C.1 · IB Physics
Simple harmonic motion
Release a mass on a spring and watch it oscillate. The period depends on the mass and the spring constant, not on the amplitude. Add damping to see the amplitude decay, then increase it until the oscillation stops.
Predict first
Start from the default settings (m = 1.0 kg, k = 8 N m⁻¹, no damping). Double the mass to 2.0 kg. What happens to the period T?
T = 2π√(m/k), so T is proportional to √m. Doubling m multiplies T by √2: from 2.22 s to 3.14 s.
Check it in the model: Move the Mass m slider from 1.0 to 2.0 kg and compare the Period T readout (2.22 s then 3.14 s).
motion energy / envelope
T = 2π√(m/k), ω₀ = √(k/m)
undamped: x = A cos ω₀t
damped: x = A e^(−γt) cos(ω_d t + φ), γ = b ÷ 2m
ω_d = √(ω₀² − γ²), E = ½ m v² + ½ k x²
Try this
- Set b = 0 and double the mass from 1.0 to 2.0 kg. Compare the new period with the old one. Then change the amplitude and check that the period does not move.
- Choose the Energy view and watch KE and PE swap while the total stays constant. At what displacement are KE and PE equal?
- Raise b from 0 to 6.0 kg s⁻¹. At about what value does the oscillation disappear, and does the ratio ζ match b ÷ 2√(km)?
Common error: the period T = 2π√(m/k) does not depend on amplitude, so a bigger swing takes the same time, but the maximum speed Aω does grow with amplitude.