Use ultrasound echoes for non-destructive material testing
Locate a reflecting defect using a round-trip delay while keeping depth separate from the travelled distance.
Exam scope and assessment notes
Extended only (Supplement) · Physics 0972, 2026–2028. Follow the selected spectrum or sound section. Diagrams illustrate stated ideal models and supplied data. Written practice is self-checked. Human subject-teacher review, supervised practical competence and cumulative assessment remain pending. Cambridge states that IGCSE 0972 has the same subject content as IGCSE 0625, with 9–1 grading. This StudyForA pathway maps the 2026–2028 syllabus only. This 9–1 graded qualification is available only in a limited number of administrative zones. Confirm that your centre can enter you for this code and exam series with Cambridge International. Cambridge has published a separate syllabus for 2029. StudyForA has not reconciled that version yet; this page supports exams in 2026, 2027 and 2028 only.
Before you begin
- Define ultrasound and an echo.
- Convert microseconds and use distance = speed × time.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What does an early internal ultrasound echo suggest?
What you will learn
- Describe ultrasound in non-destructive testing.
- Calculate a reflector distance from a supplied speed and round-trip time.
Send and detect a reflected pulse
A transducer sends an ultrasound pulse into a material and detects returned echoes. An internal boundary or defect can reflect part of the disturbance before it reaches the back surface. Arrival times provide information about reflector locations without cutting the object open.
A useful measurement requires coupling into the material, a known sound speed and identified returns. An unexpected early echo can indicate an internal reflector, but amplitude alone does not uniquely identify every defect or establish the object's structural safety.
Halve the round-trip path
When the transmitter and receiver are effectively at the same surface position and the pulse follows the same direct path out and back, a reflector depth is d = vt/2. At 5000 m/s and 12 μs, the pulse travels 0.060 m in total, giving a depth of 0.030 m.
The factor two belongs to the path geometry, not to ultrasound itself. An electronic delay or a different reflected route must be accounted for if supplied. Written calculations are explanatory checks; actual instrument interpretation and material evaluation need specialist practice.
Pause and explain
For 5000 m/s and 12 μs round trip, what is the one-way depth?
Worked example
A pulse returns from a reflector after 12 μs in a material with sound speed 5000 m/s. Find depth for the direct out-and-back model.
Show the worked solution
- 12 μs = 12 × 10⁻⁶ s.
- Total sound path = 5000 × 12 × 10⁻⁶ = 0.060 m.
- Depth = 0.060/2 = 0.030 m.
Answer 0.030 m, or 3.0 cm.
Try a topic test
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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.
Describe ultrasound non-destructive testing.
Give me a hint
Include pulse launch, internal reflection and arrival measurement.
Compare my reasoning
- A transducer sends ultrasound into the material.
- A boundary or defect reflects a pulse.
- The receiver measures the return and uses the stated speed to locate the reflector.
Pulse echoes reveal reflector locations without cutting the object open.
Look for these in your work
- Name ultrasound.
- Include reflection and timing.
Find reflector depth for 20 μs round trip at 5000 m/s.
Give me a hint
Use one half of the travelled path.
Compare my reasoning
- t = 20 × 10⁻⁶ s.
- vt = 0.100 m total.
- d = 0.100/2 = 0.050 m.
0.050 m, or 5.0 cm.
Look for these in your work
- Convert microseconds.
- Halve the round trip.
A 14 μs recorded delay includes a known 2 μs electronic delay. Find depth at 5000 m/s.
Give me a hint
Remove the non-propagation part before halving the path.
Compare my reasoning
- Sound round-trip time = 14 − 2 = 12 μs.
- Sound path = 5000 × 12 × 10⁻⁶ = 0.060 m.
- Depth = 0.030 m.
0.030 m after correcting the stated electronic delay.
Look for these in your work
- Separate system and propagation delays.
- Use the corrected round-trip time.
Common mistakes
- The recorded delay is round trip.
- Convert microseconds.
- An echo is evidence of a reflector, not a complete safety judgement.
What can you explain now?
Why divide vt by two in this model?
Compare with the explanation
The pulse travels to the reflector and back.
The travelled distance is twice the one-way depth for the stated geometry.
After trying it yourself, choose your next review. This is your self-assessment.
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Make it stick
Explain non-destructive testing, calculate a reflector depth and state the limits of an echo interpretation.
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.