Cambridge International A-Level Physics · topic 7
Waves
Connect the ideas, make your working visible, and test the result. Use the hints when you need a starting point; open the full solution after your attempt.
Extra challenging questions
Work through the linked stages and explain what each result means.
Reconstructing a travelling wave and an overlapping pulse
At one instant a transverse wave has the supplied displacement–position profile. Adjacent points are joined smoothly as a sinusoid. A fixed sensor observes five complete cycles in 2.0 s. Later an upward 3.0 cm pulse and a downward 2.0 cm pulse overlap at the sensor in a linear medium. Supplied data (Position / m; Displacement / cm): (0, 0), (0.025, 0.5853), (0.05, 1.1481), (0.075, 1.6667), (0.1, 2.1213), (0.125, 2.4944), (0.15, 2.7716), (0.175, 2.9424), (0.2, 3), (0.225, 2.9424), (0.25, 2.7716), (0.275, 2.4944), (0.3, 2.1213), (0.325, 1.6667), (0.35, 1.1481), (0.375, 0.5853), (0.4, 0), (0.425, -0.5853), (0.45, -1.1481), (0.475, -1.6667), (0.5, -2.1213), (0.525, -2.4944), (0.55, -2.7716), (0.575, -2.9424), (0.6, -3), (0.625, -2.9424), (0.65, -2.7716), (0.675, -2.4944), (0.7, -2.1213), (0.725, -1.6667), (0.75, -1.1481), (0.775, -0.5853), (0.8, 0). (a) Find amplitude, wavelength, period, frequency and propagation speed. Explain why the snapshot does not determine the direction of travel. [3 marks] (b) Find the displacement during pulse overlap and explain what happens to the pulses and their transported energy afterward. [3 marks]
Wave snapshot
Supplied question data. Read the labelled axes and use the observations to test the physical model.
Vertical axis (y): Displacement / cm
Horizontal axis (x): Position / m
Read the exact data as a table
| Position / m | Displacement / cm |
|---|---|
| 0 | 0 |
| 0.025 | 0.5853 |
| 0.05 | 1.1481 |
| 0.075 | 1.6667 |
| 0.1 | 2.1213 |
| 0.125 | 2.4944 |
| 0.15 | 2.7716 |
| 0.175 | 2.9424 |
| 0.2 | 3 |
| 0.225 | 2.9424 |
| 0.25 | 2.7716 |
| 0.275 | 2.4944 |
| 0.3 | 2.1213 |
| 0.325 | 1.6667 |
| 0.35 | 1.1481 |
| 0.375 | 0.5853 |
| 0.4 | 0 |
| 0.425 | -0.5853 |
| 0.45 | -1.1481 |
| 0.475 | -1.6667 |
| 0.5 | -2.1213 |
| 0.525 | -2.4944 |
| 0.55 | -2.7716 |
| 0.575 | -2.9424 |
| 0.6 | -3 |
| 0.625 | -2.9424 |
| 0.65 | -2.7716 |
| 0.675 | -2.4944 |
| 0.7 | -2.1213 |
| 0.725 | -1.6667 |
| 0.75 | -1.1481 |
| 0.775 | -0.5853 |
| 0.8 | 0 |
These are the supplied values, not a worked answer. Lines join the supplied values; they are not a fitted model.
Show a hint
Begin with part (a): Find amplitude, wavelength, period, frequency and propagation speed. Explain why the snapshot does not determine the direction of travel.
Reveal the worked solution
- One complete spatial cycle spans 0.80 m, and maximum displacement from equilibrium is 3.0 cm.
- Frequency is cycles/time = 5/2.0 = 2.5 Hz; T = 1/f = 0.40 s; propagation speed is fλ = 2.0 m s⁻¹.
- A single spatial profile can move either left or right. A later snapshot or local displacement–time information is needed to establish direction.
- Superposition adds signed displacements: +3.0+(−2.0) = +1.0 cm upward.
- Partial cancellation describes displacement at that location and instant. It does not mean the waves permanently destroy each other.
- The pulses continue in the ideal linear model. Energy is redistributed between kinetic and elastic forms; displacement alone is insufficient to determine all local energy during overlap.
Answer (a) A = 3.0 cm; λ = 0.80 m; T = 0.40 s; f = 2.5 Hz; v = 2.0 m s⁻¹. (b) Resultant displacement is +1.0 cm; in the ideal linear medium both pulses continue through each other.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Finding a hidden polariser axis
Unpolarised light of intensity 80 W m⁻² passes through an ideal first polariser and then an ideal analyser. The analyser data are supplied for angles relative to the first axis. Use I = I0 cos²θ for linearly polarised light at an analyser. Supplied data (Relative angle / °; Transmitted intensity / W m⁻²): (0, 40), (30, 30), (45, 20), (60, 10), (90, 0). (a) Explain the first intensity reduction and predict the intensity when a third polariser at 45° is inserted between crossed first and last polarisers. [3 marks] (b) Explain what the extinction supports about light and give one limitation when using this model to reduce glare. [3 marks]
Analyser transmission
Supplied question data. Read the labelled axes and use the observations to test the physical model.
Vertical axis (y): Transmitted intensity / W m⁻²
Horizontal axis (x): Relative angle / °
Read the exact data as a table
| Relative angle / ° | Transmitted intensity / W m⁻² |
|---|---|
| 0 | 40 |
| 30 | 30 |
| 45 | 20 |
| 60 | 10 |
| 90 | 0 |
These are the supplied values, not a worked answer. Lines join the supplied values; they are not a fitted model.
Show a hint
Begin with part (a): Explain the first intensity reduction and predict the intensity when a third polariser at 45° is inserted between crossed first and last polarisers.
Reveal the worked solution
- An ideal polariser transmits half the average intensity of unpolarised light, giving 80/2 = 40 W m⁻².
- The middle axis at 45° transmits 40cos²45° = 20 W m⁻² and defines a new output polarisation direction.
- The last axis is another 45° from the middle, so it transmits 20cos²45° = 10 W m⁻². Without the middle element, the crossed pair transmits zero in the ideal model.
- A transverse oscillation has a selectable direction perpendicular to propagation. A simple longitudinal wave has no equivalent independent transverse axis to filter.
- The evidence supports the transverse nature of light, not every microscopic detail of a classical-wave picture.
- Reflected glare may be only partially polarised, and real filters leak light and absorb other components. Measured minima therefore need not be exactly zero.
Answer (a) First output = 40 W m⁻²; with the middle polariser, final output = 10 W m⁻². (b) Polarisation supports transverse electric-field oscillations; real glare and real filters need not reach complete extinction.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.