An engineer is checking a metal pipeline for internal cracks using ultrasound. The ultrasound emitter is placed against the outside of the pipe and sends pulses into the metal. Explain how this method can detect a crack inside the pipeline, and why ultrasound is more suitable for this purpose than visible light.
Written & reviewed by James Millett — Biology (Imperial College London), PGCE Science (University of Cambridge).
Ultrasound waves have a frequency higher than 20 kHz, above 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 pulses are sent into the metal pipe. Where the pulse meets a crack, the acoustic impedance changes (metal to air), so part of the wave is reflected back to the surface. A detector next to the emitter receives this reflected pulse. Because the crack is closer to the surface than the far wall of the pipe, the reflected pulse from the crack arrives earlier than the pulse reflected from the far wall. By measuring the time between emission and reception and using the constant speed of sound in the metal, the distance to the crack can be calculated. Visible light cannot penetrate the metal to reach the crack, so it cannot be used for this inspection, whereas ultrasound can travel through the solid metal.
Examiner tips
- Mention the impedance mismatch at the crack and the partial reflection.
- Explain the time‑of‑flight difference between crack and far wall reflections.
- State that sound speed in metal is constant for distance calculation.
- Contrast light’s inability to penetrate metal with ultrasound’s ability.
Common mistakes
- Failing to note that the crack is detected by an earlier reflected pulse.
- Confusing the speed of light with the speed of sound.
- Omitting the reason why visible light cannot be used.
Mark scheme (5 marks)
- Ultrasound waves are partially reflected at the boundary between the metal and the crack (air gap)
- The reflected waves are detected by a receiver / detector placed next to the emitter
- The crack is detected because the reflected pulse arrives earlier than the pulse reflected from the far wall of the pipe
- The speed of ultrasound is constant in the metal, so the time taken for the pulse to return indicates the distance to the crack
- Visible light cannot pass through the metal to reach the internal crack, whereas ultrasound can travel through solid metal
Related
- All OCR A-Level Physics B: Advancing Physics (H557) revision notes →
- How to answer a "Explain" question →
- Decode the mark scheme abbreviations →
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