Simple Harmonic Motion
Introduction
Simple harmonic motion is the mathematics of repeated motion when the restoring acceleration always points back toward equilibrium and grows with displacement. The IB guide uses it as a bridge between mechanics, waves, resonance, and molecular models.
Guide Focus
- Recognize the conditions for SHM and use a = -omega^2 x.
- Use period equations for mass-spring systems and simple pendulums.
- Apply HL equations for displacement, velocity, phase, and energy.
Key Concepts
1. Defining condition
A particle undergoes SHM when its acceleration is proportional to displacement from equilibrium and directed toward equilibrium: a = -omega^2 x. The minus sign is essential because the acceleration is restoring.
2. Oscillation quantities
Key quantities include amplitude, displacement, equilibrium position, time period T, frequency f, angular frequency omega, and, at HL, phase angle phi. T = 1/f = 2pi/omega.
3. Standard systems
For a mass-spring system, T = 2pi sqrt(m/k). For a simple pendulum at small amplitude, T = 2pi sqrt(l/g).
4. HL equations and energy
At HL, x = x0 sin(omega t + phi), v = omega x0 cos(omega t + phi), and v = +/- omega sqrt(x0^2 - x^2). Total energy is ET = (1/2)m omega^2 x0^2 and potential energy is Ep = (1/2)m omega^2 x^2.
Common Mistakes
- Forgetting that acceleration and displacement have opposite signs.
- Using degrees instead of radians for phase calculations.
- Assuming pendulum period depends on mass.
Exam Tips
- At equilibrium, speed is maximum and acceleration is zero.
- At maximum displacement, speed is zero and acceleration magnitude is maximum.
- Use energy conservation to connect speed and displacement quickly in HL problems.
Practice Questions
Question 1 (Multiple Choice)
For a mass undergoing SHM, where is the kinetic energy maximum?
A. At equilibrium. B. At maximum positive displacement. C. At maximum negative displacement. D. At all positions equally.
Solution Architecture
The speed is greatest at equilibrium, so kinetic energy is maximum there.
Question 2 (Structured Paper 2 Style)
A 0.20 kg mass attached to a spring of spring constant 80 N m-1 oscillates with small amplitude.
(a) Calculate the period. [2 marks]
(b) Calculate omega. [2 marks]
Markscheme Breakdown
Part (a) Solution:
T = 2pi sqrt(m/k) = 2pi sqrt(0.20 / 80) = 0.31 s.
Part (b) Solution:
omega = 2pi / T = 20 rad s-1, also sqrt(k/m) = sqrt(80/0.20).