Question

The acceleration due to gravity on the surface of the moon is $1.7\, m\, s^{-2}$. The time period of a simple pendulum on the moon if its time period on the earth is $3.5\, s$ is (Given, $g = 9.8\, m\, s^{-2})$.

• A. $2.2\,s$
• B. $4.4\,s$
• C. $8.4\,s$
• D. $16.8\,s$

Answer 1

Answer: (C)

For moon, $g_m = 1.7\, m\, s^{-2}$
For earth, $g_e = 9.8\, m\, s^{-2}, T_e = 3.5\, s$
But, $T_{m} = 2\pi\sqrt{\frac{l}{g_{m}}}$ and $T_{e} = 2\pi\sqrt{\frac{l}{g_{e}}} $
$\therefore \frac{T_{m}}{T_{e}} = \sqrt{\frac{g_{e}}{g_{m}}}$ or
$T_{m} = \sqrt{\frac{g_{e}}{g_{m}}}\times T_{e} $
$ = \sqrt{\frac{9.8}{1.7}}\times 3.5 $
$= 8.4\, s$