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Study Guide: AP Physics – Simple Harmonic Motion (Spring-Mass, Pendulum)
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AP Physics – Simple Harmonic Motion (Spring-Mass, Pendulum)

By Fatskills Exam Guides Team — the exam nerds behind 28,500+ quizzes and 2.1M practice questions across 500+ global exams.

⏱️ ~5 min read

AP Physics – Simple Harmonic Motion (Spring?Mass, Pendulum)

AP Physics: Simple Harmonic Motion (Spring-Mass & Pendulum) – Exam-Ready Study Guide


What This Is

Simple Harmonic Motion (SHM) describes the repetitive back-and-forth motion of systems like springs and pendulums, where the restoring force is proportional to displacement. It’s a must-know for the AP Physics exam because it appears in multiple-choice, FRQs, and lab-based questions. Real-world example: A grandfather clock’s pendulum keeps time using SHM—its period depends only on the pendulum’s length, not the mass or amplitude (for small angles).


Key Terms & Concepts

  • Simple Harmonic Motion (SHM): Motion where the restoring force is directly proportional to displacement (F--x) and acts opposite to the direction of motion.
  • Hooke’s Law: F = -kx
  • F = restoring force (N), k = spring constant (N/m), x = displacement from equilibrium (m).
  • The negative sign shows the force opposes displacement.
  • Amplitude (A): Maximum displacement from equilibrium (m). Determines energy but not period in SHM.
  • Period (T): Time for one full cycle (s). For springs: T = 2(m/k); for pendulums: T = 2(L/g).
  • m = mass (kg), k = spring constant, L = pendulum length (m), g = gravitational acceleration (9.8 m/s²).
  • Frequency (f): Number of cycles per second (Hz). f = 1/T.
  • Angular Frequency (?): ? = 2?f = ?(k/m) (springs) or ? = ?(g/L) (pendulums). Units: rad/s.
  • Energy in SHM: Total energy is conserved and equals ½kA² (max potential energy at amplitude).
  • At equilibrium: all energy is kinetic (½mv²).
  • At max displacement: all energy is potential (½kA²).
  • Pendulum SHM: For small angles (? < 15°), a pendulum approximates SHM. The period does not depend on mass or amplitude.
  • Damping: Real-world SHM loses energy over time (e.g., a swinging door slows down). AP exams ignore damping unless specified.
  • Phase Constant (?): Determines the starting point of motion in equations like x(t) = A cos(?t + ?).
  • Graphs of SHM: Displacement vs. time is a sine or cosine wave; velocity and acceleration graphs are 90° and 180° out of phase, respectively.

Step-by-Step: Solving SHM Problems

For Spring-Mass Systems:

  1. Identify given quantities (mass m, spring constant k, amplitude A, displacement x).
  2. Determine what’s asked (period, frequency, max velocity, energy, etc.).
  3. Choose the right formula:
  4. Period: T = 2(m/k)
  5. Max velocity: v_max = A? = A?(k/m)
  6. Energy: E_total = ½kA²
  7. Plug in values and solve. Check units (e.g., k must be in N/m, m in kg).
  8. For energy problems, remember:
  9. At equilibrium: KE = ½mv² = ½kA² (all energy is kinetic).
  10. At max displacement: PE = ½kA² (all energy is potential).

For Pendulums:

  1. Confirm small-angle approximation (? < 15°). If not, SHM formulas don’t apply.
  2. Use the period formula: T = 2(L/g).
  3. L = length of pendulum (m), g = 9.8 m/s² (or 10 m/s² for quick estimates).
  4. For energy, use mgh (potential) and ½mv² (kinetic), where h is height above equilibrium.
  5. If asked for frequency or angular frequency, use f = 1/T or ? = ?(g/L).

General Tips:

  • Draw a diagram (e.g., spring stretched/compressed, pendulum at angle).
  • Label equilibrium position (where net force = 0).
  • For graphs, remember:
  • Displacement vs. time: cosine/sine wave.
  • Velocity vs. time: 90° out of phase (peaks at equilibrium).
  • Acceleration vs. time: 180° out of phase (peaks at max displacement).

