IGCSE Physics 3.4 Sound MCQs – 30 Practice Questions with Answers (Cambridge 0625)

Free practice MCQs for Cambridge IGCSE Physics (0625) Section 3.4, Sound. Work through 30 original questions on how sound is produced and travels, echoes and measuring the speed of sound, ultrasound, and pitch and loudness – then open each Show Answer toggle to check the worked explanation.

Formula recap: speed = distance / time  |  wave speed v = f × λ  |  echo distance = (speed × echo time) / 2  |  period T = 1 / f. Revise the full list on the IGCSE Physics Formula Sheet.

Sound Basics & Speed

Q1. A sound is produced when an object

  • (A) is heated
  • (B) vibrates
  • (C) gives out light
  • (D) carries an electric current
Show Answer

Answer: B – Every sound starts with a vibrating object, such as a guitar string, a loudspeaker cone or vocal cords. The vibrations pass to the surrounding air and travel outwards as a sound wave.

Q2. An electric bell rings inside a glass bell jar. As the air is pumped out of the jar, the ringing gets quieter until it cannot be heard at all, although the bell hammer can still be seen striking. This experiment shows that sound

  • (A) travels faster in a vacuum
  • (B) is stopped by glass
  • (C) needs a material medium to travel
  • (D) is produced only when air is present at the source
Show Answer

Answer: C – With no air left in the jar there is no medium to carry the vibrations to the outside, so no sound is heard. Sound cannot travel through a vacuum; light can, which is why the hammer is still visible.

Q3. Sound in air is a longitudinal wave. This means the air particles vibrate

  • (A) parallel to the direction in which the wave travels
  • (B) at right angles to the direction in which the wave travels
  • (C) in circles around the source
  • (D) only up and down
Show Answer

Answer: A – In a longitudinal wave the vibrations are along the same line as the direction of travel. This produces a series of compressions (particles pushed together) and rarefactions (particles spread apart) moving through the air.

Q4. In which material does sound travel the fastest?

  • (A) air
  • (B) water
  • (C) a vacuum
  • (D) steel
Show Answer

Answer: D – Sound travels fastest in solids (about 5000 m/s in steel), slower in liquids (about 1500 m/s in water) and slowest in gases (about 330-340 m/s in air). It cannot travel in a vacuum at all.

Q5. The approximate speed of sound in air at room temperature is

  • (A) 3.0 m/s
  • (B) 33 m/s
  • (C) 330 m/s
  • (D) 3.0 × 108 m/s
Show Answer

Answer: C – Sound in air travels at roughly 330-340 m/s. The value 3.0 × 108 m/s is the speed of light, which is why lightning is seen almost instantly but the thunder arrives later.

Q6. During a storm, a student sees a lightning flash and hears the thunder 6.0 s later. Taking the speed of sound in air as 330 m/s, how far away was the lightning?

  • (A) 55 m
  • (B) 1980 m
  • (C) 990 m
  • (D) 336 m
Show Answer

Answer: B – distance = speed × time = 330 × 6.0 = 1980 m, about 2.0 km. The light arrives almost instantly, so the 6.0 s is the time taken by the sound alone; no halving is needed because the sound travels one way only.

Q7. Sound travels slightly faster in warm air than in cold air because in warm air the air particles

  • (A) move faster on average, so vibrations are passed on more quickly
  • (B) are pushed much closer together
  • (C) become lighter
  • (D) stop moving completely
Show Answer

Answer: A – At a higher temperature the particles have more kinetic energy and collide more often, so each compression is handed on to the next layer of particles more quickly.

Q8. In a sound wave, a compression is a region where the air particles are

  • (A) furthest apart and the pressure is lowest
  • (B) stationary
  • (C) vibrating at right angles to the wave
  • (D) closest together and the pressure is highest
Show Answer

Answer: D – A compression is where the vibrating source has pushed the air particles together, raising the local pressure. Between compressions are rarefactions, where particles are spread out and the pressure is lower.

Q9. A sound wave passes from air into water. What happens to the speed of the sound?

  • (A) It decreases to about 330 m/s
  • (B) It increases to about 1500 m/s
  • (C) It stays exactly the same
  • (D) It becomes zero
Show Answer

Answer: B – Sound travels faster in liquids than in gases because the particles are much closer together, so vibrations are passed on more quickly. In water the speed is about 1500 m/s.

Q10. Two students stand a long distance apart on a still, open field. When a steady wind blows from student X towards student Y, a shout from X is heard by Y slightly sooner than on a calm day because

  • (A) the wind makes the air particles vibrate faster
  • (B) the wind pushes the echo back
  • (C) the moving air carries the sound along, increasing its effective speed towards Y
  • (D) the wind lowers the frequency of the sound
Show Answer

Answer: C – The sound still travels at about 330 m/s relative to the air, but the air itself is moving towards Y, so the total speed of the sound relative to the ground is greater downwind (and smaller upwind).

