Answers · Echoes and ultrasound

Echoes and the speed of sound — answer key

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

1. EQ1 · Practice

Model answer. An echo is the reflection of sound waves. The shout travels to the hard wall, much of it is reflected, and the reflected sound travels back to the pupil. It arrives later than the shout because it has gone further. The wall makes no new sound of its own, and the second sound is not a vibration left in the ear.

A full answer includes

I chose “The sound waves reflect from the wall and travel back to the pupil.”
I can say that an echo is the reflection of sound waves.

Also accept

I chose “The sound waves reflect from the wall and travel back to the pupil.”
I can say that an echo is the reflection of sound waves.

Watch for

“The wall starts to vibrate and makes a new sound of its own.” — The wall is not a new source here: it sends back the sound that reached it. That reflected sound is the echo.
“The sound waves return from the far side of the wall.” — A hard, thick wall reflects most of the sound at the surface facing the pupil. The echo comes from that side.
“The second sound is the ear still vibrating.” — The second sound comes after a clear gap, and the gap depends on how far away the wall is. So it came from outside the ear.

2. EQ2 · Practice

Model answer. In 2.5 s the sound travels 340 m/s × 2.5 s = 850 m. That is the path to the wall and back, so the wall is 850 m ÷ 2 = 425 m away.

Checked automatically

Distance to the far wall: 425 m (within ±5 m)
- Common wrong answer 850 : This is the path to the wall and back. Halve it.
- Common wrong answer 136 : You divided 340 by 2.5. Distance = speed × time.

A full answer includes

I calculated the path: 340 m/s × 2.5 s = 850 m.
I halved the path because the sound went to the wall and back.
I wrote 425 m.

Also accept

I calculated the path: 340 m/s × 2.5 s = 850 m.
I halved the path because the sound went to the wall and back.
I wrote 425 m.

Watch for

“The distance to the wall is speed × echo time.” — Speed × echo time gives the whole path, out and back. Halve it to find the distance to the wall.

3. EQ3 · Practice

Model answer. Row C. A hard, flat brick wall reflects sound well, and from 60 m the echo returns 120 m ÷ 340 m/s ≈ 0.35 s after the clap, long enough to hear it separately. Curtains (rows A and D) absorb much of the sound, so their echo is weak. From 10 m (rows B and D) the echo returns after about 0.06 s and merges with the clap.

A full answer includes

I chose row C.
I wrote that a hard surface such as brick reflects sound well.
I wrote that the surface must be more than about 17 m away, so the echo arrives about 0.1 s or more after the clap.

Also accept

I chose row C.
I wrote that a hard surface such as brick reflects sound well.
I wrote that the surface must be more than about 17 m away, so the echo arrives about 0.1 s or more after the clap.

Watch for

“Any surface gives a clear echo if it is far enough away.” — Soft materials such as curtains absorb much of the sound, so their echo is weak even from far away.
“A nearer wall gives a clearer echo.” — A nearer wall reflects well, but its echo returns within about 0.1 s and merges with the clap instead of being heard on its own.
“Row D works because curtains are close.” — Row D fails both checks: curtains absorb sound, and at 10 m any reflection merges with the clap.

4. EQ4 · Practice

Model answer. The sound travels to the building and back: 2 × 119 m = 238 m. Time = distance ÷ speed = 238 m ÷ 340 m/s = 0.70 s.

Checked automatically

Time from clap to echo: 0.70 s (within ±0.01 s)
- Common wrong answer 0.35 : This is the time to reach the building. Add the journey back.
- Common wrong answer 2.86 : You divided 340 by 119. Time = distance ÷ speed.

A full answer includes

I used the whole path: 2 × 119 m = 238 m.
I used time = distance ÷ speed.
I wrote 238 m ÷ 340 m/s = 0.70 s.

Also accept

I used the whole path: 2 × 119 m = 238 m.
I used time = distance ÷ speed.
I wrote 238 m ÷ 340 m/s = 0.70 s.

Watch for

“Time = 119 ÷ 340.” — That is the time for the sound to reach the building only. The echo has to come back too, so use 238 m.

