A security engineer is testing whether cracks are present inside a metal railway track. Ultrasound pulses are sent into the metal from one end. Explain how ultrasound can be used to detect a crack inside the track, and why ultrasound is preferred over visible light for this purpose.
Written & reviewed by James Millett — Biology (Imperial College London), PGCE Science (University of Cambridge).
Ultrasound waves have a frequency higher than 20 kHz, the upper limit of human hearing. When ultrasound reaches a boundary between two media, the waves are partially reflected back, while the remainder continue through.
Model answer (5 marks)
Ultrasound waves are partially reflected when they reach a boundary between two different media.
A crack inside the metal creates such a boundary, so part of the pulse is reflected back to the transducer before the pulse reaches the far end of the track.
Because the speed of sound in the metal is essentially constant, the time interval between the transmitted pulse and the reflected echo can be measured; using (d= frac{1}{2}v,t) this gives the distance to the crack.
Visible light cannot penetrate an opaque metal, so it cannot reveal internal defects.
Ultrasound can travel through the solid metal, making it a non‑invasive method that can detect internal features such as cracks without damaging the track.
A crack inside the metal creates such a boundary, so part of the pulse is reflected back to the transducer before the pulse reaches the far end of the track.
Because the speed of sound in the metal is essentially constant, the time interval between the transmitted pulse and the reflected echo can be measured; using (d= frac{1}{2}v,t) this gives the distance to the crack.
Visible light cannot penetrate an opaque metal, so it cannot reveal internal defects.
Ultrasound can travel through the solid metal, making it a non‑invasive method that can detect internal features such as cracks without damaging the track.
Examiner tips
- Use the command word ‘Explain’ – give a clear sequence of cause and effect. Mention the constant speed of sound and the time‑distance calculation. State why light fails and why ultrasound succeeds. Keep each point concise and use exact terminology (e.g. ‘reflected echo’, ‘speed of sound’).
Common mistakes
- Confusing the reflected pulse with the transmitted pulse. Assuming light can pass through metal. Omitting the time‑distance relationship or the constant speed of sound.
Mark scheme (5 marks)
- Ultrasound waves are partially reflected when they reach a boundary between two different media
- A crack creates a boundary inside the metal, so some ultrasound is reflected back earlier / before reaching the far end
- The speed of ultrasound in the metal is constant, so the time between emission and detection of the reflected pulse indicates the distance to the crack
- Visible light cannot pass through an opaque metal, so cannot be used to detect internal cracks
- Ultrasound can travel through the solid metal, making it suitable for detecting internal features / it is non-invasive and does not damage the track
Related
- All OCR A-Level Physics A (H556) revision notes →
- How to answer a "Explain" question →
- Decode the mark scheme abbreviations →
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