AS 91524 · 2025 paper · simple harmonic motion
Question 3(c) Merit to Excellence
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° |
After the ball is released from the starting position, calculate the time it takes to reach a point 3.30 cm before its next stationary position. A reference circle may help.
Work it out
Try it on paper first. Stuck? Take one hint at a time.
-
1 Where does the ball start, and where does it next stop?
It is let go from rest at one end of its swing, y = +0.117 m. It next stops at the other end, y = −0.117 m. -
2 Where is the point 3.30 cm short of that stop, as a displacement?
y = −0.117 + 0.0330 = −0.084 m: past the middle, 0.084 m on the far side. -
3 It starts at the end of its swing. Which SHM equation fits that start?
y = A cos ωt, because at t = 0 it is at y = +A. On a reference circle, the arrow starts pointing straight at +A and turns through ωt. -
4 Solve for the angle ωt. Is your calculator in radians?
cos ωt = −0.084 / 0.117 = −0.7179, so ωt = cos⁻¹(−0.7179) = 2.3716 rad (135.9°). -
5 Turn that angle into a time.
t = 2.3716 / 2.86 = 0.829 s. Check it a second way: a quarter period to reach the middle (2.1975 / 4 = 0.549 s), then sin⁻¹(0.084 / 0.117) / 2.86 = 0.280 s more to reach −0.084 m. 0.549 + 0.280 = 0.829 s.
What the marker wanted
- Achieved One of: the displacement, −0.084 m; a reference circle with the arrow’s start and finish in the right places; the angle past the middle (0.801 rad, or 45.9°); or the whole angle turned (2.3716 rad, or 135.9°).
- Merit The right method with one error in it: a wrong amplitude or displacement, the right angle with a wrong time, or one of the slips under “Common mistake”.
- Excellence The right time, 0.829 s, by this route or another complete one, such as 135.9° / 360° of the 2.1975 s period, or a quarter period plus the 0.280 s from the middle.
In our words, from NZQA’s 2025 marking schedule.
✕Common mistake
Using sin instead of cos. sin⁻¹ gives 0.280 s, which is only the time from the middle of the swing to the point, and misses the quarter period it took to reach the middle. The schedule holds that at Merit, along with dropping the minus sign (0.269 s) and leaving the calculator in degrees (47.5 s). NZQA’s 2025 report says wrong substitutions in exactly this kind of timing question were common.How did it go?
Got it moves it further out. Review brings it back tomorrow.