First slide

Applications of SHM

Question

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

Easy

A

$T = 2 π \sqrt{\frac{l}{g}}$
B

$T = 2 π \sqrt{\frac{l}{2 g}}$
C

Zero
D

Infinite

Solution

This is the case of freely falling lift and in free fall of lift effective g for pendulum will be zero. So $T = 2 π \sqrt{\frac{l}{g - a}} = 2 π \sqrt{\frac{l}{0}} = \infty (∵ a = g)$

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