Add a damper and the characteristic equation m·r² + c·r + k = 0 has a discriminant c² − 4mk whose sign sorts every oscillator into one of three kinds. Positive, and there are two real decays and no oscillation. Zero, and the motion is critically damped, the borderline. Negative, and it still oscillates inside a shrinking envelope. The damping that makes the discriminant zero is c_crit = 2√(mk).
Core ideas
Three ideas: the sign that decides, the three solution forms it chooses between, and the damping at which the choice changes.
Substitute x = e^(rt) into m·x″ + c·x′ + k·x = 0 and divide by e^(rt): m·r² + c·r + k = 0. The quadratic formula gives r = (−c ± √(c² − 4mk))/(2m). The sign of c² − 4mk says whether the roots are two real numbers, one repeated real number, or a complex pair, exactly as in “The Characteristic Equation” in the previous course. Here every root has a negative real part, so every motion dies away.
Overdamped: two negative real roots r₁ and r₂, and x = C₁·e^(r₁t) + C₂·e^(r₂t). Critical: one repeated root r = −c/(2m), and x = (C₁ + C₂·t)·e^(rt). Underdamped: a complex pair α ± β·i with α = −c/(2m), and an oscillation inside the decay e^(αt). The previous course derived all three; this lesson reads them as statements about a damper.
Set c² − 4mk = 0 and c = 2√(mk). Below it the damper is too weak to stop the oscillation, and above it the damper is strong enough that the mass never swings through rest and back. For m = 1 and k = 4 the critical value is c = 2√4 = 4. The lab calls it c_crit and prints it exactly, as a surd when m·k is not a perfect square.
The material
The characteristic equation again, the three cases with a lab preset each, the borderline value, and the reason the strongest damping is not the quickest.
A damper turns the free oscillator into the general equation m·x″ + c·x′ + k·x = 0. It is a second-order linear equation with constant coefficients, and the previous course solved all of them: look for e^(rt), and the roots of m·r² + c·r + k decide the rest. What this lesson adds is what the three outcomes mean for a spring.
With c > 0, the motion is overdamped when c² − 4mk > 0, critically damped when c² − 4mk = 0, and underdamped when c² − 4mk < 0.
The lab also calls c = 0 undamped; there the roots are ±ω₀·i and the previous module applies.
Keep m = 1 and vary the damper. The three presets of the lab are the three kinds, and the discriminant tile says which is which.
c² − 4mk = 25 − 16 = 9, positive. The roots are −1 and −4, so x = C₁·e^(−t) + C₂·e^(−4t). Neither term oscillates, and both decay. The mass slides back toward rest and may pass through it once at most.
The lab's c_crit for m = 1, k = 4 is 4, and c = 5 is above it. That is the meaning of overdamped: more damping than the critical amount.
c² − 4mk = 4 − 20 = −16, negative. The roots are −1 ± 2i, so x = e^(−t)·(C₁cos(2t) + C₂sin(2t)): the oscillation of the earlier lessons inside a decay. The critical value for this mass and spring is 2√5, about 4.47214 and rounded, and c = 2 is well below it.
The pattern in the three examples is one comparison: c against c_crit. Below it, the mass oscillates; at it, the roots merge; above it, they separate into two real decays.
It is natural to expect that a bigger damper brings the mass to rest sooner. It does not, once the damping passes the critical value. In the overdamped case the two roots are r₁ and r₂, and the slow one controls how long the motion takes. For c = 5 it is −1, giving e^(−t). The critical case, c = 4, has the repeated root −2, giving (C₁ + C₂·t)·e^(−2t). Eventually e^(−2t) is far smaller than e^(−t), even with the factor t in front of it.
The lab's fourth preset pushes this further. For c = 10 and k = 4 the discriminant is 100 − 16 = 84, the roots are −5 ± √21, and the slow root −5 + √21 is between −1 and 0, because 4 < √21 < 5. Ten units of damper make the return slower than five. The damper holds the mass so firmly that it creeps.
Interactive
Free, damped and forced motion, with every exact quantity
Method
Four moves. Compute the discriminant as a number before you name the class.
m·r² + c·r + k = 0, with the three numbers read from the equation. For x″ + 5x′ + 4x = 0 it is r² + 5r + 4 = 0.
Square the damping, subtract four times the product of mass and spring constant. Do it as a number: 25 − 16 = 9. It is the first number the lab prints.
Positive: overdamped. Zero: critically damped. Negative: underdamped. Write the solution form for that class and the roots that go into it.
c_crit = 2√(mk). Compare c with it: smaller is underdamped, equal is critical, larger is overdamped. This check must agree with the sign of the discriminant.
Take m = 1 and k = 4. First find the critical damping, then classify c = 5, and then take a lighter damper c = 2 with the spring changed to k = 5 to see the third kind.
The two checks agree. c = 5 exceeds c_crit = 4 and has a positive discriminant. For k = 5 the critical damping is 2√5, which is larger than 2, and the discriminant is negative. Whenever the comparison with c_crit and the sign of c² − 4mk disagree, one of them has a slip in it.
You should be able to compute c² − 4mk exactly, name the motion overdamped, critical or underdamped from its sign, write the solution form and the roots for each, compute c_crit = 2√(mk) as an integer or a surd, and say why more damping is not always a quicker return.
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