Answers · Sound waves

Sound needs a medium — answer key

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

1. SQ13 · Practice

Model answer. The hammer keeps the bell vibrating, so the source has not stopped. Removing air reduces the medium carrying sound through the jar; letting air back in restores that path and the sound becomes louder.

A full answer includes

Hammer motion shows the source continues vibrating.
Less air remains to carry the sound.
Air is a medium: sound needs it to travel.
Letting air back in makes the sound louder again.

Also accept

Hammer motion shows the source continues vibrating.
Less air remains to carry the sound.
Air is a medium: sound needs it to travel.
Letting air back in makes the sound louder again.

Watch for

“Removing air switches off the bell.” — The visible hammer still hits; it is the sound-transmission path through the jar that changes.

2. SQ14 · Practice

Model answer. Sea water (a liquid), a wooden table (a solid) and helium (a gas) are all material media: their particles can pass a vibration on to their neighbours, so all three carry sound. A perfect vacuum has no particles, so there is nothing to pass the vibration on and sound cannot cross it.

A full answer includes

I chose sea water.
I chose a wooden table.
I chose helium gas.
I did not choose a perfect vacuum.
I explained that a perfect vacuum has no particles to pass the vibration on.

Also accept

I chose sea water.
I chose a wooden table.
I chose helium gas.
I did not choose a perfect vacuum.
I explained that a perfect vacuum has no particles to pass the vibration on.

Watch for

“Only air can carry sound.” — Sound needs a medium, and solids and liquids can serve as media as well as gases.
“A solid table is too stiff to carry sound, because its particles cannot move about.” — The particles of a solid vibrate about fixed places and pass the vibration to their neighbours. Knock on a table with your ear on it and you hear it.

3. SQ15 · Practice

Model answer. The light arrives almost at once, so the 1.5 s delay is the time the sound takes. Distance = speed × time = 340 m/s × 1.5 s = 510 m. The sound makes one journey, from the firework to the girl, so the distance is not halved.

Checked automatically

Distance: 510 m (within ±5 m)
- Common wrong answer 255 : This is one journey from the firework to the girl; do not halve the distance.
- Common wrong answer 227 : You divided 340 by 1.5. Sound covers 340 m in each second, so in 1.5 s it goes 340 × 1.5.

A full answer includes

I used distance = speed × time.
I wrote 340 × 1.5 = 510 m.
I did not halve the distance: the sound makes one journey.

Also accept

I used distance = speed × time.
I wrote 340 × 1.5 = 510 m.
I did not halve the distance: the sound makes one journey.

Watch for

“Halve the distance because the flash and the bang are two journeys.” — The flash marks the start time; the sound makes one journey to the girl. Do not halve it.

4. SQ16 · Practice

Model answer. Time = distance ÷ speed = 680 m ÷ 340 m/s = 2.0 s. The distance is twice what sound covers in one second.

Checked automatically

Travel time: 2.0 s (within ±0.02 s)
- Common wrong answer 231200 : You multiplied 680 by 340. To find a time, divide the distance by the speed: 680 ÷ 340.
- Common wrong answer 0.5 : You divided the speed by the distance. Time = distance ÷ speed, so 680 goes on top.

A full answer includes

I used time = distance ÷ speed.
I wrote 680 ÷ 340 = 2.0 s.

Also accept

I used time = distance ÷ speed.
I wrote 680 ÷ 340 = 2.0 s.

Watch for

“Multiply distance by speed to find time.” — To find seconds, divide metres by metres per second.

5. SQ17 · Practice

Options as numbered on the pupil sheet: 1 — Aluminium; 2 — Carbon dioxide; 3 — Sea water.

Model answer. The speeds rise from carbon dioxide (about 260 m/s) to sea water (about 1530 m/s) to aluminium (about 6400 m/s), so the order is carbon dioxide, sea water, aluminium. Over the same distance the greatest speed gives the shortest time, so aluminium gives the earliest arrival.

A full answer includes

I ordered carbon dioxide, sea water, aluminium.
I wrote that aluminium gives the earliest arrival.
I gave the reason: over the same distance, a greater speed means a shorter time.

Also accept

I ordered carbon dioxide, sea water, aluminium.
I wrote that aluminium gives the earliest arrival.
I gave the reason: over the same distance, a greater speed means a shorter time.

Watch for

“Carbon dioxide is fastest because gas particles move around freely.” — Sound speed is the speed of the travelling disturbance, not how freely a particle flows. Use the speed column.

