Question 1(d) Merit to Excellence

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.

An underwater microphone monitors noise from jet ski engines. Martyn's engine makes a sound at 574.0 Hz. As the jet ski travels away from the microphone at 27.3 m s⁻¹, the microphone measures 563.6 Hz from the engine.

Calculate the velocity of sound in water.

The 2025 paper, page 4 ↗ Marking schedule, page 3 ↗

Work it out

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

1. 1 A source moving away should lower the frequency. Which sign goes in the bottom of the fraction?

   A plus: f′ = f × v_w / (v_w + v_s), where v_w is now the speed of sound in water, the unknown.

2. 2 Put in what you know.

   563.6 = 574.0 × v_w / (v_w + 27.3)

3. 3 Multiply both sides by the bracket, then gather the v_w terms on one side.

   563.6 v_w + 563.6 × 27.3 = 574.0 v_w, so 15 386.28 = (574.0 − 563.6) v_w = 10.4 v_w

4. 4 Solve, and check the size makes sense.

   v_w = 15 386.28 / 10.4 = 1479.45 m s⁻¹. Sound in water is several times faster than in air, and the shift is small (574.0 → 563.6 Hz) because 27.3 m s⁻¹ is a small fraction of that speed.

What the marker wanted

• Achieved Picks the plus sign for a source moving away, or puts the values into the formula correctly.
• Merit The calculation with one minor slip, or the right value without enough working to show how it came.
• Excellence The right answer, about 1480 m s⁻¹, with the working.

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

How did it go?

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