First slide

Applications of SHM

Question

The period of a simple pendulum measured inside a stationary lift is found to be T. If the lift starts accelerating upwards with acceleration of $g / 3,$ then the time period of the pendulum is

Moderate

A

$\frac{T}{\sqrt{3}}$
B

$\frac{T}{3}$
C

$\frac{\sqrt{3}}{2} T$
D

$\sqrt{3} T$

Solution

For stationary lift $T_{1} = 2 π \sqrt{\frac{l}{g}}$
For ascending lift with acceleration a, $T_{2} = 2 π \sqrt{\frac{l}{g + a}}$
$\Rightarrow \frac{T_{1}}{T_{2}} = \sqrt{\frac{g + a}{g}} \Rightarrow \frac{T_{1}}{T_{2}} = \sqrt{\frac{g + \frac{g}{3}}{g}} = \sqrt{\frac{4}{3}} \Rightarrow T_{2} = \frac{\sqrt{3}}{2} T$

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