A-LevelPhysicsWaves9702/22 · May/June 2018 · Q47 Marks

A loudspeaker producing sound of constant frequency is placed near the open end of a pipe, as shown in Fig. 4.1. [Figure 4.1] A movable piston is at distance x from the open end of the pipe. Distance x is increased from x = 0 by moving the piston to the left with a constant speed of 0.75cms⁻¹. The speed of the sound in the pipe is 340ms⁻¹.

(a)(i)Define the wavelength of a progressive wave.[1]
(a)(ii)State what is meant by an antinode of a stationary wave.[1]
(b)(i)A much louder sound is first heard when x = 4.5cm. Assume that there is an antinode of a stationary wave at the open end of the pipe. Determine the frequency of the sound in the pipe.[3]
(b)(ii)After a time interval, a second much louder sound is heard. Calculate the time interval between the first louder sound and the second louder sound being heard.[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 (May/June 2018). The mark or mark type is shown in brackets.

[B1] distance moved by wavefront/energy during one cycle/oscillation/period (of source) OR minimum distance between two wavefronts OR distance between two adjacent wavefronts
[B1] (position where) maximum amplitude
[C1] λ = 4 × 0.045 (= 0.18 (m) or 18 (cm))
[C1] v = fλ
[A1] f = 340/0.18 = 1900 Hz
[C1] distance = λ/2 (= 0.09 (m) or 9 (cm))
[A1] time = 0.09/0.0075 = 12 s
OR
(C1) t = 4.5/0.75 and t = 13.5/0.75
(A1) time = 18 – 6 = 12 s

On Oracle Prep you can write or photograph your own answer and have it marked by AI against this mark scheme, with feedback on what you missed.