A simple pendulum oscillates in a lift accelerating UPWARD with acceleration $a$. The effective time period becomes:
A$T = 2\pi\sqrt{L/g}$ (unchanged)
B$T = 2\pi\sqrt{L/(g+a)}$ (faster — effective gravity larger)
C$T = 2\pi\sqrt{L/(g-a)}$ (slower)
DPendulum stops oscillating
Answer & Solution
Correct answer: B. $T = 2\pi\sqrt{L/(g+a)}$ (faster — effective gravity larger)
In the lift's accelerated frame, effective gravity = g + a (downward). Pendulum sees stronger restoring force ⇒ shorter period. If lift accelerates DOWNWARD with $a$: effective g = g − a (longer period). If lift in free-fall ($a = g$): pendulum doesn't oscillate (T → ∞).
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