6. SQ18 · Practice

Model answer. The explosions shake the Sun’s gases, so there are vibrations at the source. Light can travel through a vacuum, so it reaches Earth. The space between the Sun and Earth is almost a vacuum, with too few particles to pass a vibration on, and sound needs a medium, so no sound wave reaches us.

A full answer includes

I wrote that light can travel through a vacuum.
I wrote that the space between the Sun and Earth is almost a vacuum.
I wrote that sound needs a medium to pass the vibration on.
I did not write that the explosions make no vibrations.

Also accept

I wrote that light can travel through a vacuum.
I wrote that the space between the Sun and Earth is almost a vacuum.
I wrote that sound needs a medium to pass the vibration on.
I did not write that the explosions make no vibrations.

Watch for

“We cannot hear the explosions only because the Sun is so far away that the sound dies out.” — Distance alone would make a sound fainter. Here no sound wave can cross the near-vacuum at all, however loud the source.
“The explosions make no vibrations because the Sun is made of gas.” — Gases vibrate too: air is a gas, and it carries sound. What is missing is a medium between the Sun and Earth.

7. SQ19 · Practice

Model answer. The 2.7 s delay is approximately the sound travel time. Distance = 340 m/s × 2.7 s = 918 m, which is 920 m to 2 significant figures, for this single source-to-listener journey.

Checked automatically

Distance: 9.2×10^2 m (2 significant figures)
- Common wrong answer 130 : You divided 340 by 2.7. In 2.7 s sound covers 2.7 times 340 m.

A full answer includes

I used distance = speed × time.
I wrote 340 × 2.7 = 918 m.
I gave 920 m, to 2 significant figures.
I did not divide 340 by 2.7.

Also accept

I used distance = speed × time.
I wrote 340 × 2.7 = 918 m.
I gave 920 m, to 2 significant figures.
I did not divide 340 by 2.7.

Watch for

“Distance = 340 ÷ 2.7.” — Speed is distance per second; multiply by the number of seconds to obtain distance.

8. SQ20 · Practice

Model answer. (a) 0.52 s is far from the other four times, so it is anomalous (perhaps the watch was stopped too early) and is left out. Mean = (0.74 + 0.73 + 0.75 + 0.74) ÷ 4 = 2.96 ÷ 4 = 0.74 s. (b) Speed = distance ÷ time = 250 m ÷ 0.74 s = 338 m/s, which is 340 m/s to 2 significant figures. (c) The speed of sound in air is approximately 330–350 m/s; 340 m/s lies inside this range, so the result agrees.

Checked automatically

(a) Mean of the other four times: 0.74 s (within ±0.005 s)
- Common wrong answer 0.696 : You included the anomalous 0.52 s. Leave it out and average the other four times.
(b) Speed of sound: 3.4×10^2 m/s (2 significant figures)
- Common wrong answer 360 : This speed comes from a mean that includes the anomalous 0.52 s. Leave it out.
- Common wrong answer 0.0030 : You divided the time by the distance. Speed = distance ÷ time = 250 ÷ 0.74.

A full answer includes

(a) I identified 0.52 s as anomalous.
(a) I left 0.52 s out of the mean.
(a) I wrote (0.74 + 0.73 + 0.75 + 0.74) ÷ 4 = 0.74 s.
(b) I used speed = distance ÷ time.
(b) I wrote 250 ÷ 0.74 = 338 m/s.
(b) I gave 340 m/s, to 2 significant figures.
(c) I wrote that the speed of sound in air is about 330–350 m/s.
(c) I wrote that my value lies inside this range, so it agrees.

Also accept

(a) I identified 0.52 s as anomalous.
(a) I left 0.52 s out of the mean.
(a) I wrote (0.74 + 0.73 + 0.75 + 0.74) ÷ 4 = 0.74 s.
(b) I used speed = distance ÷ time.
(b) I wrote 250 ÷ 0.74 = 338 m/s.
(b) I gave 340 m/s, to 2 significant figures.
(c) I wrote that the speed of sound in air is about 330–350 m/s.
(c) I wrote that my value lies inside this range, so it agrees.

Watch for

“Every reading must go into the mean, so the mean is 0.696 s.” — A reading far from the others is anomalous: leave it out. Including 0.52 s gives a mean of 0.696 s and a speed of about 360 m/s, outside 330–350 m/s.

