Pendulum in Elevator – Rankers Physics
Topic: Oscillation
Subtopic: Angular SHM and Simple Pendulum

Pendulum in Elevator

Assertion (A): A simple pendulum is attached on a roof of a elevator. Time period of SHM is \( T \) when elevator is at rest. Time period of SHM must be greater than \( T \) if elevator start moving upward.
Reason (R): Time period of simple pendulum does not depend on acceleration due to gravity.
 
Both (A) & (R) are true and the (R) is the correct explanation of the (A)
Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
(A) is true but (R) is false
Both (A) and (R) are false

Solution:

The time period of a simple pendulum is \( T = 2\pi \sqrt{\frac{L}{g}} \). If the elevator accelerates upward with \( a \), the effective gravity becomes \( g_{eff} = g + a \). The new period is \( T' = 2\pi \sqrt{\frac{L}{g+a}} \). Since \( g+a > g \), then \( T' < T \). So (A) is false. The time period *does* depend on gravity, so (R) is false. Both (A) and (R) are false.

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