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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.

Extra challenging · 01

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]

Explore the question data

Restoring force

Supplied question data. Read the labelled axes and use the observations to test the physical model.

Vertical axis (y): Force / N

Restoring forceSupplied question data. Read the labelled axes and use the observations to test the physical model. Horizontal axis: Displacement / m. Vertical axis: Force / N. Exact values are available in the data table below.-2-1012-0.1-0.0500.050.1xy

Horizontal axis (x): Displacement / m

Reading 1Displacement / m: -0.1Force / N: 2
Read the exact data as a table
Restoring force — supplied values
Displacement / mForce / N
-0.12
-0.051
00
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
  1. The graph gives F = −20x, so a = F/m = −40x: acceleration is proportional to displacement and directed toward equilibrium.
  2. Comparing a = −ω²x gives ω = √40 rad s⁻¹ and T = 2π/ω ≈0.993 s.
  3. The amplitude is 0.10 m. From energy conservation or vmax = ωA, the maximum speed is 0.632 m s⁻¹ at equilibrium.
  4. At x = 0.060 m, acceleration is −40(0.060) = −2.4 m s⁻², toward equilibrium.
  5. 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.
  6. 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.

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Extra challenging · 02

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]

Explore the question data

Wave snapshot

Supplied question data. Read the labelled axes and use the observations to test the physical model.

Vertical axis (y): Displacement / cm

Wave snapshotSupplied question data. Read the labelled axes and use the observations to test the physical model. Horizontal axis: Position / m. Vertical axis: Displacement / cm. Exact values are available in the data table below.-3-1.501.5300.20.40.60.8xy

Horizontal axis (x): Position / m

Reading 1Position / m: 0Displacement / cm: 0
Read the exact data as a table
Wave snapshot — supplied values
Position / mDisplacement / cm
00
0.0250.5853
0.051.1481
0.0751.6667
0.12.1213
0.1252.4944
0.152.7716
0.1752.9424
0.23
0.2252.9424
0.252.7716
0.2752.4944
0.32.1213
0.3251.6667
0.351.1481
0.3750.5853
0.40
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
0.80

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
  1. One complete spatial cycle spans 0.80 m, and maximum displacement from equilibrium is 3.0 cm.
  2. 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.
  3. A single spatial profile can move either left or right. A later snapshot or local displacement–time information is needed to establish direction.
  4. Superposition adds signed displacements: +3.0+(−2.0) = +1.0 cm upward.
  5. Partial cancellation describes displacement at that location and instant. It does not mean the waves permanently destroy each other.
  6. 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.

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Extra challenging · 03

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]

Explore the question data

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)

Air-to-material refractionSupplied question data. Read the labelled axes and use the observations to test the physical model. Horizontal axis: sin(refracted angle). Vertical axis: sin(incident angle). Exact values are available in the data table below.00.2250.450.6750.900.150.30.450.6xy

Horizontal axis (x): sin(refracted angle)

Reading 1sin(refracted angle): 0sin(incident angle): 0
Read the exact data as a table
Air-to-material refraction — supplied values
sin(refracted angle)sin(incident angle)
00
0.20.3
0.40.6
0.60.9

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
  1. Snell's law gives nair sin i = nmaterial sin r. The graph gradient sin i/sin r is 1.50.
  2. 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°.
  3. The internal angle 45° exceeds θc and light travels from higher to lower index, meeting both conditions for total internal reflection.
  4. 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.
  5. 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.
  6. 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.

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Extra challenging · 04

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]

Explore the question data

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

Fringe spacing versus screen distanceSupplied question data. Read the labelled axes and use the observations to test the physical model. Horizontal axis: D / m. Vertical axis: Mean fringe spacing / mm. Exact values are available in the data table below.0123400.511.52xy

Horizontal axis (x): D / m

Reading 1D / m: 0.5Mean fringe spacing / mm: 1
Read the exact data as a table
Fringe spacing versus screen distance — supplied values
D / mMean fringe spacing / mm
0.51
12
1.53
24

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
  1. For two-slit interference, w = λD/d. The slope w/D is 2.0 mm m⁻¹ = 0.0020.
  2. λ = slope×d = 0.0020(0.00030) = 6.0×10⁻⁷ m.
  3. Doubling d halves w at fixed λ and D. At 1.5 m, the original 3.0 mm spacing becomes 1.5 mm.
  4. 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.
  5. 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.
  6. 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.

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