A quality control engineer uses ultrasound to check for internal cracks in a large metal casting. The engineer places an ultrasound emitter and receiver on the surface of the metal. A pulse is emitted, and two reflected signals are detected: the first arrives back after a short time and the second arrives back after a longer time. Explain how this technique detects the crack and why two reflected signals are received rather than one.

OCR A-Level Physics B: Advancing Physics (H557) — 5.2 Circular motion and gravitational fields · Explain · 5 marks · View as Markdown

Written & reviewed by James Millett — Biology (Imperial College London), PGCE Science (University of Cambridge).

Ultrasound pulses are transmitted into a metal casting from one surface. The speed of ultrasound in the metal is constant throughout.

Model answer (5 marks)

Ultrasound waves are partially reflected at any interface where the acoustic impedance changes. When the pulse is sent into the casting, part of the wave is reflected at the internal crack because the crack creates a boundary between metal and air. This reflected wave returns to the receiver quickly – the first, short‑time signal. The remainder of the pulse continues through the metal, reaches the far surface of the casting, is reflected there, and returns later – the second, longer‑time signal. The constant speed of ultrasound in the metal means that the time interval between emission and reception directly gives the distance to the reflecting surface, so the engineer can locate the crack.

Thus two signals are received: one from the crack and one from the back surface, because the pulse is only partially reflected at the crack and the rest reaches the back surface before returning.

Examiner tips

  • Use the term ‘acoustic impedance’ or ‘boundary’ to show understanding of partial reflection.
  • Explain that the first signal comes from the crack and the second from the back surface – this shows the two‑step process.
  • Mention the constant speed to justify using time to infer depth.
  • Show the pulse is partially reflected at the crack, so the remainder reaches the back surface – this explains why two signals appear.

Common mistakes

  • Saying the crack reflects all the energy, ignoring the second signal.
  • Confusing the order of signals (claiming the back surface signal is first).
  • Failing to mention the constant speed of ultrasound in the metal.

Mark scheme (5 marks)

  1. Ultrasound is partially reflected at each boundary between two different media (or at any interface/surface it meets)
  2. The first (earlier) reflected signal comes from the crack inside the metal, because the crack creates a boundary within the casting
  3. The second (later) reflected signal comes from the far/back surface of the metal casting
  4. The speed of ultrasound is constant, so the time between emission and detection of each signal indicates the depth/distance at which the reflection occurred
  5. Two signals are received because the ultrasound pulse is only partially reflected at the crack, so the remainder continues through the metal and is then reflected from the back surface

Key terms in this question

ultrasound

Related

More Circular motion and gravitational fields questions

▶ Try answering this question with AI marking (free) →