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

Answer: (A)

$2\,\pi \sqrt{\frac{3L}{2g}}$

Answer 2

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 $\therefore T=2 \pi \sqrt{\frac{L}{g_{\text{eff}}}}$ $=2 \pi \sqrt{\frac{L}{2 g / 3}}$ $=2 \pi \sqrt{\frac{3 L}{2 g}}$