Time period of a simple pendulum in a stationary lift is ' T '. If the lift accelerates with $\frac{\mathrm{g}}{6}$ vertically upwards then the time period will be : (where $\mathrm{g}=$ acceleration due to gravity)

(C) $\sqrt{\frac{6}{7}} \mathrm{~T}$

(A) $\sqrt{\frac{6}{5}} \mathrm{~T}$

(B) $\sqrt{\frac{5}{6}} \mathrm{~T}$

(D) $\sqrt{\frac{7}{6}} \mathrm{~T}$

JEE Main · 2022 · 26 Jun 2022 · Shift 1 · Elasticity & SHM

Time period of a simple pendulum in a stationary lift is ' T '. If the lift accelerates with

\frac{g}{6}

vertically upwards then the time period will be :
(where

g =

acceleration due to gravity)

Answer: (C)

The time period of a simple pendulum is

T = 2 π \frac{ℓ}{g _{eff}}

. In a stationary lift,

g_{eff} = g

. When the lift accelerates upward with

a = g /6

, the effective gravity becomes

g_{eff} = g + a = g + g /6 = \frac{7 g}{6}

. The new time period is

T^{'} = 2 π \frac{ℓ}{7 g /6} = 2 π \frac{6 ℓ}{7 g} = \frac{6}{7} \times 2 π \frac{ℓ}{g} = \frac{6}{7} T

JEE Main · 2022 · 26 Jun 2022 · Shift 1 · Single correct option