Time period of a simple pendulum of length L as measured in an elevator descending with acceleration

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13.Oscillations

medium

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

$2\pi \sqrt {\frac{{3L}}{g}} $

$\pi \sqrt {\left( {\frac{{3L}}{g}} \right)} $

$2\pi \sqrt {\left( {\frac{{3L}}{{2g}}} \right)} $

$2\pi \sqrt {\frac{{2L}}{{3g}}} $

Solution

(c)The effective acceleration in a lift descending with acceleration $\frac{g}{3}$ is ${g_{eff}} = g – \frac{g}{3} = \frac{{2g}}{3}$
$T = 2\pi \sqrt {\left( {\frac{L}{{{g_{eff}}}}} \right)} $$ = 2\pi \sqrt {\left( {\frac{L}{{2g/3}}} \right)} $$ = 2\pi \sqrt {\left( {\frac{{3L}}{{2g}}} \right)} $

Standard 11

Physics

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medium

(KVPY-2010)

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easy

(AIIMS-1999)

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(KVPY-2014)

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hard

(JEE MAIN-2023)