A simple pendulum is suspended from the ceiling of a lift. When the lift is at rest its time period is T . With what acceleration should the lift be accelerated upwards in order to reduce its period t

Question

A simple pendulum is suspended from the ceiling of a lift. When the lift is at rest its time period is T . With what acceleration should the lift be accelerated upwards in order to reduce its period to T / 2 ? ( g is the acceleration due to gravity)

TestHub · Accepted Answer

Time period of a simple pendulum is given by,

$T = 2 π \sqrt{\frac{l}{g}}$ ...(i)

When the lift is moving up with an acceleration $a$ , then time period becomes

$T^{'} = 2 π \sqrt{\frac{l}{g + a}}$

Here, $T^{'} = \frac{T}{2}$

$\Rightarrow$ $\frac{T}{2} = 2 π \sqrt{\frac{l}{g + a}}$ ...(ii)

Dividing Eq.(ii) by Eq. (i), we get

$a = 3 g$

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