Answers · Echoes and ultrasound

Ultrasound and sonar — answer key

Check meaning, not exact wording. These are teaching points, not official Cambridge marks.

1. EQ11 · Practice

Options as numbered on the pupil sheet: 1 — Ultrasound; 2 — Audible to humans.

Model answer. Compare each with 20 kHz = 20 000 Hz. The tone is 12 kHz = 12 000 Hz: audible. The cleaning bath, 25 000 Hz, is higher than 20 000 Hz: ultrasound. The probe is 0.5 MHz = 500 000 Hz: ultrasound. The whistle, 19 000 Hz, is just below 20 000 Hz: audible to a young person.

A full answer includes

I matched the 12 kHz tone (12 000 Hz) to audible.
I matched the 25 000 Hz cleaning bath to ultrasound.
I matched the 0.5 MHz test probe (500 000 Hz) to ultrasound.
I matched the 19 000 Hz whistle to audible.

Also accept

I matched the 12 kHz tone (12 000 Hz) to audible.
I matched the 25 000 Hz cleaning bath to ultrasound.
I matched the 0.5 MHz test probe (500 000 Hz) to ultrasound.
I matched the 19 000 Hz whistle to audible.

Watch for

“A frequency in MHz is small, because 0.5 is less than 20.” — Change the units first: 0.5 MHz = 500 000 Hz, far above 20 000 Hz.
“Any frequency given in kHz is ultrasound.” — “k” means thousand: 12 kHz = 12 000 Hz is below 20 000 Hz, so it is audible.

2. EQ12 · Practice

Model answer. (a) 1 kHz = 1000 Hz, so 45 kHz = 45 × 1000 = 45 000 Hz. (b) 1 MHz = 1 000 000 Hz, so 2.5 MHz = 2 500 000 Hz. (c) 30 000 Hz ÷ 1000 = 30 kHz.

Checked automatically

(a) 45 kHz: 45000 Hz (exact value)
- Common wrong answer 45 : You kept the number in kHz. 1 kHz = 1000 Hz.
- Common wrong answer 4500 : 1 kHz = 1000 Hz, so multiply 45 by 1000.
(b) 2.5 MHz: 2500000 Hz (exact value)
- Common wrong answer 2500 : That is 2.5 MHz in kHz. 1 MHz = 1 000 000 Hz.
(c) 30 000 Hz: 30 kHz (exact value)
- Common wrong answer 30000000 : Divide by 1000 to change Hz to kHz.

A full answer includes

(a) I multiplied by 1000: 45 000 Hz.
(b) I multiplied by 1 000 000: 2 500 000 Hz.
(c) I divided by 1000: 30 kHz.

Also accept

(a) I multiplied by 1000: 45 000 Hz.
(b) I multiplied by 1 000 000: 2 500 000 Hz.
(c) I divided by 1000: 30 kHz.

Watch for

“Change kHz to Hz by dividing by 1000.” — A hertz is smaller than a kilohertz, so there are more of them: multiply by 1000.

3. EQ13 · Practice

Model answer. Ultrasound is defined by its frequency: sound higher than 20 kHz. Loudness depends on amplitude, not frequency. Its speed is set by the medium, the same as for audible sound in that medium. Like all sound it needs a medium, so it cannot cross a vacuum.

A full answer includes

I chose “It is sound with a frequency higher than 20 kHz.”

Also accept

I chose “It is sound with a frequency higher than 20 kHz.”

Watch for

“Ultrasound is sound too loud to hear.” — Loudness depends on amplitude. Ultrasound may be quiet or loud; we cannot hear it because its frequency is above 20 kHz.
“Ultrasound travels faster than audible sound.” — The speed of sound is set by the medium. In the same water, ultrasound and an audible sound travel at the same speed.
“Ultrasound can travel through a vacuum.” — Ultrasound is sound, a vibration passed on by particles, so it needs a medium.

4. EQ14 · Practice

Model answer. (a) 23 kHz = 23 000 Hz, higher than 20 kHz, so the whistle makes ultrasound, above the range of human hearing. A dog can hear higher frequencies than a human. (b) No. The buzzer still vibrates, but ultrasound is still sound: it needs a medium to pass the vibration on, so it cannot cross the vacuum inside the jar to reach the glass and the dog.

A full answer includes

(a) I wrote that 23 kHz = 23 000 Hz is higher than 20 kHz.
(a) I wrote that this is ultrasound, above the range of human hearing.
(a) I wrote that a dog can hear higher frequencies than a human.
(b) I wrote “No”.
(b) I wrote that ultrasound is sound, which needs a medium and cannot cross a vacuum.

