IB Physics SL C.1 Simple harmonic motion Notes

Simple harmonic motion is not just any back-and-forth motion. First, when the object is displaced from equilibrium, the restoring force or acceleration must point back toward equilibrium. Second, the magnitude of that restoring effect must be directly proportional to the displacement from equilibrium. This is why SHM is often written as F ∝ -x or a ∝ -x: the minus sign shows the force or acceleration is opposite to the displacement. If either condition fails, the object may still oscillate, but it is not undergoing simple harmonic motion.

A system oscillates about an equilibrium position. State the two additional conditions needed for the motion to be simple harmonic.

The equation a = -ω^2x is the mathematical test for simple harmonic motion. The displacement x must be measured from the equilibrium position. The factor ω^2 is positive, so the minus sign shows that acceleration always points back toward equilibrium, opposite to the displacement. Because acceleration is directly proportional to x, an a-x graph is a straight line through the origin with gradient -ω^2. The acceleration is not constant: as the oscillator moves, x changes, so the acceleration changes too.

A graph of acceleration a against displacement x for an oscillator is a straight line through the origin with gradient -25 s^-2. Explain why this is SHM and determine ω.

SHM is described relative to the equilibrium position. Displacement x is a signed distance from equilibrium, so it can be positive or negative depending on the chosen direction. Amplitude A is the largest value of |x|, measured from equilibrium to an extreme position, not from one extreme to the other. The period T is the time for one complete oscillation, such as crest to crest on a displacement-time graph. Frequency f tells how many complete oscillations occur each second, and angular frequency ω tells how quickly the oscillator moves through phase.

A displacement-time graph for an oscillator has a maximum displacement of +0.040 m and a minimum displacement of -0.040 m. Consecutive maxima occur 0.50 s apart. State the amplitude and period.

Period, frequency, and angular frequency describe the same repeated motion in different units. The period T is seconds per oscillation. Frequency f is oscillations per second, so f = 1/T. Angular frequency ω counts phase angle per second; one full cycle is 2π radians, so ω = 2πf = 2π/T. Standard SHM systems have useful period formulas: a mass-spring oscillator has T = 2π√(m/k), and a simple pendulum has T = 2π√(l/g) when the amplitude is small enough for the restoring acceleration to be proportional to displacement.

A pendulum of length l is modelled as simple harmonic for small amplitudes. State the period formula and explain one condition for using it.

In undamped SHM, energy is not lost from the oscillator; it is exchanged between kinetic energy and potential energy. At the extremes, the oscillator momentarily stops, so speed and kinetic energy are zero. All the mechanical energy is stored as potential energy, and the restoring acceleration has maximum magnitude. At equilibrium, displacement and acceleration are zero, speed is maximum, and kinetic energy is maximum. For a mass-spring oscillator the total energy is E = 1/2 kA^2, so increasing amplitude increases total energy by the square of the change in amplitude.

A mass-spring oscillator has amplitude A and negligible damping. Explain how the kinetic and elastic potential energies change as the mass moves from an extreme position to equilibrium.

SHM graphs are linked by both calculus and the defining equation. The velocity at any instant is the gradient of the displacement-time graph, so velocity is zero at the turning points and largest at equilibrium. Acceleration is not read from area; use a = -ω^2x, so acceleration is always opposite to displacement and has its largest magnitude at the extremes. If displacement is modelled as x = A cos(ωt), velocity is a quarter-cycle out of phase with displacement and acceleration is half a cycle out of phase with displacement.

An SHM displacement-time graph has amplitude A and starts at x = +A. Describe the velocity and acceleration at t = 0 and at the first equilibrium crossing.

Uniform circular motion is a powerful model for SHM. Imagine a point moving around a circle at constant angular speed while its shadow is projected onto a diameter. The shadow moves back and forth exactly like an SHM oscillator. The circle radius is the amplitude A, and the angular speed is the angular frequency ω. The phase angle θ tells where the oscillator is within its cycle. Depending on where θ = 0 is chosen, the projected displacement is x = A cos θ or x = A sin θ. The projection of centripetal acceleration always points toward the centre line and is proportional to the projected displacement, giving a = -ω^2x.

A point moves in a circle of radius 0.050 m with angular speed 12 rad s^-1. Its horizontal projection is used to model SHM. State the amplitude and angular frequency of the SHM, and explain why the projected acceleration is restoring.

Core C.1 is secure when the student can start from the restoring condition, write a = -ω²x, and then move cleanly between equations, graphs, and energy statements. The equilibrium position is the reference for displacement. The extremes give maximum potential energy and acceleration magnitude; equilibrium gives maximum speed and kinetic energy. Period, frequency, and angular frequency describe the same cycle in different units, and the circular-motion projection explains why the motion is sinusoidal.