Q171 / 180 · COMEDK 2018 · Physics

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
    2g
  • Correct answer
    B
    3g
  • C
    4g
  • D
    g

Solution

Time-period of simple pendulum is given as $$ T=2 \pi \sqrt{\frac{l}{g}} $$ If $a$ be the acceleration of the lift, so that time-period of simple pendulum is reduced to $T / 2$. $\begin{aligned} \text { i.e. } & T^{\prime}=T / 2 \\ \Rightarrow \quad 2 \pi \sqrt{\frac{l}{g+a}} &=\frac{1}{2} \times 2 \pi \sqrt{\frac{l}{g}} \end{aligned}$ $\Rightarrow \quad \frac{l}{g+a}=\frac{1}{4}, \frac{l}{g}$ $\Rightarrow \quad g+a=4 g$ $\Rightarrow \quad a=3 g$

Hence, the correct answer is option: 'B'.

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