper on a car of this size is typically somewhere between b_\text{crit} and 0.5 b_\text{crit} — engineers prefer slight underdamping (b \approx 0.7 b_\text{crit}, so Q \approx 0.7) because a critically-damped car can feel slow to respond to small bumps, and a bit of controlled ringing is subjectively more comfortable.

Step 3. Time to return to equilibrium at critical damping.

For a critically-damped system, x(t) = (A + Bt)e^{-\gamma t} with \gamma = \omega_0. After one 1/\gamma = 1/10 = 0.1 s, the envelope has dropped to 1/e \approx 37\%; after three time constants (0.3 s), it has dropped to 5\%. The full settling time is about 3/\gamma = 0.3 s.

Why: at critical damping, \gamma = \omega_0 = 10 /s, and the pure exponential part e^{-\gamma t} decays with time constant 1/\gamma = 0.1 s. The (A + Bt) prefactor grows only linearly and is overwhelmed by the exponential for t \gg 1/\gamma.

Step 4. Underdamped comparison.

If the damper were half as strong, b = 3000 N·s/m, then \gamma = b/(2m) = 3000/600 = 5 /s. The damped frequency would be \omega' = \sqrt{\omega_0^2 - \gamma^2} = \sqrt{100 - 25} \approx 8.66 rad/s, so the car would oscillate with period T' = 2\pi/\omega' \approx 0.73 s (about 14% longer than undamped). Each bounce would be e^{-5 \times 0.73} \approx 2.7\% of the previous — so after 3 bounces (\approx 2 s) the motion is essentially gone. Noticeably more wallowing than a critically-damped ride.

Result: b_\text{crit} = 6000 N·s/m per wheel. A critically-damped car settles after a bump in about 0.3 seconds, with no bouncing.

What this shows: The critical damping coefficient is b_\text{crit} = 2\sqrt{mk}, and it depends on both the mass of the vehicle and the stiffness of its springs. A heavier SUV needs stronger dampers than a light hatchback to stay critically damped with the same springs; a sports car with stiffer springs also needs stronger dampers. Tuning these to the right balance is most of what a suspension engineer does.

Common confusions

If you came here to understand damping, solve problems in the three regimes, and know the quality factor, you have what you need. What follows is for readers who want to see the energy-dissipation calculation done carefully, the connection to the LCR circuit, and why critical damping is an "isolated" point in parameter space.

Energy dissipation — rigorous cycle-average

The instantaneous rate of energy loss is dE/dt = -b \dot x^2. For the underdamped solution x(t) = A_0 e^{-\gamma t}\sin(\omega' t + \varphi),

\dot x(t) = A_0 e^{-\gamma t}\bigl[-\gamma\sin(\omega't + \varphi) + \omega'\cos(\omega't + \varphi)\bigr].

For light damping (\gamma \ll \omega_0), the velocity is dominated by the \omega'\cos term: \dot x \approx A_0 \omega_0 e^{-\gamma t}\cos(\omega_0 t + \varphi). Squaring and time-averaging over one cycle (using \langle \cos^2 \rangle = 1/2):

\langle \dot x^2 \rangle = \tfrac{1}{2} A_0^2 \omega_0^2 e^{-2\gamma t}.

So the cycle-averaged dissipation rate is

\bigl\langle \tfrac{dE}{dt} \bigr\rangle = -b\cdot \tfrac{1}{2} A_0^2 \omega_0^2 e^{-2\gamma t} = -2\gamma \cdot \tfrac{1}{2} k A_0^2 \cdot e^{-2\gamma t} = -2\gamma \, E(t),

which integrates to E(t) = E_0 e^{-2\gamma t} — confirming equation (7) of the main article. The factor of 2 in the exponent comes directly from the amplitude-to-energy squaring.

Physical picture. In each half-cycle, the oscillator passes through equilibrium at peak speed \sim A\omega_0. The damping force at that instant is b \cdot A\omega_0, and the work it does over the half-cycle is of order b A \omega_0 \cdot 2A = 2bA^2\omega_0. Over a full period T = 2\pi/\omega_0, the energy drained is \sim 2\pi \cdot 2\gamma E, so \Delta E/E \sim 4\pi\gamma/\omega_0 = 2\pi/Q. The oscillator loses a fraction \sim 2\pi/Q of its energy each cycle — which is another way of interpreting the quality factor.

The LCR circuit — the same ODE dressed differently

A series circuit of an inductor L, a resistor R, and a capacitor C, disconnected from any source, obeys Kirchhoff's voltage law:

L\frac{di}{dt} + Ri + \frac{q}{C} = 0, \qquad i = \frac{dq}{dt}.

Substituting for i in terms of the charge q(t):

L\ddot q + R\dot q + \frac{1}{C} q = 0.

This is exactly equation (1) of this article, with the substitutions

m \leftrightarrow L, \qquad b \leftrightarrow R, \qquad k \leftrightarrow 1/C.

So all the analysis — natural frequency, damping parameter, three regimes, quality factor, exponential decay — transfers verbatim. The natural frequency of an undamped LC circuit is \omega_0 = 1/\sqrt{LC}; adding resistance dampens the oscillation, critical damping happens at R_\text{crit} = 2\sqrt{L/C}, and the quality factor is Q = \omega_0 L / R = (1/R)\sqrt{L/C}.

This mechanical-electrical analogy runs deep. Every mechanical SHM has an electrical analogue, and vice versa. Engineers use this freely — circuit designers think in terms of mechanical oscillators for complex resonant systems, and mechanical engineers use circuit simulators to model vibration problems. The mathematics is one subject.

Why critical damping is a boundary, not a regime

The three regimes — underdamped, critically damped, overdamped — are distinguished by the sign of the discriminant \gamma^2 - \omega_0^2. Underdamped and overdamped are two-dimensional open regions in the (\gamma, \omega_0) parameter plane; critical damping is a one-dimensional boundary curve \gamma = \omega_0 between them.

If you pick random parameter values, you will almost never hit critical damping exactly — the set \gamma = \omega_0 has measure zero. That is the mathematical content of "critical damping is a knife-edge condition." In engineering practice, you aim close to critical from one side — typically slightly underdamped (say 0.5 \gamma/\omega_0 < 1), because slight underdamping is preferable to slight overdamping for most applications.

The fact that critical damping produces a repeated root of the characteristic equation is itself a mathematical marker of the transition. In the underdamped regime, the two roots are distinct complex conjugates. In the overdamped regime, they are distinct real numbers. Critical damping is where both pairs degenerate into a single number — a bifurcation point in the language of dynamical systems.

The phase-space picture

A damped oscillator's trajectory in phase space (x, \dot x) is a spiral inward, not a closed ellipse. The spiral spirals toward the origin, which is a stable focus (for underdamping) or a stable node (for critical / overdamping). The energy of the oscillator corresponds to its "radial distance" in phase space, weighted by mass and stiffness — and that distance shrinks monotonically because dE/dt = -b\dot x^2 \le 0.

At critical damping, the spiral degenerates into trajectories that approach the origin directly (along characteristic directions determined by the repeated root). Above critical (overdamped), the trajectories approach along two different exponential directions.

Draw enough of these phase portraits and you see the full picture: the origin is always a stable attractor, and the damping controls how trajectories approach it — spiralling (underdamped), radially converging with a tangent (critical), or following two different exponential directions (overdamped).

Where this leads next