One substitution, and everything from the last two pages applies unchanged.
Sweep the damping and say nothing.
Let it run from zero to seven. The time graph goes from a permanent oscillation, through a fading wobble, to a silent slide home; the phase portrait goes from a closed loop, through a spiral, to a node; the eigenvalues go from imaginary to complex to real.
Three representations of one change, moving together. Then ask where the changeover happened, and bring out a² = 4b.
Let y = dx/dt. The equation becomes dx/dt = y and dy/dt = −bx − ay, which is a coupled system, which they can already classify and already step through with Euler.
Teach it as a reduction to something known rather than as new content, and it takes half the time.
| x″ + ax′ + bx = 0 | Matrix [[0, 1], [−b, −a]]. |
| x″ + 3x′ + 2x = 0 | λ = −1 and −2. Stable node, overdamped. |
| x″ + 2x′ + 5x = 0 | λ = −1 ± 2i. Stable spiral, underdamped. |
| x″ + 4x = 0 | λ = ±2i. Centre, undamped. |
| The changeover | a² = 4b, which is a = 4 when b = 4: critically damped. |
| 1. The value of p | −6. |
| 2. −1 ± 2i | B, an oscillation that fades. |
| 3. No damping | B, a centre with closed loops. |
1 markStating the substitution y = dx/dt.
1 markWriting both first order equations, with correct signs.
1 markFinding the eigenvalues and classifying.
1 markSaying what it means physically: oscillating or not, fading or not.
| They give | What it means |
|---|---|
| p = 6 rather than −6 | Forgot that the terms move across the equals sign. The bottom row is minus b and minus a. |
| Matrix [[0, 1], [b, a]] | Same sign error, both entries. Always rearrange to x″ = … first, on its own line. |
| "Oscillation that never fades" (Q2) | Ignored the real part. The 2i says it turns; the minus 1 says it shrinks. |
| "Stable node" for no damping (Q3) | Expected everything to settle. With no damping there is nothing to remove energy, and the real part is exactly zero. |
"Why does y mean the velocity?" Because that is what dx/dt is. Naming it y is only a relabelling, which is why the substitution costs nothing.
"Can I solve it exactly?" In the real distinct eigenvalue case, yes, and that links back to the previous sub-topic. Otherwise classify it or step it numerically.
"Where would this come from?" Springs, circuits, suspension. Understanding that helps, but in an examination the equation is given, so do not spend the lesson deriving one.
| What is happening | |
|---|---|
| 1 | A real damped oscillation: a door closer, a car suspension. What do we want to predict? |
| 2 | The substitution, derived on the board, with the sign rearrangement done slowly. |
| 3 | The worked example, classified using last lesson's method. |
| 4 | The damping sweep. Three representations changing together, and a² = 4b. |
| 5 | Questions 1 to 3. |
| 6 | Euler applies here unchanged, which is worth stating. |
Do not introduce this as a new topic. It is a reduction, and a class that sees it as new will not reach for the classification they already know.
Do not skip the sign rearrangement. Writing x″ = −bx − ay on its own line prevents the one error that appears in nearly every first attempt.
Three tiers, ramping the way practice should: the method on its own, then the method inside something real, then a challenge. Set the tier the class in front of you needs rather than one undifferentiated sheet. Answers are given so these can go straight onto a board.
The method on its own, with friendly numbers. Set these first and move on quickly once they are secure.
The same skill inside a real situation, where the first job is working out what is being asked.
Reasoning, working backwards, or spotting an error. These are where the top grades are decided.
Works on a phone. Nothing is loaded from any other site, so it runs behind a school firewall, and nothing a student does is saved or sent anywhere.