Physics • Year 11 • Module 3 • Lesson 2
Wave Properties and the Wave Equation
Apply the wave equation and graph-reading skills to real data, wave boundary problems and a displacement-time graph analysis.
1. Complete the wave data table
Use v = fλ and T = 1/f to find each missing quantity. Show your working below the table. 10 marks (1 per missing cell)
| Wave | Wave speed v (m/s) | Frequency f (Hz) | Wavelength λ (m) | Period T (s) |
|---|---|---|---|---|
| Sound in air (room temperature) | 340 | 440 | ||
| Wave on a guitar string | 280 | 0.35 | ||
| Seismic P-wave in granite | 5.0 | 1200 | ||
| AM radio signal (electromagnetic) | 3.0 × 108 | 300 | ||
| Ripple in water tank | 12 | 0.040 |
Working space:
2. Interpret a displacement-time graph — sonar pulse return
A sonar device emits a pulse and detects its return. The graph below shows the displacement of a receiver membrane over time as the echo arrives. 8 marks
Figure 2.1. Displacement-time graph for a sonar receiver membrane. Sound speed in water = 1500 m/s. Illustrative data.
2.1 Use the graph to determine the period of the wave. Show how you read this from the graph. 2 marks
2.2 Calculate the frequency of the sonar wave. 2 marks
2.3 The speed of sound in water is 1500 m/s. Calculate the wavelength of this sonar wave in water. 2 marks
2.4 A student reads this displacement-time graph and states: “The horizontal spacing between crests is the wavelength.” Explain the error and write the corrected statement. 2 marks
3. Compare displacement-distance and displacement-time graphs
Complete the two-column table below. For each feature, write a concise description contrasting the two graph types. 10 marks (1 per cell)
| Feature | Displacement-distance graph | Displacement-time graph |
|---|---|---|
| What the horizontal axis represents | ||
| Horizontal axis units | ||
| What horizontal spacing between crests gives you | ||
| What the vertical axis shows | ||
| What “snapshot” or “history” it represents |
4. Predict and justify — sound crossing from air into water
A sound wave of frequency 500 Hz travels through air (v = 340 m/s) and then crosses into water (v = 1500 m/s). 6 marks
4.1 Calculate the wavelength in air and the wavelength in water. Show full working. 3 marks
4.2 Predict and justify what happens to the frequency when the sound enters the water. 2 marks
4.3 A marine engineer claims that sound “slows down” when it travels from air into the ocean. Is this claim correct? Explain. 1 mark
Q1 — Wave data table
Sound in air: λ = v/f = 340/440 ≈ 0.77 m; T = 1/f = 1/440 ≈ 2.3×10−3 s.
Guitar string: f = v/λ = 280/0.35 = 800 Hz; T = 1/800 = 1.25×10−3 s.
Seismic P-wave: v = fλ = 5.0×1200 = 6000 m/s; T = 1/5.0 = 0.20 s.
AM radio: f = v/λ = 3.0×108/300 = 1.0×106 Hz (1 MHz); T = 1×10−6 s (1 µs).
Water ripple: v = fλ = 12×0.040 = 0.48 m/s; T = 1/12 ≈ 0.083 s.
Q2.1 — Period from graph (2 marks)
Three complete cycles span from t = 1.0 ms to t = 4.0 ms, a total of 3.0 ms [1]. Period T = 3.0/3 = 1.0 ms [1].
Q2.2 — Frequency (2 marks)
f = 1/T = 1/(1.0×10−3) = 1000 Hz (1.0 kHz). [1 mark for correct rearrangement; 1 mark for correct answer with unit.]
Q2.3 — Wavelength in water (2 marks)
λ = v/f = 1500/1000 = 1.5 m. [1 mark substitution; 1 mark answer with unit.]
Q2.4 — Graph error correction (2 marks)
Error: the student is confusing the horizontal axis. On a displacement-time graph the horizontal axis represents time, not distance, so crest-to-crest spacing gives the period, not the wavelength [1]. Corrected statement: on a displacement-time graph, the horizontal spacing between two crests is the period T; to find wavelength you also need the wave speed and must use λ = v/f [1].
Q3 — Compare and contrast table
Horizontal axis: d-d = position (space); d-t = time.
Units: d-d = metres (m); d-t = seconds (s).
Crest-to-crest spacing: d-d = wavelength (λ); d-t = period (T).
Vertical axis: both show displacement of the medium from equilibrium.
Snapshot/history: d-d = snapshot of the whole wave at one instant; d-t = history of one point over time.
Q4.1 — Wavelengths at boundary (3 marks)
λair = v/f = 340/500 = 0.68 m [1].
Frequency stays at 500 Hz in water because it is set by the source [1].
λwater = 1500/500 = 3.0 m [1].
Q4.2 — Frequency at boundary (2 marks)
Frequency remains 500 Hz in water [1]. Frequency is determined by the source that generates the wave; when the wave enters a different medium its speed changes but the source is still vibrating at the same rate, so frequency is unchanged [1].
Q4.3 — Sound speed in water (1 mark)
The claim is incorrect. Sound travels faster in water (1500 m/s) than in air (340 m/s) because water is denser and less compressible, so pressure disturbances propagate more quickly through it [1].