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
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
$T\sqrt 3 $
$T\sqrt 3 /2$
$T/\sqrt 3 $
$T/3$
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
$T = 2\pi \sqrt {\frac{l}{g}} $ and $T' = 2\pi \sqrt {\frac{l}{{4g/3}}} $
$[As\;g' = g + a = g + \frac{g}{3} = \frac{{4g}}{3}$]
$T' = \frac{{\sqrt 3 }}{2}T$