Investigate stationary waves and determine wavelength
Interpret string, air-column and microwave experiments and infer wavelength from node or antinode spacing.
Section 8 · Superposition
AS Level; Cambridge International Physics 9702, 2025–2027 syllabus, Topic 8. These original models and calculations are study aids; full timed assessment, supervised practical competence and human subject-teacher review remain outstanding.
Before you begin
- Use wavelength, frequency, phase difference, trigonometry and consistent SI units.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What is the distance between adjacent nodes on a stationary wave?
What you will learn
- Describe stationary-wave demonstrations with strings, air columns and microwaves
- Determine wavelength from adjacent node or antinode positions
- Use the appropriate spacing for the observed pattern
Stationary waves can be demonstrated in different systems
On a stretched string, a vibrator and a reflecting boundary can produce a stationary pattern. Adjusting the driving frequency can create clear nodes and antinodes; the node positions can be marked and measured along the string.
In air columns, sound reflects at the tube boundary and interferes with the incident sound. Resonant lengths can be found by listening or using a microphone. For microwaves, a reflector produces a stationary pattern; a detector records alternating maxima and minima as it is moved. The syllabus treats air-column end corrections as negligible, so no correction is required.
Adjacent nodes or antinodes are half a wavelength apart
In a stationary wave, adjacent nodes are separated by λ/2; adjacent antinodes are also separated by λ/2. A node and its nearest antinode are λ/4 apart, so identify which positions the apparatus measures.
Measure across several equal node-to-node or antinode-to-antinode intervals and divide by the number of intervals to estimate the mean spacing. Then double that spacing to obtain wavelength. Repeated intervals reduce the effect of reading uncertainty.
Pause and explain
A detector measures 4.0 cm between adjacent antinodes. What is λ?
Worked example
Adjacent microwave minima are 1.5 cm apart. Estimate the wavelength.
Show the worked solution
- Adjacent minima correspond to adjacent nodes and are separated by λ/2.
- λ = 2 × 1.5 cm.
- State wavelength in centimetres or metres.
Answer λ = 3.0 cm = 0.030 m.
Try a topic test
Choose Easy, Normal or Difficult. Save your answers and open worked review after finishing.
Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
On a string, three adjacent nodes span 0.48 m from the first node to the third. Estimate wavelength.
Give me a hint
There are two node-to-node intervals in this measured distance.
Compare my reasoning
- Number of intervals = 2.
- Mean adjacent-node spacing = 0.48/2 = 0.24 m.
- Each spacing is λ/2, so double it.
λ = 0.48 m.
Look for these in your work
- Intervals counted correctly
- Mean spacing found
- Wavelength relation applied
A stationary microwave pattern has adjacent maxima 2.8 cm apart. Find wavelength in metres.
Give me a hint
Adjacent antinodes are separated by half a wavelength.
Compare my reasoning
- λ/2 = 2.8 cm.
- λ = 5.6 cm.
- Convert 5.6 cm to metres.
λ = 0.056 m.
Look for these in your work
- Correct pattern spacing used
- Wavelength doubled
- SI conversion correct
An air-column resonance investigation records successive resonance lengths at 17 cm and 51 cm. Treat end corrections as negligible and explain how these two resonances determine the wavelength.
Give me a hint
Successive resonance lengths for the stated boundary arrangement differ by half a wavelength.
Compare my reasoning
- Length difference = 51 − 17 = 34 cm.
- For successive resonances, this difference is λ/2.
- Therefore λ = 68 cm = 0.68 m.
λ = 0.68 m; the 34 cm separation of successive resonances is half a wavelength when end corrections are negligible.
Look for these in your work
- Length difference calculated
- Successive-resonance relation used
- Negligible correction assumption stated
Common mistakes
- Using λ/4 for adjacent nodes
- Measuring from a node to the nearest antinode and doubling it as if it were λ/2
- Forgetting that a practical demonstration is not evidence of supervised practical competence
What can you explain now?
What is the distance between adjacent nodes on a stationary wave?
Compare with the explanation
Half a wavelength
A full cycle of the standing pattern spans two node spacings.
After trying it yourself, choose your next review. This is your self-assessment.
Your review choice appears on Today. Sign in to sync it across devices.
Make it stick
Why do two adjacent nodes mark half a wavelength rather than one whole wavelength?
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.