AS 91524 · 2024 paper · simple harmonic motion (watch glass)

Question 3(c) Merit

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

Eventually the ball comes to rest. Draw a labelled displacement–time graph for five complete oscillations, with values for displacement and time.

The 2024 paper, page 7 ↗ Marking schedule, page 5 ↗

Work it out

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

1 What stays the same from one oscillation to the next, and what shrinks?
The period stays at 0.882 s: light damping barely changes it. The amplitude shrinks a little every swing as energy is lost.
2 Where does the graph start?
At +0.0400 m when t = 0, because the ball is let go from rest at the end of its swing: a cosine shape, not a sine.
3 What values go on the time axis for five complete oscillations?
One period each: 0.882, 1.764, 2.646, 3.528 and 4.410 s. Each peak sits on one of those times and each trough halfway between, and the curve crosses the axis at evenly spaced quarter periods.
4 How should the peaks change?
Each one lower than the last by the same fraction, so they fall along a smooth curve towards zero (an exponential decay), mirrored by the troughs. Label the axes displacement (m) and time (s), and mark 0.0400 m on the displacement axis.

Answer A cosine curve starting at +0.0400 m at t = 0, with a constant period of 0.882 s (peaks at 0.882, 1.764, 2.646, 3.528 and 4.410 s) and each peak and trough smaller than the last, dying away exponentially towards zero.

What the marker wanted

Achieved One of: the amplitude decaying exponentially; the curve starting at the full amplitude; or a constant period.
Merit The right graph.

In our words, from NZQA’s 2024 marking schedule.

How did it go?

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