Common Mistakes

  • Mistake: Assuming period depends on amplitude. Correction: In ideal SHM, period is independent of amplitude (for springs and small-angle pendulums). This is a huge AP trap—watch for questions implying otherwise!

  • Mistake: Forgetting the negative sign in F = -kx. Correction: The negative sign shows the force is restoring (opposes displacement). Without it, you’ll get the wrong direction in free-body diagrams.

  • Mistake: Using pendulum period formula for large angles. Correction: The formula T = 2(L/g) only works for ? < 15°. For larger angles, the motion is not SHM, and the period increases.

  • Mistake: Mixing up angular frequency (?) and frequency (f). Correction: ? = 2?f, not the same thing! ? is in rad/s; f is in Hz.

  • Mistake: Calculating energy incorrectly (e.g., using ½mv² at max displacement). Correction: At max displacement, velocity = 0, so KE = 0 and PE = ½kA². At equilibrium, PE = 0 and KE = ½kA².


AP Exam Insights

  1. FRQs often ask for:
  2. Deriving T = 2(m/k) or T = 2(L/g) from Newton’s second law.
  3. Calculating energy at different points in the motion (e.g., "What’s the speed at x = A/2?").
  4. Graph interpretation: Given a displacement vs. time graph, find period, amplitude, or phase constant.

  5. Multiple-choice traps:

  6. Amplitude vs. period: Questions might imply that changing amplitude affects period (it doesn’t in SHM).
  7. Pendulum on the Moon: The period increases because g is smaller (T-1/?g).
  8. Spring constant units: k is in N/m, not N/cm (convert if needed!).

  9. Lab-based questions:

  10. Measuring period vs. mass for a spring (should be T-?m).
  11. Measuring period vs. length for a pendulum (should be T-?L).

  12. Tricky distinction:

  13. SHM vs. circular motion: SHM is the projection of uniform circular motion onto one axis (e.g., a shadow of a spinning ball moves with SHM).

Quick Check Questions

1. Multiple Choice

A mass-spring system has a period of 2.0 s. If the mass is doubled, the new period is: (A) 1.0 s (B) 1.4 s (C) 2.0 s (D) 2.8 s (E) 4.0 s

Answer: (D) 2.8 s Explanation: Period is proportional to ?m, so doubling m increases T by ?2-1.4.


2. Free-Response (Short)

A 0.50 kg mass is attached to a spring with k = 20 N/m. The mass is pulled 0.10 m from equilibrium and released. (a) Calculate the period of oscillation. (b) What is the maximum speed of the mass?

Answer: (a) T = 2(m/k) = 2(0.50/20) = 0.99 s-1.0 s (b) v_max = A? = A?(k/m) = 0.10?(20/0.50) = 0.63 m/s


3. Multiple Choice

A simple pendulum has a period of 1.5 s on Earth. If the same pendulum is taken to the Moon (where g = 1.6 m/s²), its period will be: (A) 0.6 s (B) 1.5 s (C) 2.5 s (D) 3.7 s (E) 5.6 s

Answer: (D) 3.7 s Explanation: Period is proportional to 1/?g, so T_Moon = T_Earth × ?(g_Earth/g_Moon) = 1.5 × ?(9.8/1.6)-3.7 s*.


Last-Minute Cram Sheet

  1. Hooke’s Law: F = -kx (restoring force proportional to displacement).
  2. Spring period: T = 2(m/k) (independent of amplitude!).
  3. Pendulum period: T = 2(L/g) (only for small angles,-< 15°).
  4. Angular frequency: ? = 2?f = ?(k/m) (springs) or ?(g/L) (pendulums).
  5. Energy in SHM: E_total = ½kA² (conserved).
  6. Max velocity: v_max = A? (occurs at equilibrium).
  7. Graphs: Displacement, velocity, and acceleration are 90° out of phase.
  8. Pendulum period does NOT depend on mass or amplitude (for small angles)!
  9. Spring period does NOT depend on amplitude!
  10. Units matter! k must be in N/m, m in kg, L in m.