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.

44 practice questions1 FRQ9 skill areas

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

Restoring force

F = −kx (spring); a is maximal at the extremes, zero at equilibrium

Mass-spring period

T = 2π√(m/k) - no g, no amplitude

Pendulum period

T = 2π√(L/g) - no mass, no (small) amplitude

Frequency

f = 1/T

Energy in SHM

E(total) = ½kA²; KE max at equilibrium, PE max at amplitude

Question bank breakdown

Multiple choice

44

Free response

1

Difficulty mix

12 easy · 23 medium · 9 hard

Skills covered

Frequency and Period of SHMRepresenting and Analyzing SHMSimple Harmonic Motion BasicsMass-Spring SystemsSimple PendulumEnergy in SHMInterpreting SHM GraphsPosition, Velocity, and Acceleration in SHMAmplitude and Period Relationships

Every question in the bank comes with a worked explanation, and most add a per-choice breakdown 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 mm oscillates on a horizontal frictionless surface attached to a spring with spring constant kk. If the mass is doubled to 2m2m 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 2\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=2πm/kT = 2\pi\sqrt{m/k}. The period depends on the square root of the mass, so doubling the mass increases the period by 2\sqrt{2}. This is a key distinction from the simple pendulum, where the period T=2πL/gT = 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.