AS 91523 · 2025 paper · Doppler effect

Question 1(a) Achieved

Martyn and Kerryn are riding jet skis on a lake. Martyn's jet ski has an air horn that emits 1043 Hz. Kerryn stops, and Martyn travels directly towards her at 27.3 m s⁻¹. The speed of sound in air is 343 m s⁻¹.

f source	1043 Hz
v source	27.3 m s⁻¹
v sound (air)	343 m s⁻¹

Calculate the frequency of the sound that Kerryn hears.

The 2025 paper, page 2 ↗ Marking schedule, page 1 ↗

See it move: The Doppler effect: a moving source and a moving observer

Work it out

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

1 Who is moving, and which way?
The source, the air horn, moves straight towards Kerryn, and Kerryn is still. So this is the moving-source formula, with the source approaching.
2 An approaching source should raise the frequency. Which sign in the bottom of the fraction does that?
A minus: f′ = f × v / (v − v_s). The smaller bottom makes the fraction bigger than 1.
3 Put the values in.
f′ = 1043 × 343 / (343 − 27.3) = 1043 × 343 / 315.7 = 1133.19 Hz

Answer 1133 Hz (1.13 × 10³ Hz to three figures): higher than the 1043 Hz the horn emits, as it should be while the jet ski comes towards her.

What the marker wanted

Achieved The right frequency, about 1130 Hz.

In our words, from NZQA’s 2025 marking schedule.

How did it go?

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