Five original Cambridge-style questions on wave features. Attempt each one in your notebook, then tap the reveal button to see the worked solution in a teacher's hand.
Reading the answers without trying first is the fastest way to learn nothing. Open your notebook, work each question out by hand, write your answer, and only then reveal the worked solution. Compare your method, not just your final answer.
State the SI unit, including its symbol, for each of the following wave quantities.
(a) wavelength (b) frequency (c) amplitude (d) period(a) wavelength → metres (m) ✓
(b) frequency → hertz (Hz) ✓
(c) amplitude → metres (m) ✓
(d) period → seconds (s) ✓
Each correct unit with its symbol is worth 1 mark. Don't lose marks by writing "Hertz" without the symbol Hz.
A sound wave produced by a tuning fork has a frequency of 250 Hz. Calculate the period of the wave.
Use: T = 1 ÷ f
T = 1 ÷ 250 T = 0.004 sT = 4 × 10⁻³ s
Either 0.004 s or 4 ms gets the full mark. Always include the unit.
A wave on a stretched rope oscillates between a peak that is 7 cm above the rest position and a trough that is 7 cm below it. A student claims the amplitude of the wave is 14 cm.
Explain why the student is incorrect and state the correct value of the amplitude.
Why the student is wrong:
Amplitude = max displacement from rest position to the peak (or trough).
The student has measured peak-to-trough distance, which is twice the amplitude.
Correct value:
amplitude = 7 cm
Common mistake. Always measure amplitude from rest, never crest to trough.
Two sound waves, A and B, are produced by different sources and displayed on the same displacement-time graph.
Wave A completes 8 oscillations in 2.0 seconds, and has an amplitude of 0.04 m.
Wave B completes 4 oscillations in 2.0 seconds, and has an amplitude of 0.08 m.
(a) Calculate the frequency of each wave. [2] (b) Calculate the period of each wave. [2] (c) Describe how the two sounds would differ to a listener. [1](a) Frequencies
Wave A: f = 8 ÷ 2.0 = 4 Hz Wave B: f = 4 ÷ 2.0 = 2 Hz(b) Periods
Wave A: T = 1 ÷ 4 = 0.25 s Wave B: T = 1 ÷ 2 = 0.50 s(c) Comparison of sound
Wave A: higher pitch, quieter
Wave B: lower pitch, louder
Higher frequency → higher pitch. Larger amplitude → louder. Two independent properties.
A small buoy floats on the surface of the sea. As waves pass, the buoy bobs up and down. A student observing the buoy from a nearby pier notes the following:
• In 30 seconds, the buoy completes 15 full up-and-down oscillations.
• At the highest point of each wave, the buoy is 0.60 m above its rest position.
(a) Frequency
f = oscillations ÷ time f = 15 ÷ 30 = 0.5 Hz(b) Period
T = 1 ÷ f = 1 ÷ 0.5T = 2.0 s
(c) Amplitude
a = 0.60 m
Amplitude here is given directly as the height above rest. No calculation needed.