Cambridge International A-Level Physics · topic 8
Superposition
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.
Identifying the missing fundamental on a string
A string fixed at both ends is 1.20 m long. Its three lowest observed resonances are 50, 75 and 100 Hz. The 50 Hz pattern has two loops, each between adjacent nodes. The supplied graph relates the identified harmonic number to resonance frequency. Wave speed is unchanged between these observations. Supplied data (Harmonic number; Frequency / Hz): (2, 50), (3, 75), (4, 100). (a) Determine the unobserved fundamental, wave speed and node spacing at 75 Hz. [3 marks] (b) The tension is quadrupled while length and mass per length remain fixed. Predict the resonances and explain the roles of nodes and antinodes. [3 marks]
Identified string resonances
Supplied question data. Read the labelled axes and use the observations to test the physical model.
Vertical axis (y): Frequency / Hz
Horizontal axis (x): Harmonic number
Read the exact data as a table
| Harmonic number | Frequency / Hz |
|---|---|
| 2 | 50 |
| 3 | 75 |
| 4 | 100 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Begin with part (a): Determine the unobserved fundamental, wave speed and node spacing at 75 Hz.
Reveal the worked solution
- For both ends fixed, L = nλ/2. Two loops identify 50 Hz as the second harmonic, not the fundamental.
- The fundamental is 50/2 = 25 Hz. Wave speed is 2Lf₁ = 2(1.20)(25) = 60 m s⁻¹.
- At 75 Hz, λ = 60/75 = 0.80 m; adjacent nodes are λ/2 = 0.40 m apart.
- String wave speed follows v = √(T/μ), so multiplying tension by four doubles speed and all resonance frequencies.
- At a node, destructive superposition keeps displacement zero; at an antinode the displacement amplitude is maximum.
- The standing pattern arises from incident and reflected waves of equal frequency travelling in opposite directions. It is not a single wave carrying the pattern along the string.
Answer (a) Fundamental = 25 Hz; wave speed = 60 m s⁻¹; node spacing at 75 Hz = 0.40 m. (b) The three observed resonances become 100, 150 and 200 Hz.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
The end correction in a resonating air column
A tube closed by water at one end and open at the other resonates at the supplied water-to-mouth lengths for a fixed 500 Hz tuning fork. Consecutive resonances correspond to odd quarter wavelengths. Treat the open-end correction e as the same for each measurement. Supplied data (Successive resonance index; Measured length / m): (1, 0.16), (2, 0.5), (3, 0.84). (a) Find wavelength, sound speed and the open-end correction. [3 marks] (b) Locate displacement nodes and antinodes and predict the next resonance length. Explain why not every harmonic occurs. [3 marks]
Resonant air-column lengths
Supplied question data. Read the labelled axes and use the observations to test the physical model.
Vertical axis (y): Measured length / m
Horizontal axis (x): Successive resonance index
Read the exact data as a table
| Successive resonance index | Measured length / m |
|---|---|
| 1 | 0.16 |
| 2 | 0.5 |
| 3 | 0.84 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Begin with part (a): Find wavelength, sound speed and the open-end correction.
Reveal the worked solution
- Consecutive closed–open resonances differ by λ/2, so λ = 2(0.50−0.16) = 0.68 m.
- Wave speed is fλ = 500(0.68) = 340 m s⁻¹.
- At the first resonance, L+e = λ/4 = 0.17 m, giving e = 0.17−0.16 = 0.010 m. The spacing method cancels the common end correction.
- The water boundary restricts longitudinal air displacement, producing a displacement node. The open end supports an antinode slightly outside the mouth.
- Allowed effective lengths are λ/4, 3λ/4, 5λ/4 and so on, so only odd harmonics of the fundamental occur for this ideal closed–open tube.
- The next length is 0.84+0.34 = 1.18 m. Pressure variation has the opposite node/antinode pattern to air displacement.
