Q138 / 180 · AIIMS 2001 · Physics

The time period of a simple pendulum is $T$ remaining at rest inside a lift. Find the time period of pendulum when lift starts to move up with an acceleration of $g / 4$.

  • A
    $T$
  • B
    $T / 2$
  • C
    $2 T / 5$
  • Correct answer
    D
    $2 T / \sqrt{5}$.

Solution

When the lift is at rest, the time period of the simple pendulum is $T=2 \pi \sqrt{\frac{l}{g}}$ Now, if the lift is moving up with an acceleration $\frac{g}{4}$ then the effective acceleration due to gravity $$ \begin{aligned} & g^{\prime}=g+\frac{g}{4}=\frac{5 g}{4} \\ & \therefore \text { Time period } T^{\prime}=2 \pi \sqrt{\frac{l}{g^{\prime}}} \\ &=2 \pi \sqrt{\frac{l}{5 g / 4}} \\ &=\frac{2}{\sqrt{5}} \times 2 \pi \sqrt{\frac{l}{g}}=\frac{2}{\sqrt{5}} T \end{aligned} $$

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

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