AS 91524 · 2024 paper · simple harmonic motion (watch glass)
Question 3(b) Merit to Excellence
Practises Simple harmonic motion (SHM)
A ball bearing is released on a watch glass — a shallow semi-circular glass bowl of radius of curvature R — and rolls back and forth with simple harmonic motion.
| released 0.0400 m from equilibrium | |
| T | 0.882 s |
|---|
The ball is released 0.0400 m to the right of equilibrium and oscillates with a period of 0.882 s. Using reference circles or otherwise, calculate its displacement after 1.20 s.
Work it out
Try it on paper first. Stuck? Take one hint at a time.
-
1 Turn the period into an angular frequency.
ω = 2π / T = 2π / 0.882 = 7.1238 rad s⁻¹. -
2 It is let go at +0.0400 m, the end of its swing. Which equation fits that start?
y = A cos ωt, since y = +A at t = 0. -
3 How far round the reference circle has it gone after 1.20 s? Use radians.
ωt = 7.1238 × 1.20 = 8.5486 rad. That is one full turn (2π = 6.2832 rad) plus 2.2654 rad (129.8°): 1.20 s is 1.36 periods. -
4 Now the displacement. Is your calculator in radians?
y = 0.0400 × cos(8.5486) = 0.0400 × (−0.6400) = −0.0256 m.
What the marker wanted
- Achieved The right angular frequency, 7.12 rad s⁻¹.
- Merit The method through to y with one small slip.
- Excellence The right answer, −0.0256 m.
In our words, from NZQA’s 2024 marking schedule.
✕Common mistake
A calculator in degrees (it gives 0.0396 m), sin in place of cos (0.0307 m), or a missing bracket round ωt. NZQA’s 2024 report names all three as the slips that held otherwise good answers at Merit.How did it go?
Got it moves it further out. Review brings it back tomorrow.