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 acceleration due to gravity).

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 acceleration due to gravity). (A) 2 g (B) 3 g (C) 4 g (D) g

Tardigrade Admin · Accepted Answer

Time period of simple pendulum is given by

T = 2 π \frac{l}{g} ... (i)

When the lift is moving up with an acceleration

a

, then time period becomes

T^{'} = 2 π \frac{l}{g + a}

Here,

T^{'} = \frac{T}{2}

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

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

a = 3 g

Q. 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 acceleration due to gravity).

2 g

3 g

4 g

g

Time period of simple pendulum is given by T=2πgl​​...(i) When the lift is moving up with an acceleration a, then time period becomes T′=2πg+al​​ Here,T′=2T​ ⇒2T​=2πg+al​​...(ii) Dividing Eq. (ii) by Eq. (i), we get a=3g

Q. 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 acceleration due to gravity).