9. SQ21 · Practice

Model answer. (a) In general, sound travels faster in solids than in liquids, and faster in liquids than in gases. (b) Copper, a solid, has 4700 m/s; ethanol, a liquid, 1160 m/s; helium, a gas, 1000 m/s. The order solid, liquid, gas agrees with the rule, even though the gap between ethanol and helium is small. (c) Copper’s particles vibrate about fixed positions and pass the vibration on to their neighbours; the disturbance travels, but the copper does not need to flow.

A full answer includes

(a) I wrote that, in general, sound is faster in solids than in liquids.
(a) I wrote that, in general, it is faster in liquids than in gases.
(b) I wrote that copper (4700 m/s) is faster than ethanol (1160 m/s).
(b) I wrote that ethanol (1160 m/s) is faster than helium (1000 m/s), so the data agree.
(c) I wrote that copper’s particles vibrate about fixed positions.
(c) I wrote that each particle passes the vibration on to the next.
(c) I did not write that the particles travel along with the sound.

Also accept

(a) I wrote that, in general, sound is faster in solids than in liquids.
(a) I wrote that, in general, it is faster in liquids than in gases.
(b) I wrote that copper (4700 m/s) is faster than ethanol (1160 m/s).
(b) I wrote that ethanol (1160 m/s) is faster than helium (1000 m/s), so the data agree.
(c) I wrote that copper’s particles vibrate about fixed positions.
(c) I wrote that each particle passes the vibration on to the next.
(c) I did not write that the particles travel along with the sound.

Watch for

“A solid cannot carry sound because its particles cannot move from place to place.” — Vibration about fixed positions is enough to pass the disturbance on.
“Helium and ethanol are so close that the table disagrees with the rule.” — The rule compares the order, not the size of the gap. 1160 m/s is still more than 1000 m/s, so the liquid is faster than the gas.

10. ST3 · Apply

Model answer. Air at each end can carry sound locally, but the intervening vacuum has no particles to pass the disturbance across. With no solid route either, there is no continuous material path from the vibrating bell to the listener.

A full answer includes

Sound needs a continuous material path.
Air at the two ends does not carry sound across the intervening vacuum.

Also accept

Sound needs a continuous material path.
Air at the two ends does not carry sound across the intervening vacuum.

Watch for

“Air near the source and listener is enough, even if a vacuum separates them.” — Check the whole path: a vacuum gap interrupts transmission of sound.

11. ST4 · Apply

Model answer. (a) 1.19 km = 1190 m, so time = distance ÷ speed = 1190 m ÷ 340 m/s = 3.5 s. (b) Supplement: 6.12 km = 6120 m, so time = 6120 m ÷ 1530 m/s = 4.0 s. (c) Supplement: sound is faster in water, but the diver is more than five times as far away, so the listener hears it first, 0.5 s earlier. Time depends on distance as well as speed.

Checked automatically

(a) Time to the listener: 3.5 s (within ±0.05 s)
- Common wrong answer 0.0035 : You divided kilometres by metres per second. Change 1.19 km to 1190 m first.
- Common wrong answer 405 : You multiplied the distance by the speed. Time = distance ÷ speed.
(b) Supplement: time to the diver: 4.0 s (within ±0.05 s)
- Common wrong answer 0.004 : You divided kilometres by metres per second. Change 6.12 km to 6120 m first.

A full answer includes

(a) I changed 1.19 km to 1190 m.
(a) I used time = distance ÷ speed.
(a) I wrote 1190 ÷ 340 = 3.5 s.
(b) Supplement: I wrote 6120 ÷ 1530 = 4.0 s.
(c) Supplement: I wrote that the listener hears it first, 0.5 s before the diver.
(c) Supplement: I wrote that the arrival time depends on the distance as well as the speed.
(c) Supplement: I did not write that the faster medium always gives the earlier arrival.

Also accept

(a) I changed 1.19 km to 1190 m.
(a) I used time = distance ÷ speed.
(a) I wrote 1190 ÷ 340 = 3.5 s.
(b) Supplement: I wrote 6120 ÷ 1530 = 4.0 s.
(c) Supplement: I wrote that the listener hears it first, 0.5 s before the diver.
(c) Supplement: I wrote that the arrival time depends on the distance as well as the speed.
(c) Supplement: I did not write that the faster medium always gives the earlier arrival.

Watch for

“The fastest medium always gives the first arrival regardless of distance.” — Time is distance divided by speed. A much longer fast path may still take longer.