First slide

Pseudo force

Question

The time period of simple pendulum of length l as measured in an elevator descending with acceleration g/3 is

Moderate

A

$2 π \sqrt{(3 l / g)}$
B

$π \sqrt{(3 l / g)}$
C

$2 π \sqrt{(3 l / 2 g)}$
D

$2 π \sqrt{(2 l / 3 g)}$

Solution

When the elevator is descending, then
$\begin{array}{l} mg - T = ma or T = m (g - a) \\ ∴ t = 2 π \sqrt{(\frac{l}{g - a})} \end{array}$
Given that a = g /3
$∴ t = 2 π \sqrt{[\frac{l}{g - (g / 3)}]} = 2 π \sqrt{(\frac{3 l}{2 g})}$

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