C.1 Simple harmonic motionIB Physics HL: Revision notes
Section 1
Conditions and defining equation of SHM
Simple harmonic motion occurs when acceleration is proportional to displacement from equilibrium and directed towards equilibrium:
a = −ω²x
This follows from a restoring force proportional to displacement (Hooke's-law spring; pendulum at small angles). From data, a constant negative a/x shows SHM.
Section 2
Describing oscillations; mass–spring and pendulum
Displacement x, amplitude x₀, period T, frequency f and angular frequency ω are linked by T = 1/f = 2π/ω.
- Mass–spring: T = 2π√(m/k)
- Simple pendulum (small angles): T = 2π√(l/g), independent of bob mass.
In SHM the period does not depend on the amplitude.
Section 3
Energy changes (qualitative)
Energy swaps between kinetic and potential twice per cycle: all potential at the extremes (speed zero), all kinetic at equilibrium (speed maximum). With no damping the total is constant; with damping, energy is dissipated and amplitude falls.
Section 4
HL: Phase angle
The phase angle ϕ describes where in its cycle an oscillator is at t = 0. With x = x₀ sin(ωt + ϕ):
- ϕ = 0: starts at equilibrium moving in the positive direction;
- ϕ = π/2: starts at +x₀;
- ϕ = π: starts at equilibrium moving in the negative direction;
- ϕ = 3π/2 (or −π/2): starts at −x₀.
A time difference Δt between two oscillators of the same period corresponds to a phase difference Δϕ = 2πΔt/T. A phase difference of π means they are in antiphase.
Leaving the calculator in degrees. ωt and ϕ are in radians.
Section 5
HL: Equations of SHM
- Displacement: x = x₀ sin(ωt + ϕ)
- Velocity: v = ωx₀ cos(ωt + ϕ)
- Speed at displacement x: v = ±ω√(x₀² − x²)
So the maximum speed is ωx₀ (at x = 0) and the maximum acceleration is ω²x₀ (at x = ±x₀). Velocity is a quarter of a cycle (π/2) ahead of displacement.
v = ±ω√(x₀² − x²) needs no time at all: use it whenever a question gives a displacement and asks for a speed.
Section 6
HL: Energy in SHM
- Total energy: E_T = ½mω²x₀² (constant without damping)
- Potential energy: E_P = ½mω²x²
- Kinetic energy: E_K = E_T − E_P = ½mω²(x₀² − x²)
Because E_P ∝ x², kinetic and potential energy are equal at x = ±x₀/√2, not at half the amplitude. Total energy ∝ x₀², so doubling the amplitude quadruples the energy.
That's the notes covered.
Carry on to the next subtopic.