Question

The time period of a simple pendulum in a lift descending with constant acceleration g is

• A. $T = 2\pi\sqrt{\frac{l}{g}}$
• B. $T = 2\pi\sqrt{\frac{l}{2g}}$
• C. Zero
• D. Infinite

Answer 1

Answer: (D)

Infinite

Answer 2

This is the case of freely falling lift and in free fall of lift effective g for pendulum will be zero. So $\Rightarrow y = 2\sin 1000\mspace{6mu} t + \sin 999\mspace{6mu} t + \sin 1001\mspace{6mu} t$