Angular SHM and Simple Pendulum: Practice Problem & Solution

Solution Explained:

To solve this problem, we apply the core principles of Angular SHM and Simple Pendulum. Understanding the underlying formula is key to arriving at the correct answer below:

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.