Also accept

(a) I wrote that 23 kHz = 23 000 Hz is higher than 20 kHz.
(a) I wrote that this is ultrasound, above the range of human hearing.
(a) I wrote that a dog can hear higher frequencies than a human.
(b) I wrote “No”.
(b) I wrote that ultrasound is sound, which needs a medium and cannot cross a vacuum.

Watch for

“The whistle makes no sound at all.” — It does make a sound: the dog responds. Its frequency is simply too high for human ears.
“Ultrasound can cross a vacuum because it is not ordinary sound.” — Ultrasound is sound at a high frequency. Like all sound, it needs a medium.

5. EQ15 · Practice

Model answer. Path = 1500 m/s × 0.36 s = 540 m, down to the seabed and back. Depth = 540 m ÷ 2 = 270 m.

Checked automatically

Depth: 270 m (within ±3 m)
- Common wrong answer 540 : This is the path down and back up. Halve it.
- Common wrong answer 4170 : You divided 1500 by 0.36. Distance = speed × time.

A full answer includes

I calculated the path: 1500 m/s × 0.36 s = 540 m.
I halved it: 270 m.

Also accept

I calculated the path: 1500 m/s × 0.36 s = 540 m.
I halved it: 270 m.

Watch for

“Depth = speed × time.” — Speed × time is the whole path, down and back. The depth is half of it.

6. EQ16 · Practice

Model answer. Path = 2 × 45 m = 90 m. Time = distance ÷ speed = 90 m ÷ 1500 m/s = 0.060 s.

Checked automatically

Time until the echo is received: 0.060 s (within ±0.001 s)
- Common wrong answer 0.030 : This is the time to reach the floor. The echo must come back up too.
- Common wrong answer 33.3 : You divided the speed by the distance. Time = distance ÷ speed.

A full answer includes

I used the path 2 × 45 m = 90 m.
I wrote 90 m ÷ 1500 m/s = 0.060 s.

Also accept

I used the path 2 × 45 m = 90 m.
I wrote 90 m ÷ 1500 m/s = 0.060 s.

Watch for

“Time = 45 ÷ 1500.” — That is the time to reach the floor only. Use the path down and back, 90 m.

7. EQ17 · Practice

Model answer. The fish swim above the seabed, so the pulse reaches them first and their echo has the shorter path: the 0.050 s echo is from the fish. (a) Depth of the shoal = 1500 m/s × 0.050 s ÷ 2 = 75 m ÷ 2 = 37.5 m. (b) Depth of the seabed = 1500 m/s × 0.120 s ÷ 2 = 180 m ÷ 2 = 90 m.

Checked automatically

(a) Depth of the shoal: 37.5 m (within ±0.5 m)
- Common wrong answer 75 : This is the path down to the fish and back. Halve it.
- Common wrong answer 90 : This is the seabed, from the later echo. The fish are nearer, so their echo comes first.
(b) Depth of the seabed: 90 m (within ±1 m)
- Common wrong answer 180 : This is the path down and back. Halve it.
- Common wrong answer 37.5 : This is the shoal, from the earlier echo. The seabed is farther down.

A full answer includes

(a) I wrote 1500 × 0.050 ÷ 2 = 37.5 m.
(b) I wrote 1500 × 0.120 ÷ 2 = 90 m.
(c) I wrote that the fish are above the seabed, so their echo has a shorter path.
(c) I wrote that the shorter path returns first, at 0.050 s.

Also accept

(a) I wrote 1500 × 0.050 ÷ 2 = 37.5 m.
(b) I wrote 1500 × 0.120 ÷ 2 = 90 m.
(c) I wrote that the fish are above the seabed, so their echo has a shorter path.
(c) I wrote that the shorter path returns first, at 0.050 s.

Watch for

“The later echo is from the fish.” — The order of the echoes depends on distance. The fish are nearer than the seabed, so their echo arrives first.

8. EQ18 · Practice

Model answer. The crack is inside the bar, nearer the probe than the back wall, so its echo returns first: P (3.0 µs) is the crack and Q (8.0 µs) the back wall. (a) Path to the back wall and back = 6000 m/s × 0.000 008 0 s = 0.048 m; thickness = 0.048 m ÷ 2 = 0.024 m = 24 mm. (b) Path to the crack and back = 6000 m/s × 0.000 003 0 s = 0.018 m; depth = 0.018 m ÷ 2 = 0.009 m = 9 mm.

Checked automatically

(a) Thickness of the bar: 24 mm (within ±0.5 mm)
- Common wrong answer 48 : This is the path to the back wall and back. Halve it.
- Common wrong answer 9 : This uses peak P. The back wall is the farthest surface, so its echo arrives last.
- Common wrong answer 0.024 : This is the thickness in metres. 1 m = 1000 mm.
(b) Depth of the crack: 9 mm (within ±0.3 mm)
- Common wrong answer 18 : This is the path to the crack and back. Halve it.
- Common wrong answer 24 : This uses peak Q, the back-wall echo.

