Echoes and ultrasound — final check key

1. EK1 · Final check

Model answer. (a) The echo is the sound of the horn reflected from the cliff back to the boat. (b) The echo travels to the cliff and back: 2 × 765 m = 1530 m. Time = 1530 m ÷ 340 m/s = 4.5 s.

Checked automatically

• (b) Time to hear the echo: 4.5 s (within ±0.05 s)
  • Common wrong answer 2.25 : This is the time for the sound to reach the cliff. It must come back too.
  • Common wrong answer 0.44 : You divided the speed by the distance. Time = distance ÷ speed.

A full answer includes

• (a) I wrote that the echo is the sound of the horn reflected from the cliff.
• (b) I used the path 2 × 765 m = 1530 m.
• (b) I wrote 1530 m ÷ 340 m/s = 4.5 s.

Also accept

• (a) I wrote that the echo is the sound of the horn reflected from the cliff.
• (b) I used the path 2 × 765 m = 1530 m.
• (b) I wrote 1530 m ÷ 340 m/s = 4.5 s.

Watch for

• “Time = 765 ÷ 340.” — That is the time for the sound to reach the cliff only. The echo has to return as well.

2. EK2 · Final check

Model answer. (a) 11.40 s is about 0.8 s longer than the other three, which agree within 0.07 s. (b) Mean = (10.62 + 10.55 + 10.58) ÷ 3 = 31.75 ÷ 3 = 10.58 s. Path = 2 × 60 m × 30 = 3600 m; speed = 3600 m ÷ 10.58 s = 340 m/s to 2 significant figures. (c) One echo from 60 m lasts only about 0.35 s, so reaction delays of 0.1–0.2 s at the start and stop would be a large part of it. Thirty intervals take about 10.6 s, and the same delays are a small part of that.

Checked automatically

• (a) Anomalous time: 11.40 s (exact value)
• (b) Speed of sound: 3.4×10^2 m/s (2 significant figures)
  • Common wrong answer 170 : You used 60 m for each interval. The sound goes to the wall and back.
  • Common wrong answer 330 : You included the anomalous time in the mean.

A full answer includes

• (a) I identified 11.40 s.
• (b) I wrote the mean (10.62 + 10.55 + 10.58) ÷ 3 = 10.58 s.
• (b) I used the path 2 × 60 m × 30 = 3600 m.
• (b) I wrote 3600 m ÷ 10.58 s = 340 m/s, to 2 significant figures.
• (c) I wrote that reaction delays at the start and stop are a large part of one short echo time.
• (c) I wrote that over 30 intervals the same delays are a small part of the total time.

Also accept

• (a) I identified 11.40 s.
• (b) I wrote the mean (10.62 + 10.55 + 10.58) ÷ 3 = 10.58 s.
• (b) I used the path 2 × 60 m × 30 = 3600 m.
• (b) I wrote 3600 m ÷ 10.58 s = 340 m/s, to 2 significant figures.
• (c) I wrote that reaction delays at the start and stop are a large part of one short echo time.
• (c) I wrote that over 30 intervals the same delays are a small part of the total time.

Watch for

• “Include every reading in the mean.” — A reading clearly out of line with the others is an anomaly. Leave it out of the mean.

3. EK3 · Final check

Model answer. 0.025 MHz = 0.025 × 1 000 000 Hz = 25 000 Hz, higher than 20 000 Hz, so it is ultrasound. 19 000 Hz, 2 kHz = 2000 Hz and 200 Hz are all below 20 000 Hz.

A full answer includes

• I chose 0.025 MHz.
• I changed it to 25 000 Hz, which is higher than 20 000 Hz.

Also accept

• I chose 0.025 MHz.
• I changed it to 25 000 Hz, which is higher than 20 000 Hz.

Watch for

• “19 000 Hz is ultrasound because it is a large number.” — 19 000 Hz is below 20 000 Hz, so it is still in the audible range.
• “0.025 MHz is too small a number to be ultrasound.” — Change the unit first: 0.025 MHz = 25 000 Hz.
• “2 kHz is ultrasound because it is given in kilohertz.” — 2 kHz = 2000 Hz, well below 20 000 Hz.

4. EK4 · Final check

Model answer. (a) Path = 1500 m/s × 0.028 s = 42 m, down and back, so the depth = 42 m ÷ 2 = 21 m. (b) Yes: 21 m is more than the 15 m needed.

Checked automatically

• (a) Depth below the sonar: 21 m (within ±0.5 m)
  • Common wrong answer 42 : This is the path down and back. Halve it.

A full answer includes

• (a) I wrote 1500 × 0.028 ÷ 2 = 21 m.
• (b) I wrote “Yes”, because 21 m is more than 15 m.

Also accept

• (a) I wrote 1500 × 0.028 ÷ 2 = 21 m.
• (b) I wrote “Yes”, because 21 m is more than 15 m.

Watch for

• “Depth = 1500 × 0.028 = 42 m.” — 42 m is the path down and back. The depth is half of it, 21 m.

5. EK5 · Final check

Model answer. (a) A technician moves a probe along the rail. It sends short pulses of ultrasound into the steel and detects their echoes. Each pulse reflects from the far surface of the rail (the back wall). A crack inside reflects part of the pulse sooner, so the screen shows an extra echo before the back-wall echo. Nothing is cut or damaged. (b) Read the time of the crack echo, change it to seconds, and use depth = speed of sound in steel × time ÷ 2, because the pulse goes to the crack and back. (c) X-rays are ionising and can damage cells, and the risk adds up with every scan. Ultrasound is not ionising, so repeated scans add no dose of ionising radiation, and at the levels used it is not known to harm cells.

A full answer includes

• (a) I wrote that a probe on the rail sends ultrasound pulses into the steel.
• (a) I wrote that the pulses reflect from the far surface (the back wall) and from any crack.
• (a) I wrote that a crack gives an extra echo that arrives before the back-wall echo.
• (b) I wrote depth = speed in steel × crack echo time ÷ 2.
• (c) I wrote that ultrasound is not ionising, so each scan does not give the baby a dose of ionising radiation as an X-ray would.

Also accept

• (a) I wrote that a probe on the rail sends ultrasound pulses into the steel.
• (a) I wrote that the pulses reflect from the far surface (the back wall) and from any crack.
• (a) I wrote that a crack gives an extra echo that arrives before the back-wall echo.
• (b) I wrote depth = speed in steel × crack echo time ÷ 2.
• (c) I wrote that ultrasound is not ionising, so each scan does not give the baby a dose of ionising radiation as an X-ray would.

Watch for

• “A crack shows up as a missing echo.” — A crack is a boundary inside the steel, so it reflects part of the pulse and gives an extra, earlier echo.