5. EQ5 · Practice

Model answer. (a) Path = 2 × 1.7 m = 3.4 m; time = 3.4 m ÷ 340 m/s = 0.010 s. (b) Hard tiles reflect sound well. But the ear hears two short sounds separately only when they arrive about 0.1 s or more apart. The reflection returns after only 0.010 s, so it merges with the clap and makes the clap seem longer.

Checked automatically

(a) Time for the reflection to return: 0.010 s (within ±0.001 s)
- Common wrong answer 0.005 : This is the time to reach the wall only. The reflection must come back too.

A full answer includes

(a) I used the path 2 × 1.7 m = 3.4 m.
(a) I wrote 3.4 m ÷ 340 m/s = 0.010 s.
(b) I wrote that the tiled wall does reflect the sound.
(b) I wrote that two short sounds are heard separately only when they are about 0.1 s or more apart.
(b) I wrote that 0.010 s is much shorter, so the reflection merges with the clap.
(b) I did not write that the tiles absorb the sound.

Also accept

(a) I used the path 2 × 1.7 m = 3.4 m.
(a) I wrote 3.4 m ÷ 340 m/s = 0.010 s.
(b) I wrote that the tiled wall does reflect the sound.
(b) I wrote that two short sounds are heard separately only when they are about 0.1 s or more apart.
(b) I wrote that 0.010 s is much shorter, so the reflection merges with the clap.
(b) I did not write that the tiles absorb the sound.

Watch for

“There is no echo because the tiles absorb the sound.” — Hard tiles reflect sound well, and the ringing is those reflections. They return too soon to be heard as a separate echo.
“A wall so close cannot reflect sound.” — Distance does not stop reflection. A near wall reflects too; its reflection simply arrives too soon to be heard on its own.

6. EQ6 · Practice

Options as numbered on the pupil sheet: 1 — The partner stops the stop-clock on the 20th clap after that.; 2 — Measure 100 m out from the wall with a trundle wheel and stand at the mark.; 3 — Calculate the speed from 2 × 100 m × 20 ÷ the stop-clock time.; 4 — A partner starts the stop-clock on one clap, counting it as zero.; 5 — Clap and adjust the rhythm until each clap is made as the echo of the one before arrives..

Model answer. Measure the distance first, so the clapper stands at a known distance from the wall. Match the rhythm next, so each interval between claps equals the echo time. Then start the stop-clock on clap zero, stop it on clap 20, and calculate v = 2dN ÷ T = 2 × 100 m × 20 ÷ T.

A full answer includes

I put measuring 100 m from the wall first.
I put matching the clap rhythm to the echoes before starting the stop-clock.
I put starting the stop-clock on clap zero before stopping it on the 20th clap.
I put the calculation last.

Also accept

I put measuring 100 m from the wall first.
I put matching the clap rhythm to the echoes before starting the stop-clock.
I put starting the stop-clock on clap zero before stopping it on the 20th clap.
I put the calculation last.

Watch for

“Start timing before the rhythm matches the echoes.” — Until each clap coincides with the echo of the one before, the time between claps is not the echo time, so the timing would not give the speed.

7. EQ7 · Practice

Model answer. In each interval the sound goes to the wall and back: 2 × 75 m = 150 m. In 30 intervals it travels 150 m × 30 = 4500 m. Speed = 4500 m ÷ 13.2 s = 340.9… m/s = 341 m/s to 3 significant figures.

Checked automatically

Speed of sound: 341 m/s (3 significant figures)
- Common wrong answer 170 : You used 75 m for each interval. The sound travels to the wall and back, 150 m.
- Common wrong answer 11.4 : You divided one interval’s path by the time for 30 intervals. Multiply by 30.

A full answer includes

I used 2 × 75 m = 150 m for each interval.
I multiplied by 30 intervals: 4500 m.
I wrote 4500 m ÷ 13.2 s = 341 m/s, to 3 significant figures.

Also accept

I used 2 × 75 m = 150 m for each interval.
I multiplied by 30 intervals: 4500 m.
I wrote 4500 m ÷ 13.2 s = 341 m/s, to 3 significant figures.

