17. Oscillations
- Syllabus
- 9702–2028–2029
- Section
- 17
- Level
- A2
Exam analysis
17.1 Simple harmonic oscillations
- Syllabus
- 9702–2028–2029
- Topic
- 17.1
- Level
- A2
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.
17.2 Energy in simple harmonic motion
- Syllabus
- 9702–2028–2029
- Topic
- 17.2
- Level
- A2
Kinetic and potential energy interchange during ideal SHM
In ideal undamped SHM, kinetic energy E_K and potential energy E_P continually interchange while total mechanical energy E=E_K+E_P remains constant.
| Position | Speed | Kinetic energy | Potential energy |
|---|---|---|---|
| x=0, equilibrium | maximum | maximum, E | minimum, taken as 0 |
| 0< | x | <x0 | between 0 and maximum |
| x=±x0, extremes | 0 | 0 | maximum, E |
EP=(1/2)mω2x2EK=(1/2)mω2(x02−x2)E=(1/2)mω2x02
Against x, E_P is an upward U-shaped parabola with minimum at x=0; E_K is a downward arch, zero at ±x0 and maximum at x=0; total E is a horizontal line.
Each energy is the same at +x and -x and reaches two maxima per oscillation, so energy-time graphs repeat with period T/2.
Energy is not created at equilibrium. It has changed form. With damping, mechanical energy decreases; with driving, energy may enter, so the ideal constant-total-energy statement needs that boundary.
Total energy in ideal SHM is one-half m omega-squared amplitude-squared
E=(1/2)mω2x02
Here m is the oscillating mass, ω is angular frequency and x0 is amplitude. In ideal SHM, E equals maximum kinetic energy at equilibrium and maximum potential energy at either extreme.
Atx=0,v=v0=ωx0,soE=EK,max=(1/2)mv02=(1/2)mω2x02.
For m=0.150 kg, ω=15.7 rad s⁻¹ and x0=0.0160 m, E=(1/2)(0.150)(15.7)²(0.0160)²=4.73×10⁻³ J.
If an oscillator travels 14 mm in one complete cycle, it covers 4x0, so x0=3.5 mm. Then ω=sqrt(2E/(mx0²)) when E and m are known.
E∝m,E∝ω2,E∝x02;doublingamplitudequadruplesE.
Use amplitude, not instantaneous displacement, peak-to-peak distance or total distance per cycle. Convert all lengths to metres before substitution.
17.3 Damped and forced oscillations, resonance
- Syllabus
- 9702–2028–2029
- Topic
- 17.3
- Level
- A2
Damping removes energy from an oscillator and reduces its amplitude
A resistive force opposing motion transfers energy from an oscillating system to other stores, causing the amplitude to decrease with time.
The damping force may depend on speed; distinguish the ideal SHM frequency from the changed response of a damped system.
Air resistance makes a pendulum’s swings gradually smaller because mechanical energy becomes thermal energy in the air.
Damping does not necessarily stop oscillation immediately, and reduced amplitude is not the same as reduced equilibrium position.
Light, critical and heavy damping describe how quickly oscillations return to equilibrium
Light damping allows oscillations with decreasing amplitude; critical damping returns to equilibrium fastest without oscillating; heavy damping returns more slowly without overshoot.
Sketch displacement against time with or without crossings of equilibrium and compare settling time, not just initial slope.
A door closer is designed near critical damping so the door settles promptly without repeated swinging.
Critical damping is not “maximum resistance” in every situation; too much damping can make return slower.
Resonance gives maximum amplitude when driving and natural frequencies are equal
| Term | Meaning |
|---|---|
| natural frequency | frequency at which the system oscillates freely after disturbance, with no periodic driving |
| driving frequency | frequency of the external periodic force |
| resonance | maximum steady oscillation amplitude when driving frequency equals natural frequency |
At resonance the driving force transfers energy to the oscillator most effectively on successive cycles. The amplitude grows until energy supplied per cycle balances energy dissipated by damping.
A graph of steady amplitude against driving frequency has one peak at the natural frequency in the syllabus model. Away from that frequency the amplitude is smaller.
| Increased damping | Change to resonance curve |
|---|---|
| more energy lost per cycle | lower maximum amplitude |
| response spread over a wider frequency range | broader, less sharp peak |
Regular pushes on a swing at its natural period add energy in step and build maximum amplitude; pushes at another frequency drift out of step and transfer less energy overall.
For the assessed definition, say driving frequency equals natural frequency—not merely that it is nearby. Resonance need not be destructive and does not require zero damping.