Ultrasound and Echo Sounding · seed 1 · A4, ink-friendly. The answer key prints on its own page for grown-ups.
En anglais

Seeing with echoes

Science · Waves, Light & Sound · ages 15-16
Name ______________________   Date ____________

  1. Why divide an echo's round-trip time by two when calculating the one-way distance at constant speed?

    • The receiver's timer records only the outward journey
    • The pulse makes an outward and a return journey
    • The pulse travels at twice its stated speed on return

  2. Which pairing gives the usual approximate hearing range of a young person and the definition of ultrasound?

    • Hearing: 20 Hz to 20 kHz; ultrasound: above 20 kHz
    • Hearing: 200 Hz to 20 kHz; ultrasound: above 20 kHz
    • Hearing: 20 Hz to 2 kHz; ultrasound: above 20 kHz
    • Hearing: 20 Hz to 20 kHz; ultrasound: any sound too faint to hear

  3. An ultrasound pulse hits a flat boundary head-on. What causes reflection in the ideal boundary model?

    • The materials on both sides have identical acoustic impedance
    • The reflected pulse arrives later than the original pulse was sent
    • The two sides have different acoustic impedance

  4. Why is gel placed between an ultrasound probe and the skin?

    • It cools the skin so that every echo travels twice as fast
    • It fills air gaps that would reflect much of the pulse
    • It changes reflected sound waves into visible colours

  5. A ship sends a pulse vertically down to the sea floor. Use a constant seawater sound speed of 1500 m/s. The echo returns after 2 s. What is the depth in metres?

    Answer: ______________

  6. Two vertical sea-floor soundings use the same constant sound speed. The second echo takes twice as long to return. How does the calculated depth change?

    • The calculated depth doubles
    • The calculated depth is halved
    • The calculated depth stays unchanged

  7. A maker claims that ever-higher ultrasound frequency will show structures ever deeper in soft tissue. What limitation does the claim overlook?

    • Higher frequencies cannot distinguish small tissue features
    • Ultrasound frequency never changes the detail visible in an image
    • Higher frequencies are usually absorbed more in tissue

  8. A pulse travels along a straight path into a large steel block and returns from a crack to the same probe after 0.001 s. Use a constant sound speed of 6000 m/s. What is the one-way distance to the crack in metres?

    Answer: ______________

LightMySky · lightmysky.comW1-mt_nE1uPfjhX--s1

Answer key

For grown-ups. Fold this page away before handing over the rest.

Seeing with echoes W1-mt_nE1uPfjhX--s1

  1. The pulse makes an outward and a return journey · The measured time includes both journeys. For equal outward and return distances at constant speed, one journey takes half that time.
  2. Hearing: 20 Hz to 20 kHz; ultrasound: above 20 kHz · The approximate hearing range is 20 Hz to 20 kHz. Ultrasound is defined by frequency above 20 kHz, not by faintness.
  3. The two sides have different acoustic impedance · An acoustic-impedance difference causes reflection in this model. Matched impedances give no boundary reflection.
  4. It fills air gaps that would reflect much of the pulse · Air and tissue have very different acoustic impedances. Gel fills air gaps, allowing more sound to enter the body.
  5. 1500 · The round-trip path is 1500 × 2 = 3000 m. The depth is 3000 / 2 = 1500 m.
  6. The calculated depth doubles · At fixed speed, d = vt/2 is proportional to return time. Twice the time gives twice the depth.
  7. Higher frequencies are usually absorbed more in tissue · Higher frequencies can improve detail, but stronger absorption usually limits penetration into soft tissue.
  8. 3 · The total path is 6000 × 0.001 = 6 m. The one-way distance to the crack is 6 / 2 = 3 m.
Worksheet · LightMySky