AS 91524 · 2023 paper · simple harmonic motion
Question 3(a) Achieved
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 |
Given the simple harmonic motion equation a = −ω²y, describe how the gradient of the graph line relates to the frequency of oscillation.
Work it out
Try it on paper first. Stuck? Take one hint at a time.
-
1 Compare a = −ω²y with the equation of a straight line through the origin.
It has the form y = mx, so plotting a against y gives a straight line through the origin whose gradient is −ω². The size of the gradient is ω². -
2 How is ω linked to the frequency?
ω = 2πf, so the size of the gradient is ω² = 4π²f²: it goes as the frequency squared. -
3 What does that mean for a steeper line, and for a straight one?
A steeper line means a higher frequency. And a straight line has a constant gradient, so the frequency of this oscillation is constant too.
What the marker wanted
- Achieved Any one of: the gradient is ω² (with ω = 2πf); it goes as f²; it equals 4π²f²; or a straight line means a constant frequency.
In our words, from NZQA’s 2023 marking schedule.
How did it go?
Got it moves it further out. Review brings it back tomorrow.