A simple pendulum is suspended from ceiling of a lift when lift is at rest its period is 'T'. With what acceleration 'a' should lift be accelerated upward in order to reduce the pe

Question

A simple pendulum is suspended from ceiling of a lift when lift is at rest its period is 'T'. With what acceleration 'a' should lift be accelerated upward in order to reduce the period to ' 2 T ​ ? (take 'g' as acceleration due to gravity)

TestHub · Accepted Answer

The time period of a simple pendulum when the lift is at rest is given by:

$T = 2 π \frac{L}{g}$

When the lift accelerates upward with acceleration $a$ , the effective acceleration due to gravity becomes $g^{'} = g + a$ . The new time period is:

$T^{'} = 2 π \frac{L}{g + a}$

Given that $T^{'} = \frac{T}{2}$ :

$2 π \frac{L}{g + a} = \frac{1}{2} \times 2 π \frac{L}{g}$

Squaring both sides:

$\frac{L}{g + a} = \frac{1}{4} \frac{L}{g}$

$g + a = 4 g$

$a = 3 g$

Answer: 3 g

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