A full answer includes

(a) I read the back-wall echo at 8.0 µs (peak Q).
(a) I wrote 6000 × 0.000 008 0 ÷ 2 = 0.024 m = 24 mm.
(b) I wrote 6000 × 0.000 003 0 ÷ 2 = 0.009 m = 9 mm.
(c) I wrote that P is the crack, because the crack is nearer than the back wall and its echo returns first.

Also accept

(a) I read the back-wall echo at 8.0 µs (peak Q).
(a) I wrote 6000 × 0.000 008 0 ÷ 2 = 0.024 m = 24 mm.
(b) I wrote 6000 × 0.000 003 0 ÷ 2 = 0.009 m = 9 mm.
(c) I wrote that P is the crack, because the crack is nearer than the back wall and its echo returns first.

Watch for

“The taller peak is the crack.” — The height of a peak shows how strong the echo is, not where it came from. Its time shows the distance: the crack echo arrives first.

9. EQ19 · Practice

Model answer. (a) Without gel there would be a thin layer of air between the probe and the skin, and that boundary would reflect most of the ultrasound before it entered the body. The gel removes the air. (b) Part of each pulse reflects at each boundary between different tissues, for example at the surface of the liver; the echo times give the depths of the boundaries. (c) X-rays are ionising and can damage cells, which matters most for a developing baby. Ultrasound is not ionising and is not known to harm cells at the levels used.

A full answer includes

(a) I wrote that the gel removes the air between the probe and the skin.
(a) I wrote that an air gap would reflect most of the ultrasound before it entered the body.
(b) I wrote that pulses reflect at boundaries between different tissues.
(c) I wrote that ultrasound is not ionising, unlike X-rays.
(c) I did not write that ultrasound is safe because it is too high to hear.

Also accept

(a) I wrote that the gel removes the air between the probe and the skin.
(a) I wrote that an air gap would reflect most of the ultrasound before it entered the body.
(b) I wrote that pulses reflect at boundaries between different tissues.
(c) I wrote that ultrasound is not ionising, unlike X-rays.
(c) I did not write that ultrasound is safe because it is too high to hear.

Watch for

“Ultrasound is safe because we cannot hear it.” — Being inaudible is not what makes it safer. The reason is that ultrasound is not ionising, unlike X-rays.
“The gel only helps the probe slide.” — The gel does let the probe slide, but its main job is to remove the air gap that would reflect the ultrasound.

10. ET3 · Apply

Model answer. (a) 45 kHz = 45 000 Hz, higher than 20 kHz, so the call is ultrasound. (b) The echo travels through air to the moth and back: path = 340 m/s × 0.020 s = 6.8 m, so the moth is 6.8 m ÷ 2 = 3.4 m away.

Checked automatically

(b) Distance to the moth: 3.4 m (within ±0.05 m)
- Common wrong answer 6.8 : This is the path to the moth and back. Halve it.
- Common wrong answer 15 : You used the speed of sound in water. The bat calls through air.

A full answer includes

(a) I wrote that 45 kHz is higher than 20 kHz, so the call is ultrasound.
(b) I used the speed of sound in air, 340 m/s.
(b) I wrote 340 × 0.020 ÷ 2 = 3.4 m.

Also accept

(a) I wrote that 45 kHz is higher than 20 kHz, so the call is ultrasound.
(b) I used the speed of sound in air, 340 m/s.
(b) I wrote 340 × 0.020 ÷ 2 = 3.4 m.

Watch for

“Ultrasound always travels at 1500 m/s.” — The speed depends on the medium. In air, ultrasound travels at the speed of sound in air, about 340 m/s.

11. ET4 · Apply

Model answer. In 0.108 s the pulse travels down and back: 2 × 82 m = 164 m. Speed = 164 m ÷ 0.108 s = 1518.5… m/s = 1520 m/s to 3 significant figures, close to the usual 1500 m/s; warmer or saltier water carries sound a little faster.

Checked automatically

Speed of sound in the sea water: 1.52×10^3 m/s (3 significant figures)
- Common wrong answer 759 : You used 82 m. The pulse travels down and back, 164 m.

A full answer includes

I used the path 2 × 82 m = 164 m.
I wrote 164 m ÷ 0.108 s = 1520 m/s, to 3 significant figures.

Also accept

I used the path 2 × 82 m = 164 m.
I wrote 164 m ÷ 0.108 s = 1520 m/s, to 3 significant figures.

Watch for

“Speed = 82 ÷ 0.108.” — In the echo time the pulse goes down and back up, so the distance is 2 × 82 m.