Echoes and ultrasound — return questions key

1. ER1 · Return

Model answer. The path to the far wall and back is 340 m/s × 0.30 s = 102 m, so the pool is 102 m ÷ 2 = 51 m long, close to the 50 m of a full-size pool.

Checked automatically

• Length of the pool: 51 m (within ±1 m)
  • Common wrong answer 102 : This is the path to the far wall and back. Halve it.

A full answer includes

• I calculated the path: 340 m/s × 0.30 s = 102 m.
• I halved it: 51 m.

Also accept

• I calculated the path: 340 m/s × 0.30 s = 102 m.
• I halved it: 51 m.

Watch for

• “Length = 340 × 0.30.” — That is the journey there and back. The length of the pool is half of it.

2. ER2 · Return

Model answer. Measure a distance d of about 100 m from a large wall with the trundle wheel and stand there. Clap, then adjust the rhythm until each clap coincides with the echo of the one before, so the interval between claps is the echo time. A partner times many intervals, for example from clap 0 to clap 20 (N = 20), which makes reaction delays small compared with the total time T. Repeat several times, leave out any anomalous time and take the mean. In each interval the sound travels 2d, so speed = 2dN ÷ T.

A full answer includes

• I wrote that the distance d to the wall is measured with the trundle wheel.
• I wrote that the claps are made in a rhythm so that each clap coincides with the echo of the one before.
• I wrote that the stop-clock times many intervals, for example from clap 0 to clap 20.
• I wrote that the timing is repeated several times.
• I wrote that any anomalous time is left out.
• I wrote that a mean of the other times is taken.
• I wrote speed = 2 × d × number of intervals ÷ time.
• I did not write speed = d ÷ echo time.

Also accept

• I wrote that the distance d to the wall is measured with the trundle wheel.
• I wrote that the claps are made in a rhythm so that each clap coincides with the echo of the one before.
• I wrote that the stop-clock times many intervals, for example from clap 0 to clap 20.
• I wrote that the timing is repeated several times.
• I wrote that any anomalous time is left out.
• I wrote that a mean of the other times is taken.
• I wrote speed = 2 × d × number of intervals ÷ time.
• I did not write speed = d ÷ echo time.

Watch for

• “Speed = d ÷ echo time.” — In one echo time the sound travels to the wall and back, 2d. Using d gives half the speed.

3. ER3 · Return

Model answer. Ultrasound is sound with a frequency higher than 20 kHz (20 000 Hz). The tone is 18 000 Hz = 18 kHz, lower than 20 kHz, so it is not ultrasound; many young people can hear it.

A full answer includes

• I wrote that ultrasound is sound with a frequency higher than 20 kHz.
• I wrote that 18 000 Hz = 18 kHz.
• I wrote that 18 kHz is lower than 20 kHz, so the tone is not ultrasound.

Also accept

• I wrote that ultrasound is sound with a frequency higher than 20 kHz.
• I wrote that 18 000 Hz = 18 kHz.
• I wrote that 18 kHz is lower than 20 kHz, so the tone is not ultrasound.

Watch for

• “A sound that some people cannot hear is ultrasound.” — Ultrasound is defined by frequency: higher than 20 kHz. Many adults cannot hear 18 kHz, but it is still below the boundary.

4. ER4 · Return

Model answer. 65 µs = 0.000 065 s. Path = 1540 m/s × 0.000 065 s = 0.1001 m, there and back. Depth = 0.1001 m ÷ 2 = 0.050 05 m = 50.05 mm = 50 mm to 2 significant figures.

Checked automatically

• Depth below the probe: 50 mm (2 significant figures)
  • Common wrong answer 100 : This is the path there and back. Halve it.
  • Common wrong answer 0.050 : This is the depth in metres. 1 m = 1000 mm.

A full answer includes

• I changed 65 µs to 0.000 065 s.
• I calculated the path: 1540 × 0.000 065 = 0.1001 m.
• I halved it and changed it to millimetres: 50 mm, to 2 significant figures.

Also accept

• I changed 65 µs to 0.000 065 s.
• I calculated the path: 1540 × 0.000 065 = 0.1001 m.
• I halved it and changed it to millimetres: 50 mm, to 2 significant figures.

Watch for

• “Depth = 1540 × 65.” — Change microseconds to seconds first: 65 µs = 0.000 065 s.

5. ER5 · Return

Model answer. Row B. In the body, part of each pulse reflects at every boundary between different tissues, and the time of each echo gives the depth of that boundary. In the rail, a crack lies between the probe and the far side, so the pulse reaches it sooner and its echo returns before the back-wall echo. The gel only removes the air gap at the skin, and tissues do not make sound of their own.

A full answer includes

• I chose row B.
• I wrote that in the body each echo comes from a boundary between different tissues.
• I wrote that a crack is nearer than the far side of the rail, so its echo returns before the back-wall echo.

Also accept

• I chose row B.
• I wrote that in the body each echo comes from a boundary between different tissues.
• I wrote that a crack is nearer than the far side of the rail, so its echo returns before the back-wall echo.

Watch for

• “A crack’s echo comes after the back-wall echo.” — The crack is inside the rail, nearer to the probe than the far side. The pulse reaches it first, so its echo returns first.
• “The echoes come from the gel on the skin.” — The gel removes the air gap so that the ultrasound can enter the body. The echoes that make the picture come from tissue boundaries inside.
• “The tissues make sound of their own.” — The probe sends the pulses. Tissues only reflect part of each pulse at their boundaries; they make no ultrasound themselves.