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A-LevelPhysicsWaves9702/21 · May/June 2019 · Q59 Marks

A loudspeaker oscillates with frequency f to produce sound waves of wavelength λ. The loudspeaker makes N oscillations in time t.

  1. (a)(i)State expressions, in terms of some or all of the symbols f, λ and N, for: 1. the distance moved by a wavefront in time t 2. time t.[2]
  2. (a)(ii)Use your answers in (i) to deduce the equation relating the speed v of the sound wave to f and λ.[1]
  3. (b)The waveform of a sound wave is displayed on the screen of a cathode-ray oscilloscope (c.r.o.), as shown in Fig. 5.1. [Figure 5.1] The time-base setting is 0.20 ms cm⁻¹. Determine the frequency of the sound wave.[2]
  4. (c)Two sources S₁ and S₂ of sound waves are positioned as shown in Fig. 5.2. [Figure 5.2] The sources emit coherent sound waves of wavelength 0.85 m. A sound detector is moved parallel to the line S₁S₂ from a point X to a point Y. Alternate positions of maximum loudness L and minimum loudness Q are detected, as illustrated in Fig. 5.2. Distance S₁X is equal to distance S₂X. Distance S₂Y is 7.40 m.
  5. (c)(i)Explain what is meant by coherent waves.[1]
  6. (c)(ii)State the phase difference between the two waves arriving at the position of minimum loudness Q that is closest to point X.[1]
  7. (c)(iii)Determine the distance S₁Y.[2]

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Mark scheme

Marking points from the official mark scheme (May/June 2019). The mark or mark type is shown in brackets.

  1. [1] Nλ
  2. [1] N/f
  3. [1] v (= distance / time) = Nλ / (N/f) so v = fλ
  4. [1] T = 4.0 × 0.20 = 0.80 (ms) or 8.0 × 10⁻⁴ (s)
  5. [1] f = 1/8.0 × 10⁻⁴ = 1300 Hz
  6. [1] constant phase difference (between the waves)
  7. [1] 180°
  8. [1] path difference = 2λ or S₁Y – S₂Y = 2λ
  9. [1] distance = 7.40 + (0.85 × 2) = 9.1 m

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About this question

Topic
Waves, A-Level Physics (syllabus 9702). Structured question, 9 marks.
Source
A-Level Physics, May/June 2019 Paper 2 Variant 1, question 5.

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