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(B) The time period of a simple pendulum in a stationary lift is given by $T = 2\pi \sqrt{\frac{l}{g}}$,where $l$ is the length of the string and $g$

Vedclass · Accepted Answer

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

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(B) The time period of a simple pendulum in a stationary lift is given by $T = 2\pi \sqrt{\frac{l}{g}}$,where $l$ is the length of the string and $g$ is the acceleration due to gravity. When the lift accelerates upwards with an acceleration $a = \frac{g}{3}$,the effective acceleration due to gravity becomes $g_{eff} = g + a = g + \frac{g}{3} = \frac{4g}{3}$. The new time period $T'$ is given by $T' = 2\pi \sqrt{\frac{l}{g_{eff}}} = 2\pi \sqrt{\frac{l}{4g/3}} = 2\pi \sqrt{\frac{3l}{4g}}$. This can be rewritten as $T' = \frac{\sqrt{3}}{2} 	imes 2\pi \sqrt{\frac{l}{g}}$. Substituting $T = 2\pi \sqrt{\frac{l}{g}}$,we get $T' = \frac{\sqrt{3}}{2} T$.

$A$ person measures a time period of a simple pendulum inside a stationary lift and finds it to be $T$. If the lift starts accelerating upwards with an acceleration $\left(\frac{g}{3}\right)$,the time period of the pendulum will be

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