Answer (a) λ = 0.68 m; sound speed = 340 m s⁻¹; e = 0.010 m. (b) Next measured length = 1.18 m; displacement node at the water and antinode near the open mouth.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
A fringe investigation that tests two hypotheses
Two coherent slits of separation 0.30 mm illuminate a screen. For each screen distance D, the separation between the centres of the first and eleventh bright fringes is measured and divided by ten. These mean fringe-spacing data are supplied. The small-angle condition applies. Supplied model: Use fringe spacing w = λD/d under the small-angle approximation. Supplied data (D / m; Mean fringe spacing / mm): (0.5, 1), (1, 2), (1.5, 3), (2, 4). (a) Infer the light wavelength and predict fringe spacing if slit separation is doubled at D = 1.5 m. [3 marks] (b) Explain how the method reduces relative reading uncertainty, and design a check that the pattern is interference rather than a screen ruler artefact. [3 marks]
Fringe spacing versus screen distance
Supplied question data. Read the labelled axes and use the observations to test the physical model.
Vertical axis (y): Mean fringe spacing / mm
Horizontal axis (x): D / m
Read the exact data as a table
| D / m | Mean fringe spacing / mm |
|---|---|
| 0.5 | 1 |
| 1 | 2 |
| 1.5 | 3 |
| 2 | 4 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Begin with part (a): Infer the light wavelength and predict fringe spacing if slit separation is doubled at D = 1.5 m.
Reveal the worked solution
- For two-slit interference, w = λD/d. The slope w/D is 2.0 mm m⁻¹ = 0.0020.
- λ = slope×d = 0.0020(0.00030) = 6.0×10⁻⁷ m.
- Doubling d halves w at fixed λ and D. At 1.5 m, the original 3.0 mm spacing becomes 1.5 mm.
- First-to-eleventh fringe centres contain ten intervals, not eleven. A 30 mm span is less sensitive to a fixed endpoint error than one 3 mm interval.
- Keep wavelength and slit separation fixed while changing D; proportional spacing supports the wave model. Then change d while keeping D fixed and test the inverse relationship.
- Record repeat endpoint readings, keep the screen perpendicular to the central ray and shield stray light. A ruler artefact would not track both controlled physical changes.
Answer (a) Wavelength = 600 nm; doubled-separation fringe spacing = 1.5 mm. (b) Measure a long span over many intervals and vary a physical input with a quantitative prediction.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
A diffraction grating with an impossible order
Monochromatic light falls normally on a grating with 500 lines per millimetre. Measured sine of the diffraction angle against order is supplied below. Use c = 3.0×10⁸ m s⁻¹. Angles are measured from the central direction, not from the grating surface. Supplied model: Use d sinθ = nλ and spacing d = 1/(lines per unit length). Supplied data (Order n; sin θ): (0, 0), (1, 0.3), (2, 0.6), (3, 0.9). (a) Find grating spacing, wavelength, frequency and the largest possible order. [3 marks] (b) Explain how to distinguish an aperture's diffraction envelope from the grating's narrow interference maxima, and improve an angle measurement. [3 marks]
Diffraction orders
Supplied question data. Read the labelled axes and use the observations to test the physical model.
Vertical axis (y): sin θ
Horizontal axis (x): Order n
Read the exact data as a table
| Order n | sin θ |
|---|---|
| 0 | 0 |
| 1 | 0.3 |
| 2 | 0.6 |
| 3 | 0.9 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Begin with part (a): Find grating spacing, wavelength, frequency and the largest possible order.
Reveal the worked solution
- 500 lines mm⁻¹ is 5.0×10⁵ lines m⁻¹, so d = 1/(5.0×10⁵) = 2.0×10⁻⁶ m.
- From d sinθ = nλ, λ/d is the graph gradient, 0.30; therefore λ = 6.0×10⁻⁷ m and f = c/λ = 5.0×10¹⁴ Hz.
- The condition sinθ≤1 requires n≤d/λ = 3.33. Integer orders stop at n = 3; a fourth order would require sinθ = 1.2.
- The grating equation predicts directions of principal maxima, while finite slit width determines their broad intensity envelope.
- Measure both +n and −n peaks; half their angular separation reduces an offset in the estimated central direction.
- Repeat and use higher visible orders where practical, while avoiding weak peaks and confusion between order number and wavelength. Never view a laser directly.
Answer (a) d = 2.0 μm; λ = 600 nm; f = 5.0×10¹⁴ Hz; maximum order = 3. (b) Diffraction spreads each slit contribution; interference between the many coherent slit contributions selects the narrow orders.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.