A tube is open at both ends. A loudspeaker, emitting sound of a single frequency, is placed near one end of the tube, as shown in Fig. 3.1. The speed of the sound in the tube is 340ms⁻¹. The length of the tube is 0.60 m. A stationary wave is formed with an antinode A at each end of the tube and two antinodes inside the tube. [Figure 3.1]
- (a)State the difference between a stationary wave and a progressive wave in terms of
- (a)(i)the energy transfer along the wave,[1]
- (a)(ii)the phase of two adjacent vibrating particles.[1]
- (b)(i)State what is meant by an antinode of the stationary wave.[1]
- (b)(ii)State the distance between a node and an adjacent antinode.[1]
- (b)(iii)Determine, for the sound in the tube,
- (b)(iii) 1.the wavelength,[1]
- (b)(iii) 2.the frequency.[2]
- (b)(iv)Determine the minimum frequency of the sound from the loudspeaker that produces a stationary wave in the tube.[2]
This question uses a diagram from the original paper, which isn't shown here. Open the full paper from the past papers archive to see it.
Mark scheme
Marking points from the official mark scheme (Oct/Nov 2017). The mark or mark type is shown in brackets.
- [B1] in a stationary wave energy is not transferred or in a progressive wave energy is transferred
- [B1] in a stationary wave (adjacent) particles are in phase or in a progressive wave (adjacent) particles are out of phase/have a phase difference/not in phase
- [B1] (position where) maximum amplitude
- [B1] distance = 0.10 m
- [A1] λ = 0.60/1.5 = 0.40 m
- [C1] v = fλ
- [A1] f = 340/0.40 = 850 Hz
- [C1] λ = 2 × 0.60 or f = 850/3
- [A1] f = 280 (283) Hz
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