At critical damping the characteristic equation has one repeated root, and the motion is a straight line in t times a decaying exponential. A straight line has at most one zero, so the motion crosses its rest position at most once, and two constants fitted from the start decide whether it does. Critical damping has the fastest long-run decay of any damper, which is why it is the target for a door closer, and it is not the quickest in every sense.
Core ideas
Three ideas: the solution and its two constants, the count of zeros that the form allows, and what a designer gets by sitting on the border.
At c² = 4mk the quadratic m·r² + c·r + k has the single root r = −c/(2m), and “Repeated Roots” in the previous course gave the solution x = (C₁ + C₂·t)·e^(rt). Put t = 0 in it and in its rate and the constants follow from the start: C₁ = x₀ and C₂ = v₀ − r·x₀. The second constant is not the starting velocity. It is the velocity with the decay's own contribution taken off.
The exponential e^(rt) is never zero, so x = 0 exactly when C₁ + C₂·t = 0. That is t = −C₁/C₂ when C₂ ≠ 0, and it counts only if it is positive. An underdamped motion crosses zero again and again; a critically damped one cannot, whatever the start.
Below c_crit the motion oscillates; above it, it creeps because the slow root moves toward zero. At c_crit the slow root is as far from zero as any damper can put it, so the long-run decay is the fastest there is. A door closer is built toward this value: the door shuts without swinging back and without taking all afternoon.
The material
The solution form and its constants, the count of crossings with a proof, two starts on one damper, and the reason the damper on a door sits at the border.
“Overdamped, Critical and Underdamped” sorted dampers into three kinds by the sign of c² − 4mk, and the lesson after it followed the underdamped kind. This lesson takes the middle kind, the single value c = 2√(mk) where the sign is zero, and asks what it is good for. The answer has two parts: a form of solution simple enough to settle exactly whether the mass crosses its rest position, and a reason a designer wants to land there.
When c² = 4mk the characteristic equation m·r² + c·r + k = 0 has the double root r = −c/(2m), and the motion is x = (C₁ + C₂·t)·e^(rt). It does not oscillate.
For m = 1, c = 4, k = 4 the root is r = −2. At critical damping the root is also −ω₀, minus the natural frequency.
If x = (C₁ + C₂·t)·e^(rt) is not the zero function, there is at most one t > 0 with x = 0, and there is one exactly when C₁ and C₂ are both nonzero and have opposite signs.
The factor e^(rt) is positive for every t, so x = 0 exactly where the factor C₁ + C₂·t is zero.
That factor is a straight line. If C₂ = 0 it is the constant C₁, which is zero only if the whole motion is zero. If C₂ ≠ 0 it is zero only at t = −C₁/C₂, which is positive exactly when C₁ and C₂ have opposite signs, and a line cannot meet the axis twice.
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Take m = 1, k = 4 and c = 4, which is c_crit = 2√4. The root is r = −2, and the lab's tile for the zero crossing says when, or that there is none. The only thing that changes between the two starts is the velocity.
Release the mass at x = 1 moving outward at speed 1. The constants are C₁ = 1 and C₂ = 3, both positive, so the line 1 + 3t is positive for every t > 0 and its zero, −1/3, lies in the past. The lab reports no zero for t > 0.
The mass drifts out a little further and then decays to rest from above. It never reaches the rest position.
Release it at x = 1 moving inward at speed 5. Now C₂ = −5 + 2 = −3, which has the opposite sign to C₁ = 1. The line 1 − 3t is zero at t = 1/3, which the lab prints. The mass passes through the rest position once, goes to the far side, and creeps back from below.
Critical damping does not forbid an overshoot. It forbids a second crossing, because the form has room for only one.
Model the door as a mass, the closing spring as k, and the oil damper as c. Let x be the angle from shut, so that x = 0 is the door at its frame. Someone opens the door and lets go, which is a start from rest, v₀ = 0. Then C₂ = −r·x₀ = (c/(2m))·x₀, which has the same sign as C₁ = x₀, so by the theorem the door never crosses shut. A weaker damper would swing it through the frame and back; a stronger one would close it slowly.
The word “slowly” can be measured. For m = 1 and k = 4 the long-run decay is e^(−t) at c = 2 (the envelope rate c/(2m)), e^(−2t) at c = 4, and e^(−t) again at c = 5, whose slow root is −1. The rate rises to a maximum at c_crit and then falls. The reason is that the two roots multiply to k/m = 4: past the border, raising c pushes the fast root outward and the slow root toward zero.
That is a claim about the long run, and only about the long run. A damper a little below critical reaches the rest position in finite time, which a critically damped door released from rest never does, and it then overshoots. So the designer is trading one failure for the other, and critical damping is where neither one happens.
Interactive
Free, damped and forced motion, with every exact quantity
Method
Five moves. The third is the one where a velocity is most often used as a constant.
Compute it as a number. If it is not zero the motion is not critically damped and this solution form does not apply.
r = −c/(2m). For m = 1 and c = 4 it is −2; the solution is (C₁ + C₂·t)·e^(−2t).
C₁ = x₀, and C₂ = v₀ − r·x₀. The second is the starting velocity with r·x₀ taken off, because the derivative of e^(rt) contributes r·C₁ at t = 0. For the start (1, −5): C₂ = −5 + 2 = −3.
If C₂ = 0 there is no zero. Otherwise the candidate is t = −C₁/C₂, and it is a crossing only if it is positive: that is, only if C₁ and C₂ have opposite signs.
Put t = 0 back: x(0) = C₁ must be x₀, and C₂ + r·C₁ must be v₀. If the second check fails, the velocity was used where v₀ − r·x₀ belongs.
The lab preset with start (1, −5): m = 1, c = 4, k = 4, x(0) = 1, x′(0) = −5.
Check by substituting: x′ = (6t − 5)·e^(−2t) and x″ = (16 − 12t)·e^(−2t), and then x″ + 4x′ + 4x has the bracket 16 − 12t + 24t − 20 + 4 − 12t = 0. The lab prints the crossing time t = 1/3 in the zero-crossing tile. With the start (1, 1) the same damper gives C₂ = 3 and no crossing.
You should be able to confirm that c² − 4mk is zero, write (C₁ + C₂·t)·e^(rt) with r = −c/(2m), compute C₁ = x₀ and C₂ = v₀ − r·x₀ exactly, decide the crossing time t = −C₁/C₂ and whether it is positive, and explain why a door closer is built near critical damping.
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