IB Diploma Physics · unit C
Wave behaviour
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
A selection from this pathway. Open a topic below for its complete practice and challenge set.
Energy and acceleration in an oscillator
A 0.50 kg glider on a spring performs small, undamped oscillations. Its restoring-force measurements are supplied. At release it is 0.10 m from equilibrium and momentarily at rest. Take the equilibrium displacement as x = 0. Supplied model: Use a = −ω²x, T = 2π/ω, vmax = ωA and v² = ω²(A²−x²). Supplied data (Displacement / m; Force / N): (-0.1, 2), (-0.05, 1), (0, 0), (0.05, -1), (0.1, -2). (a) Show that the force model produces SHM. Find angular frequency, period and maximum speed. [3 marks] (b) Find speed and acceleration at x = +0.060 m. Explain how damping changes a resonance experiment. [3 marks]
Restoring force
Supplied question data. Read the labelled axes and use the observations to test the physical model.
Vertical axis (y): Force / N
Horizontal axis (x): Displacement / m
Read the exact data as a table
| Displacement / m | Force / N |
|---|---|
| -0.1 | 2 |
| -0.05 | 1 |
| 0 | 0 |
| 0.05 | -1 |
| 0.1 | -2 |
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): Show that the force model produces SHM. Find angular frequency, period and maximum speed.
Reveal the worked solution
- The graph gives F = −20x, so a = F/m = −40x: acceleration is proportional to displacement and directed toward equilibrium.
- Comparing a = −ω²x gives ω = √40 rad s⁻¹ and T = 2π/ω ≈0.993 s.
- The amplitude is 0.10 m. From energy conservation or vmax = ωA, the maximum speed is 0.632 m s⁻¹ at equilibrium.
- At x = 0.060 m, acceleration is −40(0.060) = −2.4 m s⁻², toward equilibrium.
- Energy gives v² = ω²(A²−x²) = 40(0.0100−0.0036) = 0.256; speed is √0.256 ≈0.506 m s⁻¹, with direction depending on the motion phase.
- A driving frequency near the natural frequency can produce a large response. Damping transfers mechanical energy to the surroundings and reduces and broadens the resonance peak, useful when large vibrations would cause damage.
Answer (a) ω = √40 ≈6.32 rad s⁻¹; T ≈0.993 s; vmax ≈0.632 m s⁻¹. (b) Speed magnitude ≈0.506 m s⁻¹; acceleration = −2.4 m s⁻².
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
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.
Testing a fibre from refraction data
A transparent material is surrounded by air of refractive index 1.00. Refraction measurements for light entering from air are supplied as sine values. Take c = 3.0×10⁸ m s⁻¹. A ray inside the material subsequently reaches a material–air boundary at 45° to the normal. Supplied data (sin(refracted angle); sin(incident angle)): (0, 0), (0.2, 0.3), (0.4, 0.6), (0.6, 0.9). (a) Find refractive index, speed in the material and critical angle. Predict what happens at the 45° internal incidence. [3 marks] (b) Explain which wave quantities change on entry and evaluate whether all rays in a fibre necessarily remain trapped. [3 marks]
Air-to-material refraction
Supplied question data. Read the labelled axes and use the observations to test the physical model.
Vertical axis (y): sin(incident angle)
Horizontal axis (x): sin(refracted angle)
Read the exact data as a table
| sin(refracted angle) | sin(incident angle) |
|---|---|
| 0 | 0 |
| 0.2 | 0.3 |
| 0.4 | 0.6 |
| 0.6 | 0.9 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Begin with part (a): Find refractive index, speed in the material and critical angle. Predict what happens at the 45° internal incidence.
Reveal the worked solution
- Snell's law gives nair sin i = nmaterial sin r. The graph gradient sin i/sin r is 1.50.
- Wave speed in the material is c/n = 2.0×10⁸ m s⁻¹. The critical angle obeys sin θc = 1/1.50, giving about 41.8°.
- The internal angle 45° exceeds θc and light travels from higher to lower index, meeting both conditions for total internal reflection.
- The boundary must transmit the same oscillation frequency. Since v = fλ and speed drops by a factor of 1.50, wavelength also drops by that factor.
- A ray arriving below the critical angle can refract into the surrounding lower-index material, so a transparent tube does not trap all entry directions.
- Real fibres have cladding, losses and bending limits. A tight bend can change incidence angles and allow previously guided light to escape.
Answer (a) n = 1.50; speed = 2.0×10⁸ m s⁻¹; critical angle ≈41.8°; total internal reflection occurs at 45°. (b) Frequency stays fixed; speed and wavelength fall on entry; only rays meeting the total-internal-reflection condition are trapped in the ideal model.
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.