Frequency and Period of SHM
Once you have the criterion $F = -kx$ for simple harmonic motion, the next question is how long one cycle takes. For SHM that period is set by the system, not by how the object is moving. Two formulas do the work: $T_s = 2\pi\sqrt{m/k}$ for a block on a spring, and $T_p = 2\pi\sqrt{L/g}$ for a small-angle pendulum. Neither formula has the amplitude. The pendulum formula has no mass either.
Three traps to watch for. Pull a spring back farther and it feels like the block should take longer to come back. It doesn't: a larger amplitude means more distance to travel and faster motion to travel it, and the two cancel out. Doubling the block's mass feels like it should double the period, but the real scaling is the square root, so it takes a $4\times$ mass change to double $T$. And a heavier pendulum bob feels like it should swing slower than a lighter one, but gravity and inertia both grow with mass and cancel out, so both bobs swing at the same rate. All three errors share the same instinct: assuming a linear relationship where the variable either doesn't appear at all, or only appears under a square root.
The work
3 ways in · any order
Lesson
Frequency and Period of SHM
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Two formulas, three habits to break. Work through the spring and pendulum equations, the square-root scaling, and why amplitude has no effect on the period. Finish on the ten-scenario applet that hits each trap, one at a time.
Diagnostic
10-item topic check
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Ten items mixing equation recognition, scaling questions (what changes when you double a variable), and short scenarios. Each wrong answer points to one of the three traps, so the targeted-practice card knows what to drill next.
Targeted Practice
Drill a single misconception
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Pick one of the failure modes you've missed and grind it on its own. The round is adaptive: two correct in a row clears the misconception and you move on.