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

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 = 2π√(m/k); for pendulums: T = 2π√(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 (ω): ω = 2πf = √(k/m) (springs) or ω = √(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 (θ < 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 (φ): Determines the starting point of motion in equations like x(t) = A cos(ωt + φ).
• 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 = 2π√(m/k)
5. Max velocity: v_max = Aω = 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 (θ < 15°). If not, SHM formulas don’t apply.
2. Use the period formula: T = 2π√(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 ω = √(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 = 2π√(L/g) only works for θ < 15°. For larger angles, the motion is not SHM, and the period increases.

• Mistake: Mixing up angular frequency (ω) and frequency (f).
  Correction: ω = 2πf, not the same thing! ω 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 = 2π√(m/k) or T = 2π√(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 = 2π√(m/k) = 2π√(0.50/20) = 0.99 s ≈ 1.0 s
(b) v_max = Aω = 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 = 2π√(m/k) (independent of amplitude!).
3. Pendulum period: T = 2π√(L/g) (only for small angles, θ < 15°).
4. Angular frequency: ω = 2πf = √(k/m) (springs) or √(g/L) (pendulums).
5. Energy in SHM: E_total = ½kA² (conserved).
6. Max velocity: v_max = Aω (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.