Question

On a planet a freely falling body takes $2\,\sec$ when it is dropped from a height of $8\,m$ , the time period of simple pendulum of length $1\,m$ on that planet is:

• A. 3.14 sec
• B. 16.28 sec
• C. 1.57 sec
• D. none of these

Answer 1

Answer: (A)

3.14 sec

Answer 2

Here : Time $t=2\, \sec$, height $h=8\, m$ Initial velocity $v=0$ Length of pendulum $=1\, m$ Relation for the height is given by, $h =u t+\frac{1}{2} g t^{2}$ $8 =0 \times 2+\frac{1}{2} g \times 2^{2}=2\, g$ $2\, g =8$ or $g=4\, m / s ^{2}$ Hence, the time period $T$ is given by, $=2 \pi \sqrt{\frac{l}{g}}=2 \pi \sqrt{\frac{1}{4}}=\pi$ $=3.14\, \sec$