Summary — What This Chapter Covers
- Terms: displacement, amplitude, period, frequency, angular frequency (ω = 2πf = 2π/T), phase difference.
- S.h.m.: a = −ω2x; solution x = x0 sin ωt.
- v = v0 cos ωt and v = ±ω√(x02 − x2); v0 = ωx0.
- Total energy E = ½mω2x02, exchanged between kinetic and potential.
- Damping removes energy (light, critical, heavy); resonance is maximum amplitude when driving frequency = natural frequency.
1. Simple Harmonic Oscillations
17.1
a = −ω²x
Displacement x is distance from the equilibrium position in a stated direction; amplitude x0 is the maximum displacement; period T is the time for one complete oscillation; frequency f = 1/T; angular frequency ω = 2πf; phase difference is the fraction of a cycle by which one oscillation leads another, expressed as an angle (one cycle = 2π rad).
Simple harmonic motion is defined by an acceleration that is proportional to the displacement from a fixed point and always directed towards that point. Starting at the centre and moving in the positive direction, x = x0 sin ωt.
Graphs: x, v and a are all sinusoidal. Velocity leads displacement by π/2 (a quarter cycle); acceleration is in antiphase with displacement (π out of phase). Velocity is the gradient of the x–t graph, and is maximum at x = 0 and zero at the extremes. An a–x graph is a straight line through the origin with gradient −ω2.
S.h.m. equations
a = −ω2x x = x0 sin ωt v = v0 cos ωt v = ±ω√(x02 − x2)
Worked Example
A mass oscillates with s.h.m. of amplitude 0.040 m and frequency 2.5 Hz. Find the maximum acceleration, the maximum speed and the speed when the displacement is 0.020 m.
Practice Question
For the same oscillator, timing from the centre, find the displacement at t = 0.050 s.
2. Energy in Simple Harmonic Motion
17.2
KE ⇄ PE
As the object oscillates, energy is continuously exchanged between kinetic and potential. At the equilibrium position the kinetic energy is maximum and potential energy is minimum; at the extremes the object is momentarily at rest, so kinetic energy is zero and potential energy is maximum. With no damping the total stays constant.
Since v0 = ωx0, the total energy equals the maximum kinetic energy. At displacement x, Ek = ½mω2(x02 − x2) and Ep = ½mω2x2, which are parabolas on an energy–displacement graph.
Total energy
E = ½mω2x02
Worked Example
The oscillator above has mass 0.20 kg. Find its total energy and its kinetic and potential energy at x = 0.020 m.
3. Damped and Forced Oscillations, Resonance
17.3
Losing and feeding energy
damping
light damping
critical damping
heavy damping
forced oscillation
Resonance
Worked Example
A car’s suspension has a natural frequency of 1.5 Hz. It drives over bumps spaced 6.0 m apart. At what speed does resonance occur?
Memorise