Answer key: Spring Oscillator Worksheet
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Goal: Investigate how the mass and the spring stiffness set the period of a mass on a spring, and how energy moves between the spring and the moving mass.
Completed results table
| Setting | Period | Total energy |
|---|---|---|
| Default (1 kg, 20 N/m, 0.5 m) | 1.405 s | 2.5 J |
| Stiffer spring (60 N/m) | 0.811 s | 7.5 J |
| Heavier mass (4 kg) | 2.81 s | 2.5 J |
| Small stretch (0.2 m) | 1.405 s | 0.4 J |
Answers
Compare the default and “Heavier mass” rows. How did making the mass 4 times larger change the period?
The period doubled, from 1.405 s to 2.81 s. The period is T = 2π√(m/k), so it grows with the square root of the mass, and √4 = 2.
The stiffer spring is 3 times stiffer. How did the period change, and why?
It got shorter, from 1.405 s to 0.811 s (1.405 ÷ √3). A stiffer spring pulls back harder, so the mass oscillates faster.
Did the amplitude change the period? Did it change the energy?
The period stayed 1.405 s, but the total energy dropped from 2.5 J to 0.4 J. Total energy is ½kA², so a 0.2 m amplitude instead of 0.5 m multiplies it by (0.2 ÷ 0.5)² = 0.16.
At the moment the displacement is 0 m, how fast is the mass moving? Where has the energy gone?
2.236 m/s, its top speed. At x = 0 the spring is neither stretched nor compressed, so all 2.5 J is kinetic energy: ½ × 1 × 2.236² = 2.5 J.
Use E = ½kA² to explain why the stiffer spring stores 7.5 J with the same 0.5 m amplitude.
E = ½ × 60 × 0.5² = 7.5 J, three times the 2.5 J of the 20 N/m spring, because the stored energy is proportional to k.
Challenge: what spring stiffness would give the 4 kg mass the original 1.405 s period? Predict, then test it.
80 N/m. The period depends only on m ÷ k, so 4 kg on an 80 N/m spring has the same ratio as 1 kg on 20 N/m and the same 1.405 s period.