Simple Harmonic Motion

AP Physics C Mechanics

Introduction to Simple Harmonic Motion

Simple Harmonic Motion (SHM) is a type of periodic motion where the restoring force is directly proportional to the displacement and acts in the direction opposite to that of displacement.

Examples of SHM include the oscillation of a spring-mass system, a pendulum (for small angles), and vibrations of atoms in a molecule.

Key Concepts

Characteristics of SHM

  • The acceleration is proportional to the negative of displacement
  • The motion is periodic and oscillatory
  • The restoring force follows Hooke's Law: F = -kx

Important Terms

  • Amplitude (A): Maximum displacement from equilibrium
  • Period (T): Time taken for one complete oscillation
  • Frequency (f): Number of oscillations per unit time (f = 1/T)
  • Angular Frequency (ω): ω = 2πf = 2π/T
  • Phase Constant (φ): Determines the initial position

Interactive Simulations

Spring-Mass System

1.0
10
1.0
0

Period: 1.99 s

Frequency: 0.50 Hz

Angular Frequency: 3.16 rad/s

Pendulum

1.0
1.0
15
0

Period: 2.01 s

Frequency: 0.50 Hz

Angular Frequency: 3.13 rad/s

Important Formulas

Position, Velocity, and Acceleration

Position: x(t) = A cos(ωt + φ)

Velocity: v(t) = -Aω sin(ωt + φ)

Acceleration: a(t) = -Aω² cos(ωt + φ) = -ω²x

Spring-Mass System

Angular Frequency: ω = √(k/m)

Period: T = 2π√(m/k)

Frequency: f = 1/T = (1/2π)√(k/m)

Pendulum (Small Angle Approximation)

Angular Frequency: ω = √(g/L)

Period: T = 2π√(L/g)

Frequency: f = 1/T = (1/2π)√(g/L)

Energy in SHM

Kinetic Energy: K = (1/2)mv² = (1/2)mA²ω² sin²(ωt + φ)

Potential Energy: U = (1/2)kx² = (1/2)kA² cos²(ωt + φ)

Total Energy: E = K + U = (1/2)kA² = (1/2)mA²ω²

Example Problems

Example 1: Spring-Mass System

Problem: A 0.5 kg mass is attached to a spring with spring constant k = 20 N/m. If the mass is pulled 10 cm from equilibrium and released from rest, determine:

  1. The angular frequency
  2. The period and frequency of oscillation
  3. The maximum velocity
  4. The maximum acceleration
  5. The position, velocity, and acceleration as functions of time

Solution:

  1. Angular frequency: ω = √(k/m) = √(20/0.5) = √40 = 6.32 rad/s
  2. Period: T = 2π/ω = 2π/6.32 = 0.99 s
    Frequency: f = 1/T = 1/0.99 = 1.01 Hz
  3. Maximum velocity: vmax = Aω = 0.1 × 6.32 = 0.632 m/s
  4. Maximum acceleration: amax = Aω² = 0.1 × (6.32)² = 4 m/s²
  5. Position: x(t) = 0.1 cos(6.32t) m
    Velocity: v(t) = -0.632 sin(6.32t) m/s
    Acceleration: a(t) = -4 cos(6.32t) m/s²

Example 2: Pendulum

Problem: A simple pendulum has a length of 0.8 m. Determine:

  1. The period of oscillation
  2. The frequency of oscillation
  3. The maximum speed of the pendulum bob if the maximum angle is 5°

Solution:

  1. Period: T = 2π√(L/g) = 2π√(0.8/9.8) = 2π√(0.0816) = 1.8 s
  2. Frequency: f = 1/T = 1/1.8 = 0.56 Hz
  3. Maximum angle in radians: θmax = 5° × (π/180) = 0.0873 rad
    Angular frequency: ω = √(g/L) = √(9.8/0.8) = 3.5 rad/s
    Maximum speed: vmax = Lω × θmax = 0.8 × 3.5 × 0.0873 = 0.24 m/s