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A-LevelPhysicsSuperposition9702/23 · Oct/Nov 2018 · Q412 Marks

A horizontal string is stretched between two fixed points X and Y. A vibrator is used to oscillate the string and produce a stationary wave. Fig. 4.2 shows the string at one instant in time. The speed of a progressive wave along the string is 30ms⁻¹. The stationary wave has a period of 40 ms. A particle on the string oscillates with an amplitude of 13mm. At time t, the particle has zero displacement.

  1. (a)On Fig. 4.1, complete the two graphs to illustrate what is meant by the amplitude A, the wavelength λ and the period T of a progressive wave. [Figure 4.1] Ensure that you label the axes of each graph.[3]
  2. (b)(i)Explain how the stationary wave is formed on the string.[2]
  3. (b)(ii)1.Calculate 1. the displacement of the particle at time (t + 100ms), displacement = ..................................................... mm[3]
  4. (b)(ii)2.2. the total distance moved by the particle from time t to time (t + 100ms). distance = ..................................................... mm[3]
  5. (b)(iii)1.Determine 1. the frequency of the wave, frequency = ..................................................... Hz[1]
  6. (b)(iii)2.2. the horizontal distance from X to Y. distance = ..................................................... m[3]

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Mark scheme

Marking points from the official mark scheme (Oct/Nov 2018). The mark or mark type is shown in brackets.

  1. [B1] graph with x-axis labelled 'distance' and wavelength/λ correctly shown
  2. [B1] graph with x-axis labelled 'time' and period/T correctly shown
  3. [B1] graph with y-axis labelled 'displacement' and amplitude/A correctly shown
  4. [B1] wave (moves along string and) reflects at fixed point/Y/X/end/wall/boundary
  5. [B1] the incident and reflected waves interfere/superpose
  6. [C1] 100/40 or 2.5 (cycles/periods/T)
  7. [B1] displacement = 0
  8. [A1] distance = 130 mm
  9. [A1] f = 1/40 × 10⁻³ = 25 Hz
  10. [C1] v = fλ or λ = vT
  11. [C1] λ = 30/25 or 30 × 40 × 10⁻³ (= 1.2m)
  12. [A1] distance = 1.2 x 1.5 = 1.8 m

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About this question

Topic
Superposition, A-Level Physics (syllabus 9702). Structured question, 12 marks.
Source
A-Level Physics, Oct/Nov 2018 Paper 2 Variant 3, question 4.

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