The method and the results are on the simple harmonic motion lab. Do the analysis yourself first, then check the answers below.
Quick answer: Simple harmonic motion obeys d²x/dt² = −ω²x, with solution x = A cos(ωt + φ). These answers go with the AP Physics C lab on simple harmonic motion: the graph, the calculations, the uncertainties and the conclusion explained, for all 7 analysis questions.
Model answers use the sample data in the table. Your measurements will differ a little, but the graph of T²d against d² should be a straight line with gradient close to 4.0 s²/m. State the uncertainty in g and compare your value of k² with L²/12.
Level: AP Physics C (Mechanics and E&M, calculus-based) · Curriculum: AP PC:M Unit 7
d = 0.30 m: T = 15.3 / 10 = 1.53 s and T²d = 1.53² × 0.30 = 0.702 s² m. d = 0.45 m: T = 1.60 s and T²d = 2.56 × 0.45 = 1.152 s² m. Both match the table: 0.702 and 1.152 s² m.
Question 2 Straight-line graph
T² = 4π²(k² + d²)/(g d), so T²d = (4π²/g) d² + (4π²/g)k². The gradient is 4π²/g and the intercept is 4π²k²/g.
Intercept = 0.376 − 4.03 × 0.0100 = 0.336 s² m. Then k² = intercept / gradient = 0.336 / 4.03 = 0.0834 m². The theory value is 1.00² / 12 = 0.0833 m², so they agree to 0.1 %.
Question 5 Timing uncertainty
Timing 10 swings gives ΔT = 0.2 / 10 = 0.02 s. The percentage is 0.02 / 1.58 = 1.3 %. Timing many swings reduces the percentage uncertainty.
Question 6 Shape of the data
Near the centre the restoring torque mgd is small, so T is large. Far from the centre, I = M(k² + d²) grows faster than mgd, so T rises again. The minimum lies at d = k = 0.289 m.
Question 7 Amplitude check
The derivation uses sinθ ≈ θ. For amplitude θ₀ the period grows by about θ₀²/16, so at 40° (0.70 rad) it is about 3 % longer, which would bias the value of g.
