AQA GCSE Physics (8463), Higher tier · Waves › Waves in air, fluids and solids
Practise Properties of waves. 19 exam-style questions plus unlimited generated ones on this subtopic, at up to four difficulty levels, with full mark schemes and a progress tracker. Free, no account needed.
Required practical:Waves (method, variables and exam tips)
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The student measures the distance from the first bright line to the eleventh bright line. The distance is 27 cm. Calculate the wavelength of the waves.
2.7 cm
Write down the equation that links wave speed, frequency and wavelength.
wave speed = frequency × wavelength
Sample questions
Written for this site in the style of AQA exam questions. They are not taken from real past papers.
Question 1Easy6 marks
A student makes waves in a tray of water by dipping a ruler into the water at regular intervals.
(a) The student dips the ruler into the water 15 times in 5.0 seconds. Each dip makes one wave. Calculate the frequency of the waves.[2]
(b) Calculate the period of the waves. Use the equation: \(\text{period} = \dfrac{1}{\text{frequency}}\)[2]
(c) The student now dips the ruler into the water more often. What happens to the frequency and to the period of the waves?[2]
Show the answer and mark scheme
(a)Answer: 3.0 Hz
15 ÷ 5.0
3.0 (Hz)
(b)Answer: 0.33 s
period = 1 ÷ 3.0
0.33 (s)
(c)Answer: The frequency increases and the period decreases.
the frequency increases
the period decreases
Question 2Medium9 marks
A student uses a ripple tank to investigate water waves. A lamp above the tank makes a pattern of bright lines on a sheet of white paper below the tank. The pattern moves across the paper as the waves travel.
(a) The student measures the distance from the first bright line to the eleventh bright line. The distance is 27 cm. Calculate the wavelength of the waves.[2]
(b) Explain why the student measured across several waves rather than measuring one wavelength.[1]
(c) Write down the equation that links frequency, wavelength and wave speed.[1]
(d) The student counts 32 waves passing a point on the paper in 20 s. Calculate the speed of the waves in m/s.[4]
(e) Counting the waves is difficult because the pattern moves quickly. Suggest one way the student could improve the measurement of the frequency.[1]
Show the answer and mark scheme
(a)Answer: 2.7 cm
27 ÷ 10
2.7 (cm)
(b)Answer: It reduces the uncertainty in the measured wavelength.
one wavelength is small so its (percentage) uncertainty would be large / measuring many waves and dividing reduces the uncertainty in the wavelength
(c)Answer: wave speed = frequency × wavelength
wave speed = frequency × wavelength
(d)Answer: 0.043 m/s (0.0432 m/s) m/s
frequency = 32 ÷ 20 = 1.6 (Hz)
wavelength = 0.027 m
v = 1.6 × 0.027
0.043 (m/s)
(e)Answer: Film the pattern and count the waves in slow motion (or use a stroboscope).
use a stroboscope to ‘freeze’ the pattern / film the waves and play the video back in slow motion / count the waves for a longer time
Question 3Hard8 marks
Two students measure the speed of sound in air outdoors, using two different methods.
(a)Method 1 Student A stands 45 m from a tall wall and claps. Each time student A hears the echo of a clap, student A claps again. Student B uses a stopwatch to time 20 intervals between claps. The time is 5.4 s. Calculate the speed of sound given by method 1.[3]
(b) Explain why the students timed 20 intervals rather than one interval.[2]
(c)Method 2 Two microphones are placed 1.50 m apart in a straight line with a loudspeaker. The microphones are connected to a data logger. The data logger records that a sound pulse reaches the second microphone 4.4 ms after it reaches the first microphone. Calculate the speed of sound given by method 2.[2]
(d) Suggest why method 2 is likely to give a more accurate value than method 1.[1]
Show the answer and mark scheme
(a)Answer: 333 m/s
time for one interval = 5.4 ÷ 20 = 0.27 (s)
distance travelled by the sound = 2 × 45 = 90 (m)
v = 90 ÷ 0.27 = 333 (m/s)
(b)Answer: One interval is so short that reaction time would cause a large uncertainty; timing 20 intervals reduces this.
one interval is very short / similar to human reaction time
timing many intervals reduces the effect of reaction time / reduces the (percentage) uncertainty
(c)Answer: 341 m/s
v = 1.50 ÷ 0.0044
341 (m/s)
(d)Answer: The data logger's timing is not affected by human reaction time.
the data logger does not depend on human reaction time / it can measure very short time intervals precisely