AP Physics 1 / Topic
Oscillations
Unit 7 covers simple harmonic motion: the restoring-force condition, mass-spring systems and pendulums, period and frequency relationships, SHM graphs, and energy exchange between kinetic and potential forms.
What the exam asks
- Identify the SHM condition (restoring force proportional to displacement)
- Compute periods of mass-spring systems and simple pendulums and reason about parameter changes
- Read amplitude, period, and phase off position/velocity/acceleration-time graphs
- Locate where speed, acceleration, KE, and PE are maximal or zero in a cycle
- Explain why SHM period is independent of amplitude (and pendulum period of mass)
Key formulas and rules
F = −kx (spring); a is maximal at the extremes, zero at equilibrium
T = 2π√(m/k) — no g, no amplitude
T = 2π√(L/g) — no mass, no (small) amplitude
f = 1/T
E(total) = ½kA²; KE max at equilibrium, PE max at amplitude
Question bank breakdown
44
1
12 easy · 23 medium · 9 hard
Skills covered
Every question in the bank comes with a per-choice explanation, so you learn why each wrong answer is wrong — not just the key.
Sample question
From the free tier of the ScoreMint AP Physics 1 bank — try it, then check the answer.
A block of mass $m$ oscillates on a horizontal frictionless surface attached to a spring with spring constant $k$. If the mass is doubled to $2m$ while the spring constant remains the same, how does the period of oscillation change?
- A. The period doubles.
- B. The period increases by a factor of $\sqrt{2}$.
- C. The period is halved.
- D. The period remains the same.
Show answer and explanation
Answer: B. The period of a mass-spring system is $T = 2\pi\sqrt{m/k}$. The period depends on the square root of the mass, so doubling the mass increases the period by $\sqrt{2}$. This is a key distinction from the simple pendulum, where the period $T = 2\pi\sqrt{L/g}$ does not depend on mass at all. Choice D is a common trap that confuses these two systems.
Study tip
Learn the phase picture: at maximum displacement, v = 0 and |a| is maximum; at equilibrium, |v| is maximum and a = 0. Half the SHM multiple-choice bank is that one sentence in disguise.