MCQ
The time period of a simple pendulum measured inside a stationary lift is found to be $T$. If the lift starts accelerating upwards with an acceleration $g/3$, the time period is
  • A
    $T\sqrt 3 $
  • $T\sqrt 3 /2$
  • C
    $T/\sqrt 3 $
  • D
    $T/3$

Answer

Correct option: B.
$T\sqrt 3 /2$
b
(b) $T = 2\pi \sqrt {\frac{l}{g}} $ and $T' = 2\pi \sqrt {\frac{l}{{4g/3}}} $

$[As\;g' = g + a = g + \frac{g}{3} = \frac{{4g}}{3}$]

$T' = \frac{{\sqrt 3 }}{2}T$

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