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 $\frac{T}{2}$? (g is acceleration due to gravity)

Question

Dr. Abubacker · Accepted Answer

(b) Time period of simple pendulum is given by
$T = 2\pi\sqrt{\frac{l}{g}}$ …(i)
When the lift is moving up with an acceleration $a$, then time period becomes
$T^{'} = 2\pi\sqrt{\frac{l}{g + a}}$
Here, $T^{'} = \frac{T}{2}$
$\Rightarrow$ $\frac{T}{2} = 2\pi\sqrt{\frac{l}{g + a}}$ …(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 $\frac{T}{2}$? (g is acceleration due to gravity)

Explanation

Reviewed by Dr. Abubacker

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