These twelve questions cover wave speed, amplitude, reflection, refraction, wave types and ripple diagrams from wave behaviour. All data is invented for practice and is not taken from any exam paper. They go from easier to harder.
Write your answer with its unit on paper first, then open the answer. Use v = f × λ throughout. Exact syllabus wording depends on your exam year, so check the Cambridge page for your year.
Questions
1. A wave has frequency 20 Hz and wavelength 0.15 m. Find its speed.
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v = f × λ = 20 × 0.15 = 3.0 m/s.
2. Sound travels at 340 m/s. A note has frequency 425 Hz. Find its wavelength.
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λ = v ÷ f = 340 ÷ 425 = 0.80 m. Check: 425 × 0.80 = 340.
3. A wave on a rope has period 0.25 s and wavelength 0.50 m. Find its frequency and speed.
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f = 1 ÷ 0.25 = 4.0 Hz. v = 4.0 × 0.50 = 2.0 m/s.
4. A wave on a string has wavelength 12 cm and frequency 25 Hz. Find its speed in m/s.
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Convert: 12 cm = 0.12 m. v = 25 × 0.12 = 3.0 m/s (which is 300 cm/s).
5. On a displacement graph, the distance from a crest to the next trough, measured vertically, is 14 mm. State the amplitude.
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Crest to trough is twice the amplitude. 14 ÷ 2 = 7 mm.
6. A graph of water height above a tank floor shows crests at 15.0 cm and troughs at 11.0 cm. Find the amplitude.
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Rest line = (15.0 + 11.0) ÷ 2 = 13.0 cm. Amplitude = 15.0 − 13.0 = 2.0 cm. Check: 13.0 − 11.0 = 2.0 cm.
7. A wave of frequency 10 Hz and speed 3.0 m/s has its amplitude doubled in the same medium. What happens to its speed, frequency and wavelength?
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None of them change. Speed is set by the medium, frequency by the source, and λ = v ÷ f = 3.0 ÷ 10 = 0.30 m stays the same. Only the energy carried increases.
8. Ripples of frequency 5.0 Hz have wavelength 0.060 m in deep water and 0.036 m in a shallow region. Find the speed in each region.
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Deep: v = 5.0 × 0.060 = 0.30 m/s. Shallow: v = 5.0 × 0.036 = 0.18 m/s. The frequency is unchanged, so the wave is slower in the shallow region and bends towards the normal.
9. A ray meets a plane mirror so that the angle between the ray and the mirror surface is 25°. Find the angle of incidence, the angle of reflection and the angle between the incident and reflected rays.
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Angle of incidence = 90° − 25° = 65°. Angle of reflection = 65°. The angle between the rays is 65° + 65° = 130°.
10. Two successive compressions in a sound wave are 0.85 m apart. The speed of sound is 340 m/s. Find the frequency.
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λ = 0.85 m. f = 340 ÷ 0.85 = 400 Hz. Check: 400 × 0.85 = 340.
11. A diagram states that six parallel wavefronts span 0.45 m. The frequency is 3.0 Hz. Find the wavelength and the speed.
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Six wavefronts have 5 gaps. λ = 0.45 ÷ 5 = 0.090 m. v = 3.0 × 0.090 = 0.27 m/s.
12. On a ripple diagram, adjacent wavefronts are 1.8 cm apart. The scale says 1.0 cm represents 4.0 cm on the tank, and the frequency is 2.5 Hz. Find the speed.
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Real wavelength = 1.8 × 4.0 = 7.2 cm = 0.072 m. v = 2.5 × 0.072 = 0.18 m/s.
If you got these wrong
- Questions 1 to 4: equation, units or period. Go to connecting wavelength, frequency and speed.
- Questions 5 to 7: amplitude reading or what amplitude changes. Go to reading amplitude without doubling it.
- Questions 8 and 9: which quantity changes at a boundary, or angles. Go to reflection and refraction using wavefronts.
- Question 10: compressions and wave type. Go to transverse and longitudinal motion.
- Questions 11 and 12: counting gaps or applying a scale. Go to interpreting ripple-style diagrams.
The bounds and rounding explainer and the triangle and bearings reasoning board support careful reading and angle work. Log repeated slips in the mistake log and retest queue and retry the same skill a few days later.
What next?
When the set feels comfortable, move on to light and imaging, which builds on reflection and refraction. If one error type keeps coming back, our online one-to-one Physics tuition is built for tracing it to the exact step.