First slide

Applications of SHM

Question

The time period of a simple pendulum of length L as measured in an elevator descending with acceleration $\frac{g}{3}$ is

Moderate

A

$2 π \sqrt{\frac{3 L}{g}}$
B

$π \sqrt{(\frac{3 L}{g})}$
C

$2 π \sqrt{(\frac{3 L}{2 g})}$
D

$2 π \sqrt{\frac{2 L}{3 g}}$

Solution

The effective acceleration in a lift descending with acceleration $\frac{g}{3}$ is $g_{eff} = g - \frac{g}{3} = \frac{2 g}{3}$
$∴ T = 2 π \sqrt{(\frac{L}{g_{eff}})} = 2 π \sqrt{(\frac{L}{2 g / 3})} = 2 π \sqrt{(\frac{3 L}{2 g})}$

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