Question

A seconds pendulum is suspended from the roof of a lift. If the lift is moving up with an acceleration $9.8m/s^2$, its time period is

• A. $1$ s
• B. $\sqrt{2}$ s
• C. $1/\sqrt{2}$ s
• D. $2\sqrt{2}$ s

Answer 1

The correct option is B $\sqrt{2}$ s

For seconds pendulum: $T=2\pi \sqrt{\frac{l}{g}}=$ 2 seconds
But lift is moving up with an acceleration of $9.8m/s^2$.
Here, the effective $g$ will be $g+a=g+g=2g$
Thus, $T_1=\frac{T}{\sqrt{2}}=2\pi \sqrt{\frac{l}{2g}}=\frac{2}{\sqrt{2}}=\sqrt{2}sec$