Watch for

“Speed = 75 × 30 ÷ 13.2.” — In each interval the sound travels to the wall and back, so use 2 × 75 m.
“Speed = 2 × 75 ÷ 13.2.” — The stop-clock time covers 30 intervals, not one. Multiply the path of one interval by 30.

8. EQ8 · Practice

Model answer. (a) Path = 2 × 50 m = 100 m; speed = 100 m ÷ 0.45 s = 222.2… m/s = 220 m/s to 2 significant figures. (b) The true echo time is about 100 m ÷ 340 m/s ≈ 0.29 s. Reaction delays of about 0.1–0.2 s at the start and stop are a large part of so short a time, so the reading is too long and the speed too low. Twenty intervals take about 6 s. The same delays are then a small part of the time measured, and dividing by 20 shares them out over the intervals.

Checked automatically

(a) Speed from the one timing: 2.2×10^2 m/s (2 significant figures)
- Common wrong answer 111 : You used 50 m. The sound travels to the wall and back, 100 m.

A full answer includes

(a) I used the path 2 × 50 m = 100 m.
(a) I wrote 100 m ÷ 0.45 s = 220 m/s to 2 significant figures.
(b) I wrote that reaction delays at the start and the stop are about 0.1–0.2 s.
(b) I wrote that these delays are a large part of a short echo time of about 0.3 s.
(b) I wrote that 20 intervals take about 6 s, so the same delays are a small part of the time.
(b) I wrote that dividing by 20 shares the delays out over the intervals.

Also accept

(a) I used the path 2 × 50 m = 100 m.
(a) I wrote 100 m ÷ 0.45 s = 220 m/s to 2 significant figures.
(b) I wrote that reaction delays at the start and the stop are about 0.1–0.2 s.
(b) I wrote that these delays are a large part of a short echo time of about 0.3 s.
(b) I wrote that 20 intervals take about 6 s, so the same delays are a small part of the time.
(b) I wrote that dividing by 20 shares the delays out over the intervals.

Watch for

“The value is wrong because the clap was too quiet.” — Loudness does not change the speed or the echo time. The problem is timing so short an interval by hand.
“Twenty intervals are better because the sound speeds up over a longer time.” — The sound travels at the same speed in every interval. Timing more intervals only makes the reaction delays a smaller part of the time.

9. EQ9 · Practice

Model answer. (a) 15.31 s is more than a second longer than the other four, which agree within about 0.1 s; perhaps the rhythm was lost. (b) Mean = (14.02 + 14.10 + 14.06 + 13.98) ÷ 4 = 56.16 ÷ 4 = 14.04 s. (c) Path = 2 × 120 m × 20 = 4800 m; speed = 4800 m ÷ 14.04 s = 341.9… m/s = 342 m/s to 3 significant figures.

Checked automatically

(a) Anomalous time: 15.31 s (exact value)
(b) Mean of the other four: 14.04 s (within ±0.005 s)
- Common wrong answer 14.29 : This mean includes the anomalous time. Leave it out.
(c) Speed of sound: 342 m/s (3 significant figures)
- Common wrong answer 171 : You used 120 m for each interval. The sound goes to the wall and back.
- Common wrong answer 336 : You used the mean that includes the anomaly.

A full answer includes

(a) I identified 15.31 s as the anomaly.
(b) I left out 15.31 s and wrote (14.02 + 14.10 + 14.06 + 13.98) ÷ 4 = 14.04 s.
(c) I used the path 2 × 120 m × 20 = 4800 m.
(c) I wrote 4800 m ÷ 14.04 s = 342 m/s, to 3 significant figures.
I did not include the anomaly in the mean.

Also accept

(a) I identified 15.31 s as the anomaly.
(b) I left out 15.31 s and wrote (14.02 + 14.10 + 14.06 + 13.98) ÷ 4 = 14.04 s.
(c) I used the path 2 × 120 m × 20 = 4800 m.
(c) I wrote 4800 m ÷ 14.04 s = 342 m/s, to 3 significant figures.
I did not include the anomaly in the mean.

Watch for

“Every reading must go into the mean.” — A reading clearly out of line with the others is an anomaly. Leave it out of the mean, and repeat that run if you can.
“Use the shortest time, because it is the most accurate.” — The shortest time is not more accurate than the others. Use the mean of the readings that agree.

