Question

The time period of a simple pendulum of length $L$ as measured in an elevator descending with acceleration $ \frac{g}{3} $ is

• A. $ 2\,\pi \sqrt{\frac{3L}{2g}} $
• B. $ \pi \sqrt{\frac{3L}{g}} $
• C. $ 2 \,\pi \sqrt{\left( \frac{3L}{g} \right)} $
• D. $ 2 \,\pi \sqrt{\frac{2L}{g}} $

Answer 1

The Correct Option is A

The effective acceleration in a lift descending with acceleration $\frac{g}{3}$ is $g_{\text{eff}}=g-\frac{g}{3}=\frac{2 g}{3}$ Time period of simple pendulum $\therefore T=2 \pi \sqrt{\frac{L}{g_{\text{eff}}}}$ $=2 \pi \sqrt{\frac{L}{2 g / 3}}$ $=2 \pi \sqrt{\frac{3 L}{2 g}}$