Echoes & Measuring the Speed of Sound

Q11. A girl shouts towards a cliff and hears the echo 2.0 s later. The speed of sound in air is 340 m/s. How far away is the cliff?

  • (A) 680 m
  • (B) 170 m
  • (C) 340 m
  • (D) 85 m
Show Answer

Answer: C – The sound travels to the cliff and back, so total distance = 340 × 2.0 = 680 m. The cliff distance is half of this: 680 / 2 = 340 m. (680 m is the classic trap answer – it forgets the return journey.)

Q12. A ship uses SONAR to find the depth of the sea. A pulse of ultrasound is sent down and the echo from the seabed returns after 0.80 s. The speed of sound in seawater is 1500 m/s. The depth of the sea is

  • (A) 600 m
  • (B) 1200 m
  • (C) 300 m
  • (D) 2400 m
Show Answer

Answer: A – Total distance travelled by the pulse = 1500 × 0.80 = 1200 m. This is the journey down and back up, so depth = 1200 / 2 = 600 m.

Q13. In a school experiment to measure the speed of sound using echoes, students stand about 60 m from a large wall rather than 5 m from it. The main reason for using the larger distance is that with a short distance

  • (A) no echo is produced at all
  • (B) the echo returns so quickly that it merges with the original sound and cannot be timed separately
  • (C) the sound becomes too loud to hear
  • (D) the sound changes into ultrasound
Show Answer

Answer: B – At 5 m the echo would return in about 0.03 s, far too short to time with a stopwatch, and it would overlap the original clap. A long distance gives a measurable time interval and reduces the percentage error from reaction time.

Q14. In the clap-echo method, a student claps in rhythm with the echoes from a wall so that each clap coincides with the echo of the previous one. Timing 20 claps instead of just 1 improves the measurement because it

  • (A) makes the sound travel faster
  • (B) makes the echo louder
  • (C) removes the need to know the distance to the wall
  • (D) reduces the percentage error caused by the student’s reaction time
Show Answer

Answer: D – The reaction-time error (about 0.1-0.2 s) is roughly fixed, so spreading it over the total time for 20 clap intervals makes it a much smaller fraction of the measured time. The time for one interval, and hence the speed of sound, is found more accurately.

Q15. A student stands 50 m from a wall and measures the time between a clap and its echo as 0.30 s. The calculated speed of sound is

  • (A) 167 m/s
  • (B) 333 m/s
  • (C) 15 m/s
  • (D) 667 m/s
Show Answer

Answer: B – The sound travels 2 × 50 = 100 m in 0.30 s. speed = distance / time = 100 / 0.30 = 333 m/s. (167 m/s comes from forgetting that the sound makes a return journey.)

Q16. Ultrasound is sound with a frequency

  • (A) below 20 Hz
  • (B) of exactly 20 Hz
  • (C) above 20 000 Hz
  • (D) between 20 Hz and 200 Hz
Show Answer

Answer: C – Human hearing covers roughly 20 Hz to 20 000 Hz (20 kHz). Sound waves with frequencies above 20 000 Hz are called ultrasound; they are too high-pitched for us to hear.

Q17. Which of these is a medical use of ultrasound?

  • (A) Scanning an unborn baby in the womb
  • (B) Producing a magnetic field for an MRI scan
  • (C) Sterilising surgical instruments with heat
  • (D) X-raying broken bones
Show Answer

Answer: A – In prenatal scanning, pulses of ultrasound reflect off the different layers inside the body and are used to build up an image of the developing baby, which can be checked for healthy growth.

Q18. Ultrasound is preferred to X-rays for scanning an unborn baby mainly because ultrasound

  • (A) travels faster than X-rays
  • (B) is cheaper to produce
  • (C) can be heard by the baby
  • (D) is non-ionising, so it does not damage living cells
Show Answer

Answer: D – X-rays are ionising radiation and can damage cells and DNA, which is a particular risk to a developing baby. Ultrasound uses ordinary sound waves at high frequency and is considered safe for routine scanning.

Q19. Jewellery and watch parts are cleaned in an ultrasonic bath. The cleaning works because the ultrasound

  • (A) heats the liquid until the dirt boils away
  • (B) produces very rapid vibrations that shake dirt particles loose from small gaps and surfaces
  • (C) dissolves the metal surface
  • (D) attracts the dirt using static electricity
Show Answer

Answer: B – The high-frequency vibrations pass into the cleaning liquid and agitate it violently on a tiny scale, dislodging dirt from places a brush cannot reach, without scratching the object.

Q20. A fishing boat’s echo sounder receives an echo from a shoal of fish 0.40 s after sending a pulse. The speed of sound in water is 1500 m/s. The shoal is at a depth of

  • (A) 300 m
  • (B) 600 m
  • (C) 750 m
  • (D) 150 m
Show Answer

Answer: A – Distance travelled by the pulse = 1500 × 0.40 = 600 m, covering the depth twice (down and back). Depth = 600 / 2 = 300 m.


