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: Understanding the Concept
The period of a simple pendulum is \(T=2\pi\sqrt{\dfrac{l}{g}}\).

Step 2: Effective Gravity
In a lift accelerating upward with \(a=\dfrac{g}{4}\), \(g'=g+\dfrac{g}{4}=\dfrac{5g}{4}\).

Step 3: Ratio of Periods
\(\dfrac{T'}{T}=\sqrt{\dfrac{g}{g'}}=\sqrt{\dfrac{4}{5}}=\dfrac{2}{\sqrt{5}}\).

Step 4: New Period
\(T'=\sqrt{5}\times\dfrac{2}{\sqrt{5}}=2\) s.

Final Answer:
The new time period is 2 s. \[ \boxed{\text{(D) }2\text{ s}} \]