IB›IB Physics HL›Mind mapsC.1 Simple harmonic motionIB Physics HL: Mind mapStudy pack PDFAlso for this subtopic:Revision notesFlashcardsSubtopic testCover factsDefining SHMAcceleration proportional to displacementDirected towards equilibriuma=−ω2xa = -\omega^2 xa=−ω2xComes from a restoring force, e.g. Hooke's lawOscillationsT=1f=2πωT = \frac{1}{f} = \frac{2\pi}{\omega}T=f1=ω2πMass–spring: T=2πmkT = 2\pi\sqrt{\frac{m}{k}}T=2πkmPendulum: T=2πlgT = 2\pi\sqrt{\frac{l}{g}}T=2πglPeriod independent of amplitudePendulum period independent of bob massPhasex=x0sin(ωt+ϕ)x = x_0\sin(\omega t + \phi)x=x0sin(ωt+ϕ)ϕ=0\phi = 0ϕ=0: starts at equilibrium, moving positiveϕ=π2\phi = \frac{\pi}{2}ϕ=2π: starts at +x0+x_0+x0Phase difference Δϕ=2πΔtT\Delta\phi = \frac{2\pi\Delta t}{T}Δϕ=T2πΔtDifference of π is antiphaseSHMsimple harmonic motionω\omegaωTx0x_0x0fVelocityv=ωx0cos(ωt+ϕ)v = \omega x_0\cos(\omega t + \phi)v=ωx0cos(ωt+ϕ)v=±ωx02−x2v = \pm\omega\sqrt{x_0^2 - x^2}v=±ωx02−x2Maximum speed ωx0\omega x_0ωx0 at equilibriumMaximum acceleration ω2x0\omega^2 x_0ω2x0 at the extremesEnergyTotal: ET=12mω2x02E_T = \frac{1}{2}m\omega^2 x_0^2ET=21mω2x02Potential: EP=12mω2x2E_P = \frac{1}{2}m\omega^2 x^2EP=21mω2x2Kinetic equals potential at x=±x02x = \pm\frac{x_0}{\sqrt{2}}x=±2x0Doubling amplitude quadruples energyDamping dissipates energy, amplitude fallsExam tipsCalculator in radiansGiven displacement, use v=±ωx02−x2v = \pm\omega\sqrt{x_0^2 - x^2}v=±ωx02−x2Constant negative a/x from data shows SHM