A student stands between two cliffs and claps once. She hears the first echo 0.8 s after the clap and a second echo 1.6 s after the clap. The speed of sound is 330 m/s. What is the distance between the two cliffs?
A 132 m
B 264 m
C 396 m
D 792 m
Show answer
Answer: C
Question 2Multiple choice · 1 mark
A sound wave becomes louder. What has changed?
A its amplitude has increased
B its frequency has increased
C its speed has increased
D its wavelength has increased
Show answer
Answer: A
Question 3Structured · 5 marksA* challenge
Ultrasound is used to scan unborn babies.
(a)
State what is meant by ultrasound.
[1]
(b)
Explain how ultrasound is used to produce an image.
[2]
(c)
State how a louder sound and a higher-pitched sound differ from the original sound wave.
[2]
Show answers and mark scheme
Part
Answer and marking points
Marks
(a)
frequency above 20 kHz
sound with frequency above 20 000 Hz (above the human hearing range) B1
1
(b)
reflections from boundaries timed
pulses are reflected at boundaries between tissues B1 the time taken for the echoes is used to find the depth of each boundary B1
2
(c)
louder: larger amplitude; higher pitch: higher frequency
louder: larger amplitude B1 higher pitch: higher frequency B1
2
Question 4Structured · 9 marks
A microphone is connected to an oscilloscope. Fig. 4.1 shows the trace of a sound wave. The time-base is set to 0.5 ms per division.
Fig. 4.1
(a)
Use Fig. 4.1 to determine the period of the sound wave.
[2]
(b)
Use your answer to (a) to calculate the frequency of the sound.
[2]
(c)
The sound becomes quieter but the pitch stays the same. Describe how the trace changes.
[2]
(d)
Describe the motion of air particles as the sound wave passes, and explain what is meant by a compression.
[3]
Show answers and mark scheme
Part
Answer and marking points
Marks
(a)
5.00 ms
one wave occupies 10.0 divisions C1 10.0 × 0.5 = 5.00 ms A1
2
(b)
200 Hz
f = 1/T = 1/5.00 × 10−3 s C1 200 Hz A1
2
(c)
smaller amplitude; same period (same horizontal spacing of the peaks)
height of the wave (amplitude) decreases B1 the number of waves across the screen stays the same B1
2
(d)
vibrate back and forth parallel to wave direction; compression: region where particles closer together, higher pressure
air particles vibrate backwards and forwards B1 parallel to the direction the wave travels B1 a compression is a region where the particles are closer together (higher pressure) B1
3
Question 5Structured · 3 marks
A student stands 170 m from a large wall and claps. She hears the echo 1.0 s later.
(a)
Calculate the speed of sound in air.
[2]
(b)
Explain why sound cannot travel through a vacuum.
[1]
Show answers and mark scheme
Part
Answer and marking points
Marks
(a)
340 m/s
distance = 2 × 170 = 340 m C1 speed = 340 m/s A1
2
(b)
needs particles to vibrate
Any 1 of: sound is a vibration of particles; there are no particles in a vacuum B1
1
Question 6Structured · 4 marks
A student measures the speed of sound in air using two microphones connected to a timer.
(a)
Describe how the speed is calculated.
[2]
(b)
Suggest why the microphones should be several metres apart.
[1]
(c)
State the approximate speed of sound in air.
[1]
Show answers and mark scheme
Part
Answer and marking points
Marks
(a)
distance between microphones ÷ time
measure the distance between the microphones B1 divide by the time interval recorded B1
2
(b)
longer time, smaller percentage error
the time is longer, so the percentage uncertainty is smaller B1
Riyaz Shaikh has mentored IGCSE, IB and A Level students for 15+ years, one-on-one, online and in Mumbai. The hardest questions on this site are the ones he uses to find a student's weak points.
Original Cambridge-style practice questions written for mentorriyaz.in. These are not Cambridge past paper questions, and this site is not affiliated with or endorsed by Cambridge University Press & Assessment. Updated 2026-11-04. Paper generator · mentorriyaz.in