A2 Daily A Level Physics question
In a lab, a student uses a small-angle simple pendulum as a timebase inside a lift. When the lift is stationary the period is T0. The lift then descends with steady downward acceleration equal to 0.20 g. Air resistance is negligible. Which statement must be true about the period during the accelerating phase compared with T0, and why?
Answer
The correct answer is D.
Correct: D — It increases by about 11%, because the effective gravitational acceleration is g_eff = g − 0.20 g = 0.80 g and T scales as 1/√g_eff; in the limit a → g, the restoring force vanishes and oscillations cease. A misreads the direction: downward acceleration reduces effective weight, weakening the restoring effect, so the period cannot decrease, and in the limit a → g the period does not go to zero. B assumes a linear 1/g relation; the correct scaling is with 1/√g, giving T/T0 = √(g/g_eff) = √(1/0.8) ≈ √1.25 ≈ 1.12 (about 12%), not 20%, and the limit claim is also wrong. C is false because for a pendulum T does depend on g (for fixed length), and if a → g then g_eff → 0 so SHM breaks down rather than remaining unchanged. D matches both the correct direction and the correct proportional reasoning, and the limiting-case behaviour.