AS 91524 · 2023 paper · simple harmonic motion
Question 3(c) Merit to Excellence
Practises Simple harmonic motion (SHM)
Tanya studies a toy bouncing on a spring hanging from the ceiling. The spring constant is 24.6 N m⁻¹. She draws a graph of acceleration against displacement for the toy oscillating in simple harmonic motion.
| k | 24.6 N m⁻¹ |
|---|---|
| a plotted against y | |
| T | 0.199 s |
Tanya pulls the spring (period T = 0.199 s) down 0.100 m from the equilibrium position and releases it so the toy bounces in simple harmonic motion. Using a reference circle or otherwise, calculate the time the toy would take to travel 0.140 m up from its release position.
Work it out
Try it on paper first. Stuck? Take one hint at a time.
-
1 Where does the toy start, and where is it after travelling 0.140 m up?
It starts at the bottom of its motion, 0.100 m below equilibrium. Going 0.140 m up takes it 0.100 m to the middle and 0.040 m beyond: y = +0.040 m, taking up as positive. -
2 Picture it on a reference circle. Where does the arrow start, and where must it get to?
It starts pointing straight down, at the release point. It turns a quarter turn to reach the middle (y = 0), then on until its height is +0.040 m. -
3 How far past the quarter turn is that?
sin θ = 0.040 / 0.100 = 0.400, so θ = sin⁻¹(0.400) = 23.6° past the horizontal. Total angle turned: 90° + 23.6° = 113.6°. -
4 One full turn takes one period. What fraction of a period is 113.6°?
t = (113.6 / 360) × 0.199 = 0.0628 s. (In radians: 1.9823 rad / 31.57 rad s⁻¹ = 0.0628 s.)
What the marker wanted
- Achieved One of: the right reference-circle diagram; the right angle, 23.6°; or the right ω.
- Merit One error in the working, or a minor error by another method; or a calculator left in degrees.
- Excellence The right answer, 0.0628 s.
In our words, from NZQA’s 2023 marking schedule.
✕Common mistake
Substituting the wrong values or the wrong angle. NZQA’s 2023 report says this was common even where the right SHM equation was chosen. Here the 0.140 m is measured from the release point, so the toy ends 0.040 m above the middle, and the angle turned includes the quarter turn it took to get there.How did it go?
Got it moves it further out. Review brings it back tomorrow.