Question

If the length of a seconds pendulum on a planet is 2 m, then the acceleration due to gravity on the surface of that planet is :

• A. 9.8 m $s^{-2}$
• B. 19.6 m $s^{-2}$
• C. 2 ${\pi }^2$ m $s^{-2}$
• D. 4 ${\pi }^2$ m $s^{-2}$

Answer 1

Answer: (C)

2 ${\pi }^2$ m $s^{-2}$

Time period, $T=2\pi \sqrt{\frac{l}{g}}$

Time period of a seconds pendulum is $2$ sec. One second for swing in one direction and one second for return swing.

Thus, $2=2\pi \sqrt{\frac{l}{g}}$

Squaring, $1/{\pi }^2=l/g$ or $g={\pi }^{2}l=2{\pi }^{2}m/s^2$