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Question

(c) For stationary lift ${T_1} = 2\pi \sqrt {\frac{l}{g}} $ 

For ascending lift with acceleration a, ${T_2} = 2\pi \sqrt {\frac{l}{{g + a}}} $

Vedclass · Accepted Answer

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

Vedclass · Answer

(c) For stationary lift ${T_1} = 2\pi \sqrt {\frac{l}{g}} $ 

For ascending lift with acceleration a, ${T_2} = 2\pi \sqrt {\frac{l}{{g + a}}} $

$ \Rightarrow \frac{{{T_1}}}{{{T_2}}} = \sqrt {\frac{{g + a}}{g}} $

$\Rightarrow \frac{{{T_1}}}{{{T_2}}} = \sqrt {\frac{{g + \frac{g}{3}}}{g}} = \sqrt {\frac{4}{3}}$

$\Rightarrow {T_2} = \frac{{\sqrt 3 }}{2}T$

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

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