Question

On a planet, a body released from rest is found to freely fall through a distance of 8 m in 2 s. On that planet what is the time period of a simple pendulum of length 1 m?

• A. seconds.
• B. seconds.
• C. seconds.
• D. seconds.

Answer 1

The correct option is A seconds.

For a freely falling body $s=\frac{1}{2}g{t}^2$
$\therefore 8=\frac{1}{2}g\times 4\Rightarrow g=4\frac{m}{s^2}.$
The time period of a simple pendulum is given by $T=2\pi \sqrt{\frac{l}{g}}$
Substituting, l = 1 m and $g=\frac{4m}{s^2},$ we get T = $\pi$ seconds.