Question

The time period of a simple pendulum inside a stationary lift is \(\sqrt{5}\) s. The time period when the lift moves upwards with an acceleration \(\frac{g}{4}\) is (\(g\) = acceleration due to gravity)

• A. \(\sqrt{5}\) s
• B. \(2\sqrt{5}\) s
• C. \((2+\sqrt{5})\) s
• D. \(2\) s

Hint:

In a lift moving up with acceleration a, the effective gravity is g + a. Use T proportional to 1/sqrt(g).

Answer 1

The Correct Option is D

Step 1: Formula First
$T'=2\pi\sqrt{\dfrac{l}{g+a}}$ for upward acceleration.

Step 2: Substitution
$T'=T\sqrt{\dfrac{g}{5g/4}}=\sqrt{5}\times\dfrac{2}{\sqrt{5}}$.

Step 3: Result
$T'=2$ s, option (D).

Final Answer:
The pendulum period in the accelerating lift is 2 s. \[ \boxed{\text{(D) }2\text{ s}} \]