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AS 91524 · 2025 paper · simple harmonic motion
Question 3(b) Merit
Practises Simple harmonic motion (SHM)
Rīhari swings a ball on a string as a pendulum. It is 1.20 m from the fixed point to the ball's centre of mass, and the ball is released 5.60° from the vertical. The string's mass is insignificant.
| L | 1.20 m |
|---|---|
| θ | 5.60° |
Show that the angular frequency of the ball's motion is 2.86 rad s⁻¹.
Work it out
Try it on paper first. Stuck? Take one hint at a time.
-
1 Which formula gives a pendulum’s period from its length?
T = 2π√(l / g) = 2π√(1.20 / 9.81) = 2.1975 s. -
2 How do you turn a period into an angular frequency?
ω = 2π / T = 2π / 2.1975 = 2.8592 rad s⁻¹. (It is the same as ω = √(g / l).)
What the marker wanted
- Achieved The right period (2.20 s), or ω worked out from a wrong period.
- Merit ω with the working that leads to it and the unrounded value: a “show that” needs the steps on the page.
In our words, from NZQA’s 2025 marking schedule.
✕Common mistake
Leaving out the formula you started from, or the period. In a “show that” the working is what earns the mark, and NZQA’s 2025 report says candidates who did not reach Achieved often skipped the base equation and the algebra.How did it go?
Got it moves it further out. Review brings it back tomorrow.