AS 91523 · 2024 paper · Doppler effect

Question 3(c) Merit to Excellence

Jane and her classmates stand on a railway platform, listening to a train approach them at a constant speed, blowing its whistle. The speed of sound in air is 343 m s⁻¹.

v sound	343 m s⁻¹

They hear 1270 Hz as the train approaches, and 1190 Hz after it has passed them and is moving away at the same constant speed.

Calculate the speed of the train.

The 2024 paper, page 7 ↗ Marking schedule, page 5 ↗

Work it out

Try it on paper first. Stuck? Take one hint at a time.

1 Write the Doppler equation for the train approaching, and again for it going away. What do they share?
1270 = f × 343 / (343 − v_s) and 1190 = f × 343 / (343 + v_s). The whistle’s own frequency f and the train’s speed v_s are in both, and both are unknown.
2 Make f × 343 the subject of each, and set them equal.
1270 (343 − v_s) = 1190 (343 + v_s)
3 Multiply out and collect the v_s terms.
435 610 − 1270 v_s = 408 170 + 1190 v_s, so 27 440 = 2460 v_s
4 Solve, then check it.
v_s = 27 440 / 2460 = 11.15 m s⁻¹. The whistle’s own frequency is then f = 1270 × (343 − 11.15) / 343 = 1229 Hz, and 1229 × 343 / (343 + 11.15) = 1190 Hz, as heard.

Answer 11.2 m s⁻¹ (11.15 m s⁻¹ unrounded). The whistle itself sounds at about 1229 Hz.

What the marker wanted

Achieved The right equation chosen for approaching and for going away.
Merit The calculation with one error. Taking the whistle’s frequency as 1230 Hz, halfway between, which gives 10.8 m s⁻¹, is Merit.
Excellence The right speed, 11.2 m s⁻¹.

In our words, from NZQA’s 2024 marking schedule.

Common mistake

Assuming the whistle’s own frequency is halfway between, 1230 Hz. It is not: the shift up on the way in is bigger than the shift down on the way out. That shortcut gives 10.8 m s⁻¹, which the schedule credits at Merit only.

How did it go?

Got it moves it further out. Review brings it back tomorrow.