10. EQ10 · Practice

Model answer. (a) The sound crosses the 450 m gap once: speed = 450 m ÷ 1.30 s = 346.2… m/s = 350 m/s to 2 significant figures. (b) Light travels at about 300 000 000 m/s, so it crosses 450 m in less than two millionths of a second, far less than the 1.30 s measured. The sight of the boards arrives practically at once, and the time measured is the time for the sound alone.

Checked automatically

(a) Speed of sound: 3.5×10^2 m/s (2 significant figures)
- Common wrong answer 692 : You doubled the distance. In this method the sound crosses the gap once.
- Common wrong answer 173 : You halved the distance. In this method the sound crosses the gap once, so nothing is halved.

A full answer includes

(a) I used speed = distance ÷ time with the whole gap, 450 m.
(a) I wrote 450 m ÷ 1.30 s = 350 m/s to 2 significant figures.
(b) I wrote that light crosses 450 m almost instantly, in about a millionth of a second.
(b) I wrote that the measured time is therefore the time for the sound alone.

Also accept

(a) I used speed = distance ÷ time with the whole gap, 450 m.
(a) I wrote 450 m ÷ 1.30 s = 350 m/s to 2 significant figures.
(b) I wrote that light crosses 450 m almost instantly, in about a millionth of a second.
(b) I wrote that the measured time is therefore the time for the sound alone.

Watch for

“The sound travels there and back, so double the distance.” — There is no wall in this method. The sound makes one journey, from the boards to the listener.

11. ET1 · Apply

Model answer. Nearer cliff: path = 340 m/s × 0.80 s = 272 m, so it is 272 m ÷ 2 = 136 m away. Farther cliff: path = 340 m/s × 1.40 s = 476 m, so it is 238 m away. The walker stands between them, so the cliffs are 136 m + 238 m = 374 m apart.

Checked automatically

Distance between the cliffs: 374 m (within ±4 m)
- Common wrong answer 748 : You added the two whole paths. Each echo goes to its cliff and back, so halve each one.
- Common wrong answer 204 : You used the difference of the echo times. Find each distance and add them.
- Common wrong answer 102 : This is the difference of the two distances. The walker stands between the cliffs, so add them.

A full answer includes

I found the distance to the nearer cliff: 340 × 0.80 ÷ 2 = 136 m.
I found the distance to the farther cliff: 340 × 1.40 ÷ 2 = 238 m.
I added them because the walker stands between the cliffs: 374 m.

Also accept

I found the distance to the nearer cliff: 340 × 0.80 ÷ 2 = 136 m.
I found the distance to the farther cliff: 340 × 1.40 ÷ 2 = 238 m.
I added them because the walker stands between the cliffs: 374 m.

Watch for

“Distance between the cliffs = 340 × (1.40 − 0.80).” — The difference of the echo times compares the two cliffs; it is not the gap between them. The walker is between the cliffs, so find each distance and add them.

12. ET2 · Apply

Model answer. Speed = 2dN ÷ T only works if each interval between claps equals the echo time: the time for the sound to go 100 m to the wall and 100 m back. That is why each clap is made as the echo of the one before arrives. Clapping as fast as possible makes each interval shorter than the echo time, so N is too large for the time T and 2 × 100 m × N ÷ T gives a speed that is too high. Accuracy comes from timing many matched intervals, which makes reaction delays a small part of T.

A full answer includes

I wrote that each clap must be made as the echo of the one before arrives.
I wrote that then the time between claps equals the echo time for the 200 m path.
I wrote that clapping faster makes each interval shorter than the echo time, so 2dN ÷ T gives a speed that is too high.
I wrote that accuracy comes from timing many matched intervals, not from clapping fast.

Also accept

I wrote that each clap must be made as the echo of the one before arrives.
I wrote that then the time between claps equals the echo time for the 200 m path.
I wrote that clapping faster makes each interval shorter than the echo time, so 2dN ÷ T gives a speed that is too high.
I wrote that accuracy comes from timing many matched intervals, not from clapping fast.

Watch for

“Any steady rhythm works as long as many claps are timed.” — Only a rhythm matched to the echoes makes each interval equal to the time for the 2d journey.