Free oscillations, investigation and quantities
Key idea: H2 Physics lessons on free oscillations, simple harmonic motion, energy interchange, damping, forced response and resonance.
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The core idea
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Big question: How can an experiment describe a free oscillation completely?
A free oscillation occurs after displacement and release, with no continuing periodic driver. Describe it using equilibrium position, displacement, amplitude, period, frequency and phase. In an experiment, record displacement against time with a motion sensor or video, repeat several cycles, obtain T from a long time interval and identify amplitude from the graph.
Start with equilibrium and one complete cycle
A free oscillator is displaced and released, then moves at its natural frequency without a continuing periodic driver. Displacement x is measured from equilibrium with a sign; amplitude A is the greatest magnitude of displacement, not the full peak-to-peak distance.
Period T is the time for one complete cycle and frequency f is cycles per second, so f = 1/T and angular frequency ω = 2πf. Phase locates a point within the cycle; a time separation Δt corresponds to phase difference 2πΔt/T.
Check your understanding: A trace runs from +4 cm to −4 cm. What is its amplitude?
4 cm. The 8 cm separation is peak-to-peak displacement, twice the amplitude.
Design a measurement that reduces timing uncertainty
Use a motion sensor, video analysis or a fiducial marker to obtain displacement against time. If timing manually, measure many complete oscillations between the same directional crossing, divide by the number of cycles and repeat the whole measurement.
Keep the initial amplitude small when the restoring system is only approximately linear, and control mass, spring or pendulum length while investigating one factor. A graph should show the equilibrium line, labelled amplitude and a period measured between identical phase points.
Check your understanding: Why time 20 cycles rather than one?
The same start–stop reaction uncertainty becomes a much smaller fraction of the longer total interval; repeats then reveal scatter.
Key ideas to keep
- Measure several periods and divide to reduce timing uncertainty.
- Amplitude is maximum displacement from equilibrium, not peak-to-peak distance.
- Frequency and period obey f = 1/T.
See the reasoning
Worked example
Extract oscillation quantities from timing data
Question: A mass crosses equilibrium moving upward at 0.40 s and again moving upward at 1.60 s. Its next positive maximum occurs at 1.90 s. Find T, f, ω and the phase advance from the second crossing to that maximum.
Step 1: Use identical phase points
Why: Same-position crossings count a full period only when the direction also matches.
Working: T = 1.60 − 0.40 = 1.20 s.
Step 2: Convert period
Why: Frequency and angular frequency describe the same cycling rate in different units.
Working: f = 1/1.20 = 0.833 Hz; ω = 2πf = 5.24 rad s⁻¹.
Step 3: Read the part-cycle
Why: An upward equilibrium crossing reaches positive maximum after one quarter-cycle.
Working: Δt = 0.30 s = T/4, so Δφ = 2π(1/4) = π/2 rad.
Answer: T = 1.20 s, f = 0.833 Hz, ω = 5.24 rad s⁻¹ and phase advance is π/2 rad.
Check: The maximum occurs exactly a quarter-period after the upward equilibrium crossing, consistent with the trace geometry.
Another worked model
Question
An oscillator reaches consecutive positive maxima at 0.35 s and 1.15 s. Find T, f and ω. A second oscillator reaches its maximum 0.20 s later; find its phase lag.
Check the worked solution
T = 1.15 − 0.35 = 0.80 s, f = 1.25 Hz and ω = 2π/T = 7.85 rad s⁻¹. The lag is 2π(0.20/0.80) = π/2 rad.
Use a hint if needed
Practise with support
Try this
A student records 25 cycles in 18.5 s, 18.8 s and 18.6 s. Find the mean period and angular frequency, and state why the method is preferable to one cycle.
Hint: Average the total times before dividing by 25, then use ω = 2π/T.
Check your answer
Mean total time = 18.63 s, so T = 18.63/25 = 0.745 s and ω = 2π/T = 8.43 rad s⁻¹. Multi-cycle timing reduces fractional start–stop uncertainty and repeats expose scatter.
Now work without the hint
Practise independently
Your turn
Two equal-frequency oscillators have period 1.60 s. Oscillator B passes equilibrium in the positive direction 0.60 s after A does so. Find their frequency, angular frequency and B's phase lag.
Check your answer
f = 1/1.60 = 0.625 Hz, ω = 2π/1.60 = 3.93 rad s⁻¹ and the lag is 2π(0.60/1.60) = 0.75π = 2.36 rad.
Avoid these traps
Common mistakes
Common mistake
Free oscillation means no force acts on the oscillator.
What is wrong with this reasoning?
Show better thinking
A restoring force must act. Free means no periodic driving; the ideal syllabus model also exchanges no energy with the environment.
Common mistake
Timing one oscillation repeatedly is as precise as timing many oscillations.
What is wrong with this reasoning?
Show better thinking
The start–stop uncertainty is a smaller fraction of a longer multi-cycle interval. Repeats then reveal random scatter.
Write for the examiner
Exam guidance
A method answer should name what is varied, measured and controlled, then explain how the graph yields the requested quantity.
Exam-style practice [7 marks]
Describe an experiment to determine the natural frequency of a vertical spring–mass oscillator and investigate whether period depends on amplitude. Include measurements, controls and one way to improve reliability.
Plan before you answer
- Define free motion and the measured phase point.
- Change amplitude while controlling the oscillator.
- Use repeated multi-cycle timing and a suitable graph.
Mark your answer and compare the model
Marking points
Tick each point only if your answer states it clearly.
Model answer
Attach a fixed mass to one spring, displace it by a measured small amplitude and release it without pushing. Time at least 10–20 cycles between downward crossings of one fiducial mark, divide by the number of cycles and repeat. Repeat for several amplitudes while keeping mass and spring unchanged. Plot mean T against A with timing uncertainty; constant T within uncertainty supports amplitude-independent period in the approximately linear range. Natural frequency is 1/T.
Come back in three days
Check what stayed with you
Recall question 1
What makes an oscillation free?
Check the answer
After displacement and release, there is no continuing periodic driver.
Recall question 2
How are phase and time separation related?
Check the answer
Δφ = 2πΔt/T.
Recall question 3
Where should period be measured on a trace?
Check the answer
Between identical phase points, such as successive maxima or same-direction equilibrium crossings.
Syllabus and review details
This lesson covers the listed H2 Physics 9478 outcomes. The official topic states no explicit exclusions and defines natural frequency as the frequency of a system in free oscillation. Ideal SHM uses a linear restoring relation and no environmental energy exchange; damping and steady forced response are introduced only when stated.
- GCE A-Level H2 PhysicsTopic 9(a) / Topic 9(b) / Topic 9(c) · 2027Checked against the syllabus · partial topic coverageOfficial 9478 syllabus
Course and syllabus information
- Course
- GCE A-Level H2 Physics
- Edition
- GCE A-Level H2 Physics 2027