Pitch, Loudness & Sound Waves on a CRO

Q21. The pitch of a musical note is determined by the sound wave’s

  • (A) frequency
  • (B) amplitude
  • (C) speed
  • (D) wavelength in metres only
Show Answer

Answer: A – Pitch is how high or low a note sounds, and it is set by the frequency: a higher frequency gives a higher pitch. Amplitude affects loudness, not pitch.

Q22. The loudness of a sound is determined by the sound wave’s

  • (A) frequency
  • (B) speed
  • (C) period
  • (D) amplitude
Show Answer

Answer: D – A larger amplitude means the particles vibrate through a bigger distance and carry more energy, so the sound is louder. Changing the amplitude alone does not change the pitch.

Q23. The range of frequencies that a healthy young human ear can normally detect is approximately

  • (A) 2 Hz to 200 Hz
  • (B) 20 Hz to 2000 Hz
  • (C) 20 Hz to 20 000 Hz
  • (D) 200 Hz to 200 000 Hz
Show Answer

Answer: C – The audible range is about 20 Hz to 20 000 Hz (20 kHz). Sounds below 20 Hz are infrasound and sounds above 20 000 Hz are ultrasound; neither can be heard.

Q24. A note is displayed on a cathode-ray oscilloscope (CRO). The trace then becomes twice as tall but keeps exactly the same shape and number of waves on the screen. The sound has become

  • (A) higher in pitch only
  • (B) louder, with the same pitch
  • (C) quieter, with the same pitch
  • (D) lower in pitch only
Show Answer

Answer: B – The height of the CRO trace represents amplitude, so a taller trace means a louder sound. The number of waves in the same time is unchanged, so the frequency – and therefore the pitch – is the same.

Q25. On a CRO with the same time-base setting, a second note shows twice as many complete waves on the screen as the first note, with the same trace height. Compared with the first note, the second note has

  • (A) the same frequency but a larger amplitude
  • (B) half the frequency and a lower pitch
  • (C) double the frequency and a higher pitch
  • (D) the same pitch but is louder
Show Answer

Answer: C – Twice as many waves in the same time means twice the frequency, which is heard as a higher pitch. The height (amplitude) is unchanged, so the loudness is the same.

Q26. A tuning fork vibrates at a frequency of 440 Hz. The period of one complete vibration is

  • (A) 440 s
  • (B) 0.023 s
  • (C) 2.3 s
  • (D) 0.0023 s
Show Answer

Answer: D – T = 1 / f = 1 / 440 = 0.0023 s (about 2.3 ms). The fork completes 440 vibrations every second, so each single vibration takes only a small fraction of a second.

Q27. A sound wave in air has a frequency of 170 Hz and a wavelength of 2.0 m. The speed of the sound is

  • (A) 340 m/s
  • (B) 85 m/s
  • (C) 0.012 m/s
  • (D) 172 m/s
Show Answer

Answer: A – v = f × λ = 170 × 2.0 = 340 m/s. (85 m/s comes from dividing instead of multiplying.)

Q28. In a loudspeaker, an electrical signal causes the speaker cone to vibrate in and out. The sound is produced because the vibrating cone

  • (A) gives out light waves at the same frequency
  • (B) repeatedly pushes the air in front of it, creating compressions and rarefactions
  • (C) heats the air, which then expands as a wave
  • (D) removes air from the room
Show Answer

Answer: B – As the cone moves out it squeezes the air (a compression); as it moves back it leaves a region of spread-out air (a rarefaction). This repeating pattern travels away as a longitudinal sound wave. A microphone works in reverse: incoming sound vibrates a diaphragm and produces a matching electrical signal.

Q29. Two sounds are displayed on a CRO. One trace is a smooth, regularly repeating wave; the other is jagged and irregular with no repeating pattern. The irregular trace represents

  • (A) a pure musical note
  • (B) ultrasound
  • (C) a louder pure note
  • (D) noise
Show Answer

Answer: D – A musical note produces a regular, repeating waveform on the CRO. Noise is a mixture of many unrelated frequencies and amplitudes, so its trace looks random and irregular.

Q30. Two notes have the same pitch, but the second note is louder than the first. Comparing their CRO traces (same settings), the second trace has

  • (A) more waves on the screen and the same height
  • (B) fewer waves on the screen and a greater height
  • (C) the same number of waves on the screen but a greater height
  • (D) the same height but a longer period
Show Answer

Answer: C – Same pitch means the same frequency, so the same number of waves appear in the same time. Louder means a larger amplitude, so the trace is taller.

Keep practising Physics: browse all boards and topics on the Physics MCQs page, and keep the IGCSE Physics Formula Sheet open while you revise – it collects every equation you need for Cambridge 0625 in one place.

All questions on this page are original practice questions written for QuaidianPoint. They are not Cambridge past-paper questions, and this page is not affiliated with Cambridge Assessment International Education.