17.1 Simple harmonic oscillations
- Syllabus
- 9702–2028–2029
- Topic
- 17.1
- Level
- A2
Learning objectives
17.1.1The terms displacement, amplitude, period, frequency, angular frequency and• understand and use the terms displacement, amplitude, period, frequency, angular frequency and phase difference in the context of oscillations, and express the period• In terms of both frequency and angular frequency17.1.2Simple harmonic motion occurs when acceleration is proportional to• understand that simple harmonic motion occurs when acceleration is proportional to displacement from a fixed point and in the opposite direction17.1.3A = –ω2x and recall and use, as a solution to this equation, x = x0 sin ωt• use a = –ω2x and recall and use, as a solution to this equation, x = x0 sin ωt17.1.4The equations v = v0 cos ωt and v = ± ω ()xx0 22−• use the equations v = v0 cos ωt and v = ± ω ()xx0 22−17.1.5Graphical representations of the variations of displacement, velocity and• analyse and interpret graphical representations of the variations of displacement, velocity and acceleration for simple harmonic motion
Describe an oscillation with amplitude, period, frequency and phase
| Term | Meaning | Unit |
|---|---|---|
| displacement x | signed position from equilibrium | m |
| amplitude x0 | maximum magnitude of displacement | m |
| period T | time for one complete oscillation | s |
| frequency f | oscillations per unit time | Hz |
| angular frequency ω | rate of change of phase | rad s⁻¹ |
| phase difference Δφ | difference in cycle position between two oscillations | rad or ° |
f=1/T,ω=2πf=2π/T,T=1/f=2π/ω
phasedifference=2π(Δt/T)rad=360°(Δt/T)
For f=5.0 Hz, T=0.200 s and ω=10π rad s⁻¹. A time separation of 0.050 s is one quarter-cycle, so Δφ=π/2 rad=90°.
Amplitude is half the peak-to-peak displacement and is positive. Frequency f is in hertz; angular frequency ω is in rad s⁻¹.
SHM requires acceleration proportional and opposite to displacement
Simple harmonic motion occurs when acceleration is directly proportional to displacement from a fixed equilibrium point and is always directed opposite to that displacement.
a∝−xora=−ω2x
An a-against-x graph for SHM is a straight line through the origin with negative gradient. The straight line proves proportionality; the negative gradient proves opposite direction. Its gradient is -ω².
| Position | Acceleration | Motion fact |
|---|---|---|
| x=0 | a=0 | speed is maximum |
| x=+x0 | a=-ω²x0 | acceleration points negative |
| x=-x0 | a=+ω²x0 | acceleration points positive |
Oscillation alone is not enough. Both direct proportionality and restoring direction must hold throughout the motion; constant acceleration is not SHM.
Use the acceleration law and sinusoidal displacement solution for SHM
a=−ω2xand,foranoscillatorcrossingequilibriuminthepositivedirectionatt=0,x=x0sin(ωt)
x=x0sin(ωt)v=dx/dt=ωx0cos(ωt)a=d2x/dt2=−ω2x0sin(ωt)=−ω2x
The more general form x=x0 sin(ωt+φ) uses φ to match the initial position and direction. If the object starts at positive maximum displacement, x=x0 cos(ωt) is convenient.
For x0=0.230 m and ω=1.90 rad s⁻¹, maximum acceleration magnitude is a0=ω²x0=(1.90)²(0.230)=0.830 m s⁻². At x=+0.100 m, a=-(1.90)²(0.100)=-0.361 m s⁻².
x0 is the positive amplitude, while x is the instantaneous signed displacement. The minus sign in a=-ω²x is essential: acceleration points toward equilibrium.
Choose the SHM velocity equation from time or displacement
| Known information | Velocity equation |
|---|---|
| time t for x=x0 sinωt | v=v0 cosωt=ωx0 cosωt |
| displacement x | v=±ωsqrt(x0²-x²) |
v0=ωx0:maximumspeedoccursatx=0;v=0atx=±x0
x/x0=sinωtandv/(ωx0)=cosωtUsingsin2ωt+cos2ωt=1givesv2=ω2(x02−x2).
For x0=0.035 m and ω=16 rad s⁻¹, at x=0.021 m the speed magnitude is 16sqrt(0.035²-0.021²)=0.448 m s⁻¹. Use +0.448 m s⁻¹ if moving toward increasing x and -0.448 m s⁻¹ if moving toward decreasing x.
The position equation gives two possible velocity signs because the oscillator passes most positions in both directions. Choose the sign from the stated or graphed direction of motion.
Read one SHM motion across time, acceleration–displacement and velocity–displacement graphs
| Event | x | v | a |
|---|---|---|---|
| positive extreme | +x0 | 0 | -ω²x0 |
| equilibrium moving negative | 0 | -v0 | 0 |
| negative extreme | -x0 | 0 | +ω²x0 |
| equilibrium moving positive | 0 | +v0 | 0 |
On time graphs, v leads x by one quarter-cycle (π/2 rad) for x=x0 sinωt, while a is half a cycle (π rad) out of phase with x. The gradient of an x–t graph is v; the gradient of a v–t graph is a.
An a–x graph is a straight line through the origin with gradient -ω². Its x-intercepts are only at equilibrium; at x=±x0 the acceleration magnitudes are maximum.
A v–x graph is a closed ellipse because v²/ v0² + x²/x0²=1. It crosses the x-axis at x=±x0 and the v-axis at v=±v0; upper and lower halves show opposite travel directions.
Do not force every graph to have period T. Quantities such as speed, kinetic energy and potential energy repeat twice per oscillation